CBSE Solutions For Class 10 Mathematics Chapter 15 Probability

Class 10 Maths Probability

Question 1. In any situation that has only two possible outcomes, each outcome will have Probability. Find whether it is true or false.
Solution: False.

The probability of each outcome will be \(\frac{1}{2}\), only when the two outcomes are equally likely.

Question 2. A Marble is chosen at random from 6 marbles numbered 1 to 6. Find the Probability of getting a marble having number 2 and 6 on it.
solution:

The favourable Case is to get a marble on which both numbers 2 and are written. But there is no such marble.

So, N(E) = 0 and n(S) = 6

∴ Required probability = \(\frac{n(E)}{n(S)}=\frac{0}{6}=0\)

Question 3. A marble is chosen at random from 6 marbles numbered Ito 6. Find the Probability of getting a marble having number 2 or 6 on it.
Solution:

Here N(E) = 2

and n(S) = 6

∴ Required Probability = \(\frac{2}{6}=\frac{1}{3}\)

Question 4. It is given that in a group of 3 Students, the probability of 2 students not having the Same birthday is 0.992. What is the probability that the 2 students have the Same birthday.
Solution:

The probability of 2 students not having the Same birthday = 0.992

∴ Probability of 2 students having the same birthday

= 1-0.992

= 0.008

Question 5. A bag Contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is

  1. Red?
  2. Not ved?

Solution:

Total balls = 3+5 = -8

Total possible outcomes of drawing a ball at random from the bag = 8

1. Favorable outcomes of drawing a batt at random red ball =3

∴ The probability of drawing a red ball

⇒ \(\frac{\text { Favourable outcomes of drawing a red ball }}{\text { Total possible outcomes }}\)

⇒ \(\frac{3}{8}\)

2. Probability that the ball drawn is not red = 1- probability that the ball drawn

⇒ \(1-\frac{3}{8}=\frac{5}{8}\)

6. A die is thrown once. Find the probability of getting:

  1. A prime number
  2. A number lying between 2 and 6
  3. An odd number.

Solution:

Possible outcomes in one throw of a die = {1,2,3,4,5,6}

Total possible outcomes = 6

1. Prime numbers = {2,3,5}=3

∴ Probability of getting prime number = \(\frac{3}{6}=\frac{1}{2}\)

2. Numbers lying between 2 to 6 = {3, 4, 5} = 3

∴ Probability of this event = \(\frac{3}{6}=\frac{1}{2}\)

3. Odd numbers = {1,3,5}=3

∴ Probability of this event = \(\frac{3}{6}=\frac{1}{2}\)

Question 7. 12 defective pens are accidentally mixed with 132 good to just look at a pen and tell whether or not it is taken out at random from this lot. Determine the probability Out is a good one.
Solution:

No. of good pens = 132

No. of defective pens = 12

Total pens = 132 +12 = 144

Total Favourable outcomes of drawing a pen = 144

Favourable outcomes of drawing a good pen = 132

∴ Probability of drawing a good pen = \(\frac{132}{144}=\frac{11}{12}\)

Question 8. A child has a die whose Six faces show the letters as

CBSE School For Class 10 Maths Chapter 15 Probability A Child Has A Die Whose Six Fases Show The Letters

given below; The die is thrown once. what is the probability of getting

  1. A?
  2. D?

Solution:

Total possible outcomes in a throw of die = 6

1. Favourable outcome of getting A = 2

∴ Probability of getting \(A=\frac{2}{6}=\frac{1}{3}\)

2. Favourable outcomes of getting D = 1

∴ Probability of getting D = \(\frac{1}{6}\)

Question 9. A box Contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears

  1. A two-digit number,
  2. A number divisible by 5.

Solution:

We have, n(S) = 90

1. Let A be the event of getting “a two-digit number”.

∴ Favourable cases are 10,11,12,13,14, …….., 90

∴ n(A) = 81

∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{81}{90}=\frac{9}{10}\)

2. let B be the event of getting ་་ a number divisible by 5″.

∴ Favourable Cases are 10, 15, 20, 25, 30, …… 90.

Let there be n in numbers.

∴ Tn = 90

∴ 10+ (n-1)5 = 90

⇒ (n-1)5=80

⇒ n-1 = 16

⇒ n = 17

∴ n(B) = 17

P(B) = \(\frac{n(B)}{n(S)}=\frac{17}{90}\)

Question 10. It is known that a box of 600 electric bulbs Contains 12 defective bulbs. One bulb is taken out at random from this box, what is the probability that it is a non-defective bulb?
Solution:

The number of non-defective bulbs in the box = 600-12=588

So, probability of taking out a non-defective bulb = \(\frac{588}{600}=\frac{49}{50}\)

=0.98

CBSE Solutions For Class 10 Mathematics Chapter 14 Statistics

Class 10 Maths Statistics

Question 1. Find the mean by the direct method:

CBSE School For Class 10 Maths Chapter 14 Statistics Find The Mean By Direct Method.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics Find The Mean By Direct Method

Now,

mean \(\bar{x}=\frac{\sum f_{i x_i}}{\sum f_i}=\frac{1100}{50}=22\)

Question 2. Find the mean using the Direct Method:

CBSE School For Class 10 Maths Chapter 14 Statistics CBSE School For Class 10 Maths Chapter 14 Statistics Find The Mean Using Direct Method.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics Find The Mean Using Direct Method

Now, mean \(\bar{x}=\frac{\sum f_{i x i}}{\sum f_i}=\frac{13,200}{50}=264\)

Question 3. The mean of the following frequency distribution is 25. Find the value of P using the direct method:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean Of The Following Distribution Is 25.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean Of The Following Distribution Is 25

Now, Mean

⇒ \(\bar{x}=\frac{\sum f_{i x i}}{\sum f_i}\)

⇒ \(\bar{x}=\frac{1230+15 p}{42+p}\)

⇒ \(25=\frac{1230+15 P}{42+P}\)

⇒ 1050 + 25P = 1230 + 15P

⇒ 25P – 15P = 1230 + 15P

⇒ 10P = 180

⇒ P = \(\frac{180}{10}\)

P = 18

Question 4. The mean of the following distribution is 54. Find the value of P using the direct method:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean Of The Following Distribution Is 54.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean Of The Following Distribution Is 54

Now, Mean \(\bar{x}=54\)

⇒ \(\bar{x}=\frac{\sum f_{i x_i}}{\sum f_i}\)

54 = \(\frac{2070+70 p}{41+P}\)

2214+54p=2070+70P

70P-54P = 2214-2070

16P = 144

P = \(\frac{144}{16}\)

P = 9

Question 5. Find the mean from the following table using the Short Cut Method:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean From The Following Table Using Short Cut Method.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean From The Following Table Using Short Cut Method

Now, Mean a = 22.5

Mean \(\bar{x}=a+\frac{\sum f_{i d i}}{\sum f_i}\)

⇒ \(\bar{x}=22.5+\frac{0}{67}\)

⇒ \(\bar{x}=22.5\)

Question 6. Find the mean from the following table using the step. deviation method:

CBSE School For Class 10 Maths Chapter 14 Statistics Step Deviation Method.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics Step Deviation Method

Now, a = 37.5, h = 5

Mean \(\bar{x}=a+\frac{\sum f_{i u_i}}{\sum f_i} \times h\)

⇒ \(\bar{x}=37.5+\frac{-46}{50} \times 5\)

⇒ \(\bar{x}=37.5-\frac{46}{10}\)

⇒ \(\bar{x}=\frac{375-46}{10}\)

⇒ \(\bar{x}=\frac{329}{10}\)

⇒ \(\bar{x}=32.9\)

Question 7. Find the mean from the following table using the step deviation method:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean From The Following Table Using Step Deviation Method.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean From The Following Table Using Step Deviation Method

Now, a = 27.5, h = 5

mean \(\bar{x}=a+\frac{\sum f_i u_i}{\sum f_i} \times h\)

⇒ \(\bar{x}=27.5+\frac{-20}{40} \times 5\)

⇒ \(\bar{x}=\frac{220-20}{8}\)

⇒ \(\bar{x}=\frac{200}{8}\)

⇒ \(\bar{x}=25\)

Question 8. Find the median from the following data:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median From The Following Data 8.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median From The Following Data 8

Here, N = 58

∴ For Median class \(\frac{N}{2}=\frac{58}{2}=29\)

∴ Median class = 10-13

Here l1 = 10, l2 = 13

⇒ i = 13-10 = 3

f = 0, C.f = 29

∴ Median M = \(l_1+\frac{\left(\frac{N}{2}-c\right)}{f} \times i\)

M = \(10+\frac{29-29}{0} \times 3\)

M = \(10+\frac{0}{0} \times 3\)

M = 10

Question 9. Find the Median from the following data:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median From The Following Data 9.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median From The Following Data 9

Here, N = 50

⇒ \(\frac{N}{2}=\frac{50}{2}=25\)

=25

∴ Median Class = 20-30

Now, l1 = 20, l2 = 30

i = 30-20 = 10

f = 12, C = 15

and Median M = \(l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

M = \(20+\frac{(25-15)}{12} \times 10\)

M = \(20+\frac{10}{6} \times 5\)

M = \(\frac{120+50}{6}\)

M = \(\frac{170}{6}\)

M = 28.33

Question 10. Find the median from the following data:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median From The Following Data.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median From The Following Data

Here N = 50

⇒ \(\frac{N}{2}=\frac{50}{2}=25\)

∴ The median class is 15-20

Now, l1 = 15, l2 = 20, i = 20-15 = 5, f = 15, C = 20

and Median M = \(l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

M = \(15+\frac{25-20}{15} \times 5\)

M = \(\frac{45+5}{3}\)

M = \(\frac{50}{3}\)

M = 16.67

Question 11. Find the Median for the following frequency distribution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median For The Following Frequency Distribution.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median For The Following Frequency Distribution

Now, N=340

⇒ \(\frac{N}{2}=\frac{340}{2}=170\)

∴ Median class = 39.5-46.5

∴ l1 = 39.5, l2 = 46.5, i= 46.5-39.5 = 7, f = 102, C = 199

Median M = \(l_1+\frac{\left(\frac{N}{2}-c\right)}{f} \times i\)

M = \(39.5+\frac{(170-199)}{102} \times 7\)

M = \(39.5-\frac{29}{102} \times 7\)

M = \(39.5-\frac{203}{102}\)

M = \(\frac{4029-203}{102}\)

M = \(\frac{3826}{102}\)

M = 36.5

Question 12. Find the Median for the following frequency distribution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median Of The Following Frequency Distribution.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median Of The Following Frequency Distribution

Here N = 123

⇒ \(\frac{N}{2}=\frac{123}{2}=61.5\)

∴ Median Class = 20,5-25.5

l1 = 20.5, l2 = 25.5, i = 25.5-20.5 = 5, f = 24, C = 65

Median M = \(l_1+\frac{\left(\frac{N}{2}-C\right)}{F} \times i\)

M = \(20.5+\frac{(61.5-65)}{24} \times 5\)

M = \(20.5-\frac{3.5}{24} \times 5\)

M = \(\frac{492-17.5}{24}\)

M = \(\frac{474.5}{24}\)

M = 20.1

Question 13. If the Median of the following frequency distribution is 32.5. Find the value of F.

CBSE School For Class 10 Maths Chapter 14 Statistics The Median Of The Following Frequency Distribution Is 325.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median Of The Following Frequency Distribution Is 325

Here N = 34+ P

⇒ \(\frac{N}{2}=\frac{34+P}{2}\)

Median = 325 ⇒ Median class=30-40

∴ 11 = 30, l2 =40, i = 40-30=10

f = 12, C = 17

M = \(l_1+\frac{\left(\frac{N}{2}-c\right)}{f} \times i\)

32.5 = \(30+\frac{\left(\frac{34+p}{2}-17\right)}{12} \times 10\)

32.5-30 = \(\frac{10}{12}\left(\frac{34+P}{2}-17\right)\)

2.5 = \(\frac{5}{6}\left(\frac{34+p-34}{2}\right)\)

2.5 = \(\frac{5 P}{12}\)

30 = 5P

P = \(\frac{30}{5}\)

P = 6

Question 14. Determine the median for the following income distribution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean For The Following Income Distribution.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean For The Following Income Distribution

Here N = 100

⇒ \(\frac{N}{2}=\frac{100}{2}=50\)

∴ Medion class = 300-400

∴ l1 = 300, l2 = 400, i= 400-300 = 100, f = 30, C = 33

M = \(l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

M = \(300+\frac{50-33}{30} \times 100\)

M = \(300+\frac{17}{30} \times 100\)

M = \(\frac{9000+1700}{30}\)

M = \(\frac{10700}{30}\)

M = 356.67

Question 15. Find the mode of the following frequency distribution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean Median And Mode Of The Following Data.

Solution:

Clearly, the modal class is 60-80 as it has the maximum frequency

∴ l = 60, f1 = 12, f0 = 10, f2 = 6, h = 20

Mode M = \(l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

M = \(60+\frac{12-10}{2(12)-10-6} \times 20\)

M = \(60+\frac{2}{24-16} \times 20\)

M = \(60+\frac{2}{8} \times 20\)

M = \(\frac{480+40}{8}\)

M = \(\frac{520}{8}\)

M = 65

Question 16. Given below is the frequency distribution of the heights of players in a school;

CBSE School For Class 10 Maths Chapter 14 Statistics The Frequency Distribution Of The Heights Of Players In A School.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Frequency Distribution Of The Heights Of Players In A School

Clearly, the modal class is 165.5-1685 as it has a maximum frequency

∴ l = 165.5, f1 = 142, f0 =118, f2 = 127, h=3

Mode = \(l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

M = \(165.5+\frac{142-118}{2(142)-118-127} \times 3\)

M = \(165.5+\frac{24}{284-245} \times 3\)

M = \(165.5+\frac{24}{39} \times 3\)

M = \(\frac{6454.5+72}{39}\)

M = \(\frac{6526.5}{39}\)

M = 167.35

Question 17. Find the mode of the following frequency distribution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mode Of The Frequency Distribution

Solution:

Clearly, the modal class is 50-60, as it has the maximum frequency

∴ l = 50, f1 = 11, f0 = 9, f2 = 6, h = 10

Mode = \(\ell+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

M = \(50+\frac{11-9}{2(11)-9-6} \times 10\)

M = \(50+\frac{2}{22-15} \times 10\)

M = \(50+\frac{2}{7} \times 10\)

M = \(\frac{350+20}{7}\)

M = \(\frac{370}{7}\)

M = 52.86

Question 18. The following distribution represents the height of 160 students in a class;

CBSE School For Class 10 Maths Chapter 14 Statistics The Height Of 160 Students Of A Class

Solution:

Clearly, the modal class is 155-160 as it has the maximum

∴ l = 155, f1 = 38, f0 =30, f2 =24, h=5

Mode = \(l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

M = \(155+\frac{38-30}{2(38)-30-24} \times 5\)

M = \(155+\frac{8}{76-54} \times 5\)

M = \(155+\frac{40}{22}\)

M = \(\frac{3410+40}{22}\)

M = \(\frac{3450}{22}\)

M = 156.82

Question 19. The following table gives the weekly wage of workers in a factory:

CBSE School For Class 10 Maths Chapter 14 Statistics The Weekly Wage Of Workers In A Factory.

Find (1) the mean (2) the modal class (3) the mode

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Weekly Wage Of Workers In A Factory

1. Mean:

Mean = \(\frac{\sum f_{i x i}}{\sum f_i}=\frac{5520}{80}\)

= 69

2. Modal class = 55-60

3. Mode: clearly, the Modal class is 55-60 it has the maximum frequency

l = 55, f1 = 20, f0 = 5, f2 = 10, h = 5

M = \(l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

M = \(55+\frac{20-5}{2(20)-5-10} \times 5\)

M = \(55+\frac{15}{40-15} \times 5\)

M = \(55+\frac{15}{25} \times 5\)

M = \(\frac{1375+75}{25}\)

M = \(\frac{1450}{25}\)

M = 58

Question 20. The mode of the following Series is 36. Find the missing frequency in it:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mode Of The Following Series Is 36 Find The Missing Frequency

Solution:

Clearly, 30-40 is the modal class as mode 36 lies in this class

Here l = 30, f1 = 16, f0 = x (Say), f2 = 12 and h = 10 and mode 36

Mode (M) = \(\ell+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

36 = \(30+\frac{16-x}{2(16)-x-12} \times 10\)

36 = \(30+\frac{16-x}{32-x-12} \times 10\)

36 = \(30+\frac{16-x}{20-x} \times 10\)

36 – 30 = \(\frac{16-x}{20-x} \times 10\)

6(20-x) = 10(16-x)

120-6x = 160-10x

10x-6x = 160-120

4x = 40

x = \(\frac{40}{4}\)

x = 10

Question 21. Compute the mode of the following data:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mode Of The Data.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mode Of The Data

Clearly, Modal class 29.5 – 29.5 as it has the maximum frequency

∴ l = 19.5, f1 = 23, f0 = 16, f2 = 15, h = 10

M = \(\ell+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

M = \(19.5+\frac{23-16}{2(23)-16-15} \times 10\)

M = \(19.5+\frac{7}{46-31} \times 10\)

M = \(\frac{292.5+70}{15}\)

M = 24.17

Question 22. Find the mean, Median, and mode of the following data:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mode Of The Following Frequency Distribution

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean Median And Mode Of The Following Data

let assumed mean A = 70, h = 20, Ef = 50, and Σfu = -19

Mean \(\bar{x}=A+\left[h \times \frac{\sum f_u}{\sum f}\right]\)

⇒ \(\bar{x}=70+\left[20 \times \frac{-19}{50}\right]\)

⇒ \(\bar{x}=70+[20 x-0.38]\)

⇒ \(\bar{x}=70-7.6\)

⇒ \(\bar{x}=62.4\)

Here N = 50

⇒ \(\frac{N}{2}=\frac{50}{2}\) = 25

Cumulative frequency just greater than 25 is 36 and the Corresponding class is 60-80.

∴ l1 = 60, f = 12, l2 = 80, C = 24, i = 80-60 = 20

Now, median (M) = \(l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

M = \(60+\frac{(25-24)}{12} \times 20\)

M = \(60+\frac{1}{12} \times 20\)

M = \(\frac{720+20}{12}\)

M = \(\frac{740}{12}\)

M = 61.66

Mode = 3(Median)-2(Mean)

Mode = 3(61.66)-2(62.4)

= 184.98-124.8

M = 60.18

Question 23. 100 Surnames were randomly picked from a local directory and the distribution of a number of letters of the English alphabet in the Surname was obtained as follows:

CBSE School For Class 10 Maths Chapter 14 Statistics 100 Surnames Were Randomly Picked From A Local Directony And Letter Of The English Alphabet.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics 100 Surnames Were Randomly Picked From A Local Directony And Letter Of The English Alphabet

let assumed mean A = 11.5, h = 3, Σf = 100, Σfu = -106

Mean \(\bar{x}=A+\left[h \times \frac{\sum(f u)}{\sum f}\right]\)

⇒ \(\bar{x}=11.5+\left[3 \times \frac{-106}{100}\right]\)

⇒ \(\bar{x}=11.5+[3 \times-1.06]\)

⇒ \(\bar{x}=11.5-3.18\)

⇒ \(\bar{x}=8.32\)

Here, N = 100

⇒ \(\frac{N}{2}=\frac{100}{2}\)

= 50

Cumulative frequency just greater than 50 is 76 and the corresponding class is 10-13

∴ l1 = 7, f = 40, l2 = 10, C = 36, i = 10-7 = 3

Median (M) = \(\ell+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

M = \(7+\frac{50-36}{40} \times 3\)

M = \(7+\frac{14}{40} \times 3\)

M = \(\frac{280+42}{40}\)

M = \(\frac{322}{40}\)

M = 8.05

Mode = 3(Median)-2(Mean)

= 3(8.05)-2(8-32)

= 24.15-16.64

M = 7.51

Question 24. The following table gives the daily income of such workers of a factory:

CBSE School For Class 10 Maths Chapter 14 Statistics The Daily Income Of 50 Workers Of A Factory.

Find the mean, mode, and median of the above data.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Daily Income Of 50 Workers Of A Factory

let assumed mean A = 150, h = 20, Σf = 50, Σfu = -12

Mean \(\bar{x}=A+\left[h \times \frac{\sum f_4}{\sum P}\right]\)

⇒ \(\bar{x}=150+\left[20 \times \frac{-12}{50}\right]\)

⇒ \(\bar{x}=150+[20 x-0.24]\)

⇒ \(\bar{x}=150-4.8\)

⇒ \(\bar{x}=145.2\)

Here N = 50

⇒ \(\frac{N}{2}=\frac{50}{2}=25\)

Cumulative frequency just greater than 25 is 36 and the Corresponding is 120-140

∴ l1 = 120, f = 14, l2 = 140, C = 12, i = 140-120 = 20

Now, Median = \(l_1+\frac{\left(\frac{N}{2}- C\right)}{f} \times i\)

⇒ \(120+\frac{(25-12)}{14} \times 20\)

⇒ \(120+\frac{13}{14} \times 20\)

⇒ \(\frac{1680+260}{14}\)

⇒ \(\frac{1940}{14}\)

M = 138.57

Mode = 3(Median) – 2 (Mean)

M = 3(138.57)-2((45.2)

= 415.71-290.4

= 125.31

Question 25. A Survey regarding the heights (in cm) of so girls in a class was conducted and the following data was obtained:

CBSE School For Class 10 Maths Chapter 14 Statistics A Survey Regarding The Height In Cm Of 50 Girls Of A Class Was Conducted And The Following Data.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics A Survey Regarding The Height In Cm Of 50 Girls Of A Class Was Conducted And The Following Data

Let assumed Mean A=145, h=10, Σf=50, and Σfu=24

Mean \(\bar{x}=A+\left[h \times \frac{\sum f_u}{\sum f}\right]\)

⇒ \(\bar{x}=145+\left[10 \times \frac{24}{50}\right]\)

⇒ \(\bar{x}=145+[10 \times 0.48]\)

⇒ \(\bar{x}=145+4.8\)

⇒ \(\bar{x}=149.8 \mathrm{~cm}\)

Here N = 50

⇒ \(\frac{N}{2}=\frac{50}{2}=25\)

Cumulative frequency just greater than 25 is 42 and the Corresponding is 150-160

l1 = 150, f = 20, l2 = 160, C = 22, i = 160-150 = 10

Now, Median (M) = \(l_1+\frac{\left(\frac{N}{2}-c\right)}{F} \times i\)

⇒ \(150+\frac{(25-22)}{20} \times 10\)

⇒ \(150+\frac{30}{20}\)

⇒ \(\frac{3000+30}{20}\)

⇒ \(\frac{3030}{20}\)

= 151.5 cm

Mode = 3(Median)-2(Mean)

Mode = 3(151.5)-2(149.8)

= 454.5-299.6

= 154.9

Question 26. The table below shows the daily expenditure on food of 30 households in a locality:

CBSE School For Class 10 Maths Chapter 14 Statistics The Daily Expenditure On Food Of 30 House Holds In A Locality

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Daily Expenditure On Food Of 30 House Holds In A Locality.

Let assumed mean A=225, h = 50, Σf=30, Σfu = -12

Mean \(\bar{x}=A+\left[h \times \frac{\sum f u}{\sum f}\right]\)

⇒ \(\bar{x}=225+\left[50 \times \frac{-12}{30}\right]\)

⇒ \(\bar{x}=225+[50 \times -0.4]\)

⇒ \(\bar{x}=225-20\)

⇒ \(\bar{x}=205\)

Here, N = 30

⇒ \(\frac{N}{2}=\frac{30}{2}=15\)

Median Class = 200-250

l1 = 200, f = 12, l2 = 250, C = 13, i = 250-200 = 50.

Median (M) = \(l_1+\frac{\left(\frac{N}{2}-c\right)}{F} \times i\)

M = \(200+\frac{15-13}{12} \times 50\)

M = \(\frac{2400+100}{12}\)

M = \(\frac{2500}{12}\)

M = 208.33

CBSE Solutions For Class 10 Mathematics Chapter 13 Volume And Surface Area Of Solids

Class 10 Maths Volume And Surface Area Of Solids

Question 1. A tent of cloth is Cylindrical upto I’m height and Conical above it of the Same radius of base. If the diameter of the tent is 6m and the slant height of the Conical part is 5m, find the cloth required to make this tent.
Solution:

Diameter of base 2r = 6m

CBSESE School For Class 10 Maths Chapter 10 Volume And Surface Area Of Solids A Tent Of Cloth Is Cylindrical Up To Height And Conical Of The Sum Radius Of Base

r = \(\frac{6}{2}\) = 3m

Height of Cylindrical path h = 1m

The slant height of Conical part l = 5m

Cloth required in tent = 2πrh +πrl

= πr(2h+1)

⇒ \(\frac{22}{7}\) x 3 (2×1 +5)

⇒ \(\frac{22}{7}\) x 3(10)

= 66m2

CBSE Solutions For Class 10 Mathematics Chapter 13 Volume And Surface Area Of Solids

Question 2. The volume and Surface area of a Solid hemisphere are numerically equal. What is the diameter of the hemisphere?
Solution:

We have,

Volume of hemisphere = Surface area of hemisphere

⇒ \(\frac{2}{3}\) = 3πr2

⇒ \(\frac{2}{3}\) r = 3

2r = 9

Hence, the diameter of the hemisphere = 9 units.

Question 3. 2 Cubes each of volume 64 cm3 are joined end to end. Find the Surface area of the resulting Cuboid.
Solution:

Given,

volume of cube = 64 cm3

(Side)3 = 64

(Side)3 = 43

Side = 4cm

Side of cube = 4cm

A Cuboid is formed by joining two Cubes together.

∴ For Cuboid

length l = 4+4=8cm,

breadth b = 4cm

height b = 4cm

Now, the total surface area of the cuboid.

= 2(l.b+b.h+l.h)

2(8×4+ 4×4+8×4)

= 2(32+16+32)

2(80)

= 160 cm2

Question 4. From a Solid Cylinder whose height is 2.4cm and diameter is 1.4 cm, a Conical Cavity of the Same height and diameter is hollowed out. Find the total Surface area of the remaining Solid to the nearest Cm2.
Solution:

Diameter of Cylinder 2r = 1.4cm

r = 0.7cm

∴ Radius of Cylinder = radius of cone = r= 0.70m

Height of Cylinder = height of cone

h = 2.40m

If the slant height of a cone is l, then

l2 = h2+r2 = (2.4)2 + (0.7)2

5.76 +0.49 = 6.25

1 = \(\sqrt{6.25}\) = 2.5cm

CBSESE School For Class 10 Maths Chapter 10 Volume And Surface Area Of Solids A Solid Cylinder Whose Height And Diameter The Total Surface Area Of The Remaining Solid To The Nearest Cm

The surface area of the remaining solid = area of the base of the cylinder + curved Surface of the cylinder + curved surface of the cone

πr2 + 2πrh + πrl

=πr (r+2h+1)

= \(\frac{22}{7}\) × 0.7 (0.7+2×2.4+2.5)

= \(\frac{22}{7}\) × 0.7 × (5.86) 223×0.7

= \(\frac{22}{7}\): × 4.102

= 17.6cm2

Question 5. The radius and height of a solid right Circular Cone are in the ratio of 5:12. If its volume is 314 cm3, find its total Surface area. [Take π = 3.14]
Solution:

CBSESE School For Class 10 Maths Chapter 10 Volume And Surface Area Of Solids The Radius And Height Of A Solid Right circular Cone

Let the radius of Cone = 5x

∴ Height of Cone =12x

l2 = (5x)2+ (12x)2

l2 = 25x2 +144x2

l2 = 169x2

l = \(\sqrt{169 x^2}\)

l = 13x

It is given that volume = 314 cm2

∴ \(\frac{1}{3}\)π(5x)2 (12x) = 314

⇒ \(\frac{1}{3}\) × 3.14 x 25×12 × x3 = 314

⇒  x3 = \(\frac{314 \times 3}{3.14 \times 25 \times 12}\) = 1

∴ x = 1cm

∴ Radius r = 5×1 = 5cm

Height h = 12×1 = 12cm

and slant height 1 = 13×1 = 13cm

Now, total surface area of Cone = πr(l+r) = 3.14 × 5(13+5) = 3.14×5×18

= 282.60cm2

Hence, the total surface area of cone is 282.60cm2

Question 6. The Curved Surface area of a Cone of height 8m is 188.4m2. Find the volume of Cone.
Solution:

πrl = 188.4

⇒ rl = \(\frac{188.4}{3.14}\) = 60

CBSESE School For Class 10 Maths Chapter 10 Volume And Surface Area Of Solids The Curved Surface Area Of A Cone

r2l2 = 3600

r2 (h2 +r2)=3600

r2 (64+r2)=3600

r4+64r2 = 3600 = 0

(r2 +100) (r2 – 36)=0

∴ r2 = -100 or r2=36

r = 6

(r2 = -100 is not possible)

∴ Volume of a cone = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \times 3.14 \times 36 \times 8=301.44 \mathrm{~m}^3\)

Question 7. A Conical tent is required to accommodate 157 persons, each person must have 2m2 of space on the ground and 15m3 of air to breathe. Find the height of the tent” Also Calculate the slant height.
Solution:

CBSESE School For Class 10 Maths Chapter 10 Volume And Surface Area Of Solids A Conical Tent Is Required To Accommodate Persons And Also Calculate The Slant Height

1 person needs 2m2 of Space.

∴ 157 Persons needs 2 × 157 m2 of Space on the ground.

πr2 = 2 × 157

∴ r2 = \(\frac{2 \times 157}{3.14}\)

r2 = 100

r = 10m

Also, 1 person needs 15m3 of air.

∴ 157 Persons need 15×157 m2 of air.

∴ \(\frac{1}{3}\) πr2h = 15×157

⇒ h = \(\frac{15 \times 157 \times 3}{3.14 \times 100}\) = 22.5m

∴ l2 = h2 + r2 = (225)2 + (10)2 = 606.25

∴ l = \(\sqrt{606.25}\) = 24.62m

Question 8. Three Cubes of metal whose edges are in the ratio 3:4:5 are melted down into a single cube whose diagonal is \(12 \sqrt{3}\) cm. Find the edges of the three cubes.
Solution:

The ratio in the edges = 3:4:5

Let edges be 3x, 4x and 5x respectively.

∴ Volumes of three cubes will be 27x3, 64x3, and 125x3 in cm3 respectively.

Now, the Sum of the Volumes of these three Cubes = 27x3 + 64x3+125x3

216x3 cm3

Let the edge of the new cube be a cm.

∴ Diagonal of new Cube = \(a \sqrt{3}\) cm

∴ \(a \sqrt{3}=12 \sqrt{3}\)

⇒ a = 12

∴ Volume of new Cube = (12)3 = 1728 cm3

Now by the given Condition

216x3 = 1728

x3 = 8

x = 2

∴ Edge of 1 Cube = 3×2 = 6cm

Edge of 2 Cube = 4×2 = 8 Cm

Edge of 3 Cube = 5×2 = 10cm

Question 9. A Solid is in the shape of a Cone Standing on a hemisphere with both their radii being equal to Icm and the height of the cone is equal to its radius. Find the volume of the Solid in terms of π.
Solution:

Rodius of hemisphere = radius of Cone = r = 1 cm

Height of Cone h = radius of Cone = 1 cm

Volume of hemisphere = \(\frac{2}{3}\) πr3

volume of Come = \(\frac{1}{3}\) πr2h

∴ Volume of Solid = Volume of hemisphere + volume of Come

⇒ \(\frac{2}{3} \pi r^3+\frac{1}{3} \pi r^2 h\)

⇒ \(\frac{2}{3} \pi(1)^3+\frac{1}{3} \pi(1)^2(1)\)

⇒ \(\frac{2}{3} \pi+\frac{1}{3} \pi\)

= π cm3

Question 10. A granary is in the shape of a Cuboid of a size 8m x 6m x 3m. If a bag of grain occupies a space of 0.65 m3, how many bags Can be stored in the granary?
Solution:

The Size of the granary is 8m x 6m x 3m.

∴ volume of granary =8×6×3 = 144m3

the volume of one bag of grain = 0.65m3

∴ The number of bags that can be stored in the granary

= \(\frac{\text { volume of granary }}{\text { volume of each bag }}\)

= \(\frac{144}{0.65}\)

= 221.54 or 221 bags.

Question 11. A cylindrical bucket 28cm in diameter and 72cm high is full of water. the water is emptied into a rectangular tank 66 cm long and 28cm wide. Find the height of the water level in the tank.
solution:

Let the height of the water level in the tank = xm, then according to problem

πr2h = l×b×x

Or \(\frac{22}{7} \times 14 \times 14 \times 72=66 \times 28 \times x\)

Or \(x=\frac{\frac{22}{7} \times 14 \times 14 \times 72}{66 \times 28}\)

x = 24 Cm

Question 12. A Solid Spherical ball of Iron with a radius of 6cm is melted and recast into three Solid Spherical balls. The radii of the two balls are 3cm and 4cm respectively, determining the diameter of the third ball.
Solution:

Let the radius of the third ball = r cm

∴ The volume of three balls formed = volume of the ball melted

⇒ \(\frac{4}{3} \pi(3)^3+\frac{4}{3} \pi(4)^3+\frac{4}{3} \pi(r)^3=\frac{4}{3} \pi(6)^3\)

⇒ 27 +64 +r3 = 216

⇒ r3 = 125, i.e., r = 5cm

The diameter of the third ball = 2×5cm = 10cm

Question 13. A Semicircle of radius 17.5cm is rotated about its diameter. Find the Curved Surface of So generated Solid.
Solution:

The Solid generated by a circle rotated about its diameter is a Sphere

Now, a radius of Sphere r = 17.5 cm

and its Curved Surface = 4π2

= \(4 \times \frac{22}{7} \times 17.5 \times 17.5\)

= 3850 cm2

Question 14. A sphere of radius 6cm is melted and recast into a Cone of height 6cm. Find the radius of the Cone.
Solution:

Radius of Sphere = 6cm

∴ Volume of Sphere = \(\frac{4}{3} \pi(6)^3\) = 288π cm3

Let radius of Cone = r

Height of Cone = 6cm

∴ Volume of Cone = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi r^2 \times 6=2 \pi r^2\)

Given that, Volume of Cone = volume of a sphere

2πr2 = 288π

r2 = 144

r = 12

Therefore, a radius of Cone = 12cm

Question 15. A metallic Cylinder of diameter 16 cm and height 9cm is melted and recast Into Sphere of diameter 6cm. How many such Spheres can be formed?
Solution:

For the Cylinder,

Radius = \(\frac{16}{2}\) = 8cm

Height = 9cm

∴ Volume of Cylinder = π (8)2(9) = 576 π cm3

Diameter of Sphere = 6cm

∴ Radius of Sphere = \(\frac{6}{2}\) = 3 cm

Now, Volume of one Sphere = \(\frac{4}{3} \pi(3)^3\) = 36π cm3

∴ Number of Spheres formed = \(=\frac{\text { volume of Cylinder }}{\text { volume of one Sphere }}\)

⇒ \(\frac{576 \pi}{36 \pi}\)

= 16

Question 16. The volume of a sphere is 288π Cm3, 27 Small Spheres Can be formed with this Sphere. Find the radius of the Small Sphere.
Solution:

Volume of 27 Small Spheres = Volume of One big Sphere = 288π

⇒ Volume of 1 Small Sphere = \(\frac{288}{27} \pi\)

⇒ \(\frac{4}{3} \pi r^3=\frac{22}{3} \pi\)

⇒ r3 = 8

⇒ r = 2cm

Question 17. A metallic Sphere of radius 4.2cm is melted and recast into the Shape of a Cylinder of radius 6cm, Find the height of the cylinder.
Solution:

Radius of Sphere R = 4.2cm

Radius of Cylinder r = 6cm

Let the height of the Cylinder = h

Now, the volume of the cylinder = Volume of the Sphere

⇒ \(\pi r^2 h=\frac{4}{3} \pi R^3\)

⇒ h = \(\frac{4 R^3}{3 r^2}\)

⇒ \(\frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 6 \times 6} \mathrm{~cm}\)

= 2.744 сm

∴ Height of Cylinder = 2.744cm

Question 18. Metallic Spheres of radii 6cm, 8cm, and 10 cm, respectively, are melted to form a Single Solid Sphere. Find the radius of the resulting Sphere.
Solution:

Let r1 = 6cm r2 =8cm and r3 = 10cm

let the radius of a bigger Solid Sphere = R

The volume of bigger Solid Volume = Sum of volumes of three given Spheres

⇒ \(\frac{4}{3} \pi R^3=\frac{4}{3} \pi r_1^3+\frac{4}{3} \pi r_2^3+\frac{4}{3} \pi r_3^3\)

⇒ \(R^3=r_1^3+r_2^3+r_3^3\)

⇒ \(R^3=6^3+8^3+10^3\)

⇒ R3 = 216 +512 +1000

⇒ R3 = 1728

⇒ R2 = 123

⇒ R = 12cm

∴ Radius of new Solid Sphere = 12 cm

Question 18. A 20m deep well with a diameter 7m is dug and the earth from digging is evenly Spread out to form a platform 22m of the platform.
Solution:

Radius of well, r = \(\frac{7}{2} m\)

and depth h = 20m

Let the height of the platform be H meter.

∴ The volume of platform = Volume of well

⇒ 22 x 14 x H = πr2h

⇒ \(22 \times 14 \times H=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20\)

⇒ H = \(\frac{7 \times 5}{14}\)

⇒ H = \(\frac{35}{14}\)

⇒ H = 2.5mn

Height of platform = 2.5m

Question 20. A Cone is divided into two parts by drawing a plane through the mid-point of its axis, parallel to its base, Compare the volume of the two parts.
Solution:

We Can Solve this using Similarity

Let r and h be the radius and height of a Cone OAB

CBSESE School For Class 10 Maths Chapter 10 Volume And Surface Area Of Solids A Cone Is Divided Into Two Parts By Drawing A Plane Through The Mid Point Of Its Axis

Let OE = \(\frac{h}{2}\)

As OED and OFB are Similar

∴ \(\frac{OE}{O F}=\frac{ED}{F B}\)

⇒ \(\frac{h / 2}{h}=\frac{ED}{r}\)

⇒ ED = \(\frac{r}{2}\)

Now volume of Cone OCD = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \times \pi \times\left(\frac{r}{2}\right)^2 \times \frac{h}{2}\)

⇒ \(\frac{\pi r^2 h}{24}\)

and volume of Cone DAB = \(\frac{1}{3} \times \pi x r^2 \times h\) = \(\frac{\pi r^2 h}{3}\)

∴ \(\frac{\text { volume of Part } O C D}{\text { volume of Part } C D A B}\)

⇒ \(\frac{\frac{\pi r^2 h}{24}}{\frac{\pi r^2 h}{3}-\frac{\pi r^2 h}{24}}\)

⇒ \(\frac{\frac{1}{24}}{\frac{1}{3}-\frac{1}{24}}\)

⇒ \(\frac{\frac{1}{24}}{\frac{1}{3}-\frac{1}{24}}\)

⇒ \(\frac{\frac{1}{24}}{\frac{8-1}{24}}\)

⇒ \(\frac{1}{7}\)

Question 21. A well of diameter 3m is dug lum deep. The earth taken out of it has been Spread evenly all around it in the Shape of a Circular ring of width um to form an embankment. Find the height of the embankment.
Solution:

Diameter of well 2r = 3m

r = \(\frac{2}{3}\) = 1.5m

and depth h = 14m

∴ Volume of earth taken out from well = πr2h

⇒ \(\frac{22}{7} \times 1.5 \times 1.5 \times 14=99 \mathrm{~m}^3\)

Now, the outer radius of the well, R = 1.5+4 = 5.5m

∴ Area of the ring of platform = π(R2– r2)

⇒ \(\frac{22}{7}\left[(5.5)^2-(1.5)^2\right]\)

⇒ \(\frac{22}{7}[30.25-2.25]\)

⇒ \(\frac{22}{7} \times 7 \times 4=88 \mathrm{~m}^2\)

let the height of the embankment = H

∴ 88 x H=99

⇒ \(H=\frac{99}{88}=\frac{9}{8}\)

= 1.125m

∴ Height of embankment = 1.125m

Important Questions for Class 12 Chemistry Chapter 10 Biomolecules

Biomolecules

Question 1. Glucose or succrose are soluble in water but cyclohexane or benzene (simple six-membered ring compounds) are insoluble in water. Explain.
Answer:

  • A glucose molecule contains five -OH groups while a sucrose molecule contains eight -OH groups. Thus, glucose and sucrose undergo extensive H-bonding with water.
  • Hence, they are soluble in water.
  • But, cyclohexane and benzene do not contain -OH groups. Hence, they cannot undergo II- bonding with water and as a result, they are insoluble in water.

Question 2. What are the expected products of the hydrolysis of lactose?
Answer:

Lactose is composed of β-D galactose and β-D glucose. Tims, on hydrolysis, gives β-D galactose and β-D glucose.

Biomolecules Lactose

Biomolecules Lactose And D Glucose

Question 3. How do you explain the absence of an aldehyde group in the pentaacctate of D-glucose?
Answer:

D-glucose reacts with hydroxylamine (NH2, OH) to form an oxime because of the presence of an aldehydic (-CHO) group of carbonyl carbon. This happens as the cyclic structure of glucose forms an open chain structure in an aqueous medium, which then reacts with NH2, OH to give an oxime.

Biomolecules Glucose And Oxime

But pentaacctate of D-glucose does not react with NH2, OH. This is because pentaacctate does not form an open chain structure.

Read and Learn More Class 12 Chemistry with Answers Chapter Wise

Question 4. The melting points and solubility in water of amino acids are generally higher than those of the corresponding halo acids. Explain.
Answer:

Both acidic (carboxyl) as well as basic (amino) groups are present in the same molecule of amino acids. In aqueous solutions, the carboxyl group can lose a proton and the amino group can accept, thus giving rise to a dipolar ion known as a zwitter ion.

Biomolecules Zwitter Ion

  • Due to this dipolar behaviour, they have strong electrostatic interactions within them and with water. However, halo-acids do not exhibit such dipolar behaviour.
  • For this reason, the melting points and the solubility of amino acids in water are higher than those of the corresponding halo acids.

Question 5. Where does the water present in the egg go after boiling the egg?
Answer:

When an egg is boiled, the proteins present inside the egg get denatured and coagulate. After boiling the egg. the water present in it is absorbed by the coagulated protein through H-bonding.

CBSE Class 12 Chemistry Chapter 10 Biomolecules Question And Answers

Question 6. Why cannot vitamin C be stored in our bodies?
Answer: Vitamin C cannot be stored in our body because it is water-soluble. As a result, it is readily excreted in the urine.

Question 7. What products would be formed when a nucleotide from DMA containing thymine is hydrolysed?
Answer: When a nucleotide from the DNA containing thymine is hydrolyzed, thymine. β-D-2- deoxyribose and phosphoric acid are obtained as products.

Question 8. When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?
Answer:

  • A DNA molecule is double-stranded in which the pairing of bases occurs. Adenine always pairs with thymine, while cytosine always pairs with guanine.
  • Therefore, on hydrolysis of DNA. the quantity of adenine produced is equal of that of thymine and similarly, the quantity of cytosine is equal to that of guanine.
  • But when RNA is hydrolyzed, there is no relationship among the quantities of the different bases obtained. Hence, RNA is single-stranded.

Question 9. What are monosaccharides?
Answer:

  • Monosaccharides are carbohydrates that cannot be hydrolysed further to give simpler units of polyhydroxy aldehyde or ketone. Monosaccharides are classified based on a number of carbon atoms and the functional group present in them.
  • Monosaccharides containing an aldehyde group are known as aldose and those containing a keto group are known as ketose.
  • Monosaccharides are further classified as triose, tetrose, pentose, Itexose, and heptose according to the number of carbon atoms they contain. For example, a ketose containing 3 carbon atoms is called ketotriose and an aldose containing 3  carbon atoms is called aldotriose.

Question 10. What are reducing sugars?
Answer: Reducing sugars are carbohydrates that reduce l’ehling’s solution and Tollen’s reagent. All monosaccharides and disaccharides, excluding sucrose, are reducing sugars.

Question 11. Write two main functions of carbohydrates in plants.
Answer:

Two main functions of carbohydrates in plants are:

  1. Polysaccharides such as starch serve as storage molecules.
  2. Cellulose, a polysaccharide, is used to build the cell wall.

Question 12. Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose. maltose, galactose, fructose and lactose
Answer:

  1. Monosaccharides: Ribose, 2-deoxyribose, galactose, fructose
  2. Disaccharides: Maltose, lactose

Question 13. What do you understand by the term glycosidic linkage?
Answer:

  • Glycosidic linkage refers to the linkage formed between two monosaccharide units through an oxygen atom by the loss of a water molecule.
  • For example, in a sucrose molecule, two monosaccharide units, α- glucose and β- fructose, are joined together by a glycosidic linkage.

Biomolecules Surcose

Question 14. What is glycogen? How is it different from starch?
Answer:

Glycogen is a carbohydrate (polysaccharide). In animals, carbohydrates are stored as glycogen. Starch is a carbohydrate consisting of two components – amylose (15-20%) and amylopectin (80-85%). However, glycogen consists of only one component whose structure is similar to amylopectin. Also, glycogen is more branched than amylopectin.

Question 15. What are the hydrolysis products of (1) sucrose and (2) lactose?
Answer: On hydrolysis, sucrose gives one molecule of α-D glucose and one molecule of β-D-fructose.

Biomolecules The Hydrolysis Production Of Sucrose And Lactose 1

The hydrolysis of lactose gives β-D-galactose and β-D-glucose.

Biomolecules The Hydrolysis Production Of Sucrose And Lactose 2

Question 16. What is the basic structural difference between starch and cellulose?
Answer:

Starch consists of two components – amylose and atm lopectin. Amylose is a long linear chain of α-D-(+)-glucose units joined by C1– C4 glycosidic linkage (α-link).

Biomolecules Basic Steructural Difference Between Starch And Cellulose 1

Amylopectin is a branched-chain polymer of α—D—glucose units, in which the chain is formed by C1 – C4 glycosidic linkage and the branching occurs by C1 – C6, glycosidic linkage.

Biomolecules Basic Steructural Difference Between Starch And Cellulose 2

On the other hand, cellulose is a straight-chain polysaccharide of β-D-glucose units joined by C1 – C4 glycosidic linkage (β-link).

Biomolecules Basic Steructural Difference Between Starch And Cellulose 3

Question 17. What happens when D-glucose is treated with the following reagents?

  1. HI
  2. Bromine water
  3. HNO3

Answer:

When D-glucose is heated with III for a long time, n-hexane is formed.

Biomolecules D Glucose Is Treated With HI

When D-glucose is treated with Br2 water, D-gluonic acid is produced.

Biomolecules D Glucose Is Treated With Bromine Water

On being treated with HNO3, D-glucose gets oxidised to give saeeliaric acid.

Biomolecules D Glucose Is Treated With HNO3

Question 18. Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.
Answer:

  1. Aldehydes give 2, 4-DNP test. Schiff’s test, and react with NaHSO4 to form the hydrogen sulphite addition product. However, glucose does not undergo these reactions.
  2. The pentaacetate of glucose does not react with hydroxylamine. This indicates that a free-CHO group is absent from glucose.
  3. Glucose exists in two crystalline forms -α and β. The α-form (m.p. 419 K) crystallises from a concentrated solution of glucose at 303 K and the β-form (m.p.=423 K) crystallises from a hot and saturated aqueous solution at 371 K. This behaviour cannot be explained by the open chain structure of glucose.

Question 19. What are essential and non-essential amino acids? Give two examples of each type.
Answer:

Essential amino acids are required by the human body, but they cannot be synthesised in the body. They must be taken through food. For example: valine and leucine Non-essential amino acids are also required by the human body, but they can be synthesised in the body.

For example: Glycine, and Alanine

  1. Peptide linkage
  2. Primary structure
  3. Denaturation

Question 20. Define the following as related to proteins

  1. Peptide linkage
  2. Primary structure
  3. Denaturation

Answer:

Peptide linkage: The amide for pied between the -COOH group of one molecule of an amino acid and the -NH2, group of another molecule of the amino acid by the elimination of a water molecule is called a peptide linkage.

Biomolecules Peptide Linkage

Primary structure: The primary structure of a protein refers to the specific sequence in which various amino acids are present in it. i.e. the sequence of linkage between amino acids in a polypeptide chain. The sequence in which amino acids are arranged is different in each protein. A change in the sequence creates a different protein.

Denaturation: In a biological system, a protein is found to have a unique 3- dimensional structure and a unique biological activity. In such a situation, the protein is called native protein.

  • However, when the native protein is subjected to physical changes such as changes in temperature or chemical changes such as changes in pH. its H-bonds are disturbed.
  • This disturbance unfolds the globules and uncoils the helix. As a result, the protein loses its biological activity. This loss of biological activity by the protein is called denaturation.
  • During denaturation, the secondary and the tertiary structures of the protein get destroyed, but the primary structure remains unaltered.  One of the examples of denaturation of proteins is the coagulation of egg white when an egg is boiled.

Question 21. What are the common types of secondary structures of proteins?
Answer:

There are two common types of secondary .structure of proteins:

  1. α- helix structure
  2. β-pleated sheet structure

α- Helix Structure: In this structure, the Nil group of an amino acid residue forms H-bonds with the group of the adjacent turn of the right-handed screw (α-helix).

β-pleated Sheet Structure: This structure is called so because it looks like the pleated folds of draper). In this structure, all the peptide chains are stretched out to nearly the maximum extension and then laid side by side. These peptide chains are held together by intermolecular hydrogen bonds.

Biomolecules Alpha And Beta Plated Steructure

Question 22. What type of bonding helps in stabilising the a -helix structure of proteins?
Answer: The H-bonds formed between the -Ml group of each amino acid residue and the Biomolecules Amino Acid group of
the adjacent turns of the α-helix help in stabilising the helix.

Question 23. Differentiate between globular and fibrous proteins.
Answer:

Biomolecules Fibrous Protein And Globular Protein

Question 24. How do you explain the amphoteric behaviour of amino acids?
Answer: In an aqueous solution, the carboxyl group of an amino acid can lose a proton and the amino group can accept a proton to give a dipolar ion known as a zwitter ion.

Biomolecules Zwitter Ion.

Therefore, in Zwitter ionic form, the amino acid can act both as an acid and as a base.

Biomolecules Zwitter Ion Forms The Amino Acid

Thus, amino acids show amphoteric behaviour.

Question 25. What are enzymes?
Answer:

  • Enzymes are proteins that catalyse biological reactions. They are very specific and catalyse only a particular reaction for a particular substrate. Enzymes are usually named after a particular substrate or class of substrate and sometimes after a particular reaction.
  • For example, enzymes used to catalyse the hydrolysis of maltose into glucose are named as maltase.

Biomolecules Maltose And Glucose

Again, the enzymes used to catalyse the oxidation of one substrate with the simultaneous reduction of another substrate are named oxidoreductase enzymes, the name of an enzyme ends with ’-ase’.

Question 26. What is the effect of denaturation on the structure of proteins?
Answer:

  • As a result of denaturation. globules get unfolded and helixes get uncoiled. Secondary and tertiary structures of protein are destroyed, but the primary structures remain unaltered. It can be said that during denaturation.
  • secondary and tertiary-structured proteins get converted into primary-structured proteins. Also, as the secondary and tertiary structures of a protein are destroyed, the enzyme loses its activity.

Question 27. How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
Answer:

Biomolecules Vitamins Classified

Based on their solubility in water or fat. vitamins are classified into two groups.

  1. Fat-soluble Vitamins: Vitamins that are soluble in fat and oils, but not in ‘water, belong to this group.
    • For example: Vitamins A, D, E, and K
  2. Water-soluble Vitamins: Vitamins that are soluble in water belong to this group,
    • For example: B group vitamins (B1, B2, B6, B12, etc.) and vitamin C

However, biotin or vitamin H is neither soluble in water nor fat. Vitamin K is responsible for the coagulation of blood.

Question 28. Why are vitamin A and vitamin C essential to us? Give their important sources,
Answer:

  • The deficiency of vitamin A leads to xerophthalmia (hardening of the cornea of the eye) and night blindness. The deficiency of vitamin C leads to scurvy (bleeding gums).
  • The source of vitamin A is fish liver oil, carrots, butter, and milk. The sources of vitamin C are citrus fruits. amla, and green leafy vegetables.

Question 29. What are nucleic acids? Mention their two important functions.
Answer:

Nucleic acids are biomolecules found in the nuclei of all living cells, as one of the constituents of chromosomes. There are mainly two types of nucleic acids – deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Nucleic acids are also known as polynucleotides as they are long-chain polymers of nucleotides.

Two main functions of nucleic acids are:

  1. DNA is responsible for the transmission of inherent characteristics from one generation to the next. This process of transmission is called heredity.
  2. Nucleic acids (both DNA and RNA) are responsible for protein synthesis in a cell, even though the proteins are synthesised by the various RNA molecules in a cell, the message for the synthesis of a particular protein is present in DNA.

Question 30. What is the difference between a nucleoside and a nucleotide?
Answer:

A nucleoside is formed by the attachment of a base to the 1′ position of the sugar.

Nucleoside = Sugar + Base

On the other hand, all three basic components of nucleic acids (i.e. pentose sugar, phosphoric acid, and base) are present in a nucleotide.

Biomolecules Nucleoside And Nucleotide

Question 31. The two strands in DNA are not identical bill are complementary. Explain.
Answer:

In the helical structure of DNA. the two strands are held together by hydrogen bonds between specific pairs of bases. Cytosin forms hydrogen bonds with guanine, while adenine forms hydrogen bonds with thymine. As a result, the two strands are complementary to each other.

Question 32. Write the important structure and functional difference between DNA and RNA.
Answer: The structural differences between DNA and RNA are as follows:

Biomolecules Structural Difference Between DNA And RNA

The functional differences between DNA and RNA are as follows:

Biomolecules Functional Difference Between DNA And RNA

Question 33. What are the different types of RNA found in the cell?
Answer:

  1. Messenger RNA (m-RNA)
  2. Ribosomal RNA (rRNA)
  3. Transfer RNA (t-RNA)

Question 34. What are hormones and write to its functions?
Answer:

Hormones are molecules that act as intercellular messengers. These are produced by endocrine glands in the body and are poured directly into the bloodstream which transports them to the site of action.

In terms of chemical nature, some of these are steroids, for example., estrogens and androgens: some are poly peptides for example insulin and endorphins and some others are amino acid derivatives such as epinephrine and norepinephrine.

  1. Insulin keeps the blood glucose level within a narrow limit. Hormone glucagon tends to increase the glucose level in the blood. The two hormones together regulate the glucose level in the blood.
  2. Epinephrine and norepinephrine mediate responses to external stimuli. Growth hormones and sex hormones play a role in growth and development. Thyroxine produced in the thyroid gland is an iodinated derivative of the amino acid tyrosine.
  3. Steroid hormones are produced by the adrenal cortex and gonads (testes in males and ovaries in females).
  4. Glucocorticoids control carbohydrate metabolism, modulate inflammatory reactions and are involved in reactions to stress.
  5. Testosterone is the major sex hormone produced in males. It is responsible for the development of secondary male characteristics (deep voice, facial hair, general physical constitution) and estradiol is the main female sex hormone. It’s responsible for the development of secondary female characteristics and participates in the control of the menstrual cycle.
  6. Progesterone is responsible for preparing the uterus for the implantation of a fertilised egg.

Important Questions for Class 12 Chemistry Chapter 9 Amines

Amine

Question 1. Classify the following amines as primary, secondary, or tertiary:

Amine Following Amines As Primary Secondary And Teritary

Answer:

  1. Primary: (1) and (3)
  2. Secondary: (4)
  3. Tertiary: (2)

Question 2. Write structures of different isomeric amines corresponding to the molecular formula. C4H11N

  1. Write the IUPAC names of all the isomers.
  2. What type of isomerism is exhibited by different pairs of amines?

Answer: (1),(2) The structures and their IUPAC names of different isomeric amines corresponding to the molecular formula. C4H11N are given below:

Amine Structure Of Different Isomeric Amines

(3) The pairs (1) and (2). (5) and (7) exhibit position isomerism. The pairs (1) and (3); (2) and (d) exhibit chain isomerism. The pairs (5) (6) and (7) exhibit metamerism.

All primary amines exhibit functional isomerism with secondary and tertiary amines and vice-versa.

Read and Learn More Class 12 Chemistry with Answers Chapter Wise

Question 3. How will you convert

  1. Benzene into aniline
  2. Benzene into N, N-dimethylaniline
  3. Cl—(CH2)4-Cl into hexan-1, 6-diamine?

Answer:

Amine Benzene Into Aniline And N Dimethylaniline And 6 Diamine

Question 4. Arrange the following in increasing order of their basic strength:

  1. \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2, \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2, \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2 \text { and }\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}\)
  2. \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2 \cdot\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH},\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N}, \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2\)
  3. \(\mathrm{CH}_3 \mathrm{NH}_2,\left(\mathrm{CH}_3\right)_2 \mathrm{NH},\left(\mathrm{CH}_3\right)_3 \mathrm{~N}, \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 \cdot \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2\)

Answer:

  1. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2<\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2<\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2<\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}\)
  2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2<\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2<\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N}<\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}\)
  3. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2<\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2<\left(\mathrm{CH}_3\right)_3 \mathrm{~N}<\mathrm{CH}_3 \mathrm{NH}_2<\left(\mathrm{CH}_3\right)_2 \mathrm{NH}\)

Question 5. Complete the following acid-base reactions and name the products:

  1. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NH}_2+\mathrm{HCl} \longrightarrow\)
  2. \(\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N}+\mathrm{HCl} \longrightarrow\)

Answer:

Amine Acid Base Reactions

CBSE Class 12 Chemistry Chapter 9 Amines Important Question And Answers

Question 6. Write reactions of the final alkylation product of aniline with excess methyl iodide in the presence of sodium carbonate solution.

Answer: Aniline reacts with methyl iodide to produce N. N-dimethylaniline.

Amine N Dimethylaniline

With excess methyl iodide, in the presence of Na2CO3 solution. N, N-dimethylaniline produces N, N, N-trimethylanilinum carbonate.

Amine N Trimethylanilium Carbonate

Question 7. Write the chemical reaction of aniline with benzyl chloride and write the name of the product obtained.
Answer:

Amine Chemical Reaction Of Aniline With Benozyl Chloride

Question 8. Write the structure of different isomers corresponding to the molecular formula, C3H9N. Write IUPAC names of the isomers that will liberate nitrogen gas on treatment with nitrous acid.
Answer:

The structure of different isomers corresponding to the molecular formula, C3H9N are given below:

Amine Structure Of Different Isomers

1 amine i.e., (1) propan-1-amine, (2) Propan-2-amine will liberate nitrogen gas on treatment with nitrous acid.

Amine Propan One Amine And Propan Two Amine

Question 9. Convert

  1. 3-Methylaniline into 3-nitrotoluene.
  2. Aniline into 1,3,5-tribromobenzene

Answer:

Amine Three Nitrotoluene And Aniline Tribromobenzene

Question 10. Write chemical equations for the following reactions:

  1. Reaction ofelhanolie NH3 with C2H5CI.
  2. Ammonolysis of benzyl chloride and the reaction of amine so formed with two moles of CH3CI.

Answer:

Amine Reaction Of Ethanoic And Ammonolysis Of Benzyl Chloride

Question 11. Write chemical equations for the following conversions:

  1. \(\mathrm{CH}_3 \cdot \mathrm{CH}_2 \cdot \mathrm{Cl} \text { into } \mathrm{CH}_3 \cdot \mathrm{CH}_2 \cdot \mathrm{CH}_2 \cdot \mathrm{NH}_2\)
  2. \(\mathrm{C}_6 \mathrm{H}_5 \cdot \mathrm{CH}_2 \cdot \mathrm{Cl} \text { into } \mathrm{C}_6 \mathrm{H}_5 \cdot \mathrm{CH}_2 \cdot \mathrm{CH}_2 \cdot \mathrm{NH}_2\)

Answer:

Amine Chemical Equations Of The Conversion

Question 12. Write structures and IUPAC names of

  1. The amide gives propanamide by Hoffmann bromamide reaction.
  2. The amine produced by the Hoffmann degradation of ben/amide.

Answer:

Propanamine contains three carbons. Hence, the amide molecule must contain four carbon atoms. The structure and IUPAC name of the starling amide with four carbon atoms are given below:

Amine Butanamide

Benzamide is an aromatic amide containing seven carbon atoms. Hence, the amine formed from benzamide is an aromatic primary amine-containing six carbon atoms.

Amine Aniline Or Benzenamine

Question 13. Arrange the following mg in decreasing order of their basic strength:

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 \cdot \mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2 \cdot\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH} . \mathrm{NH}_3\)

Answer: The decreasing order of basic strength of the above amines and ammonia follows the following order:

⇒ \(\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}>\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2>\mathrm{NH}_3>\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2\)

Question 14. How will you convert 4-nitrotoluene to 2-bromobenzoic acid?

Amine Four Nitrotoluene To Two Bromobenzoic Acid

Question 15. Write the IUPAC names of the following compounds and classify them into primary, secondary, and tertiary amines.

  1. \(\left(\mathrm{CH}_3\right)_2 \mathrm{CH} \mathrm{N} \mathrm{H}_2\)
  2. \(\mathrm{CH}_3\left(\mathrm{CH}_2\right)_2 \mathrm{NH}_2\)
  3. \(\mathrm{CH}_3 \mathrm{NHCH}\left(\mathrm{CH}_3\right)_2\)
  4. \(\left(\mathrm{CH}_3\right)_3 \mathrm{C} \mathrm{NH}_2\)
  5. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_3\)
  6. \(\left(\mathrm{CH}_3 \mathrm{CH}_2\right)_2 \mathrm{NCH}_3\)
  7. \(\mathrm{m}-\mathrm{BrCC}_6 \mathrm{H}_4 \mathrm{NH}_2\)

Answer:

  1. Propan – 2 – amine (p)
  2. Propan – I – amine(p)
  3. N – methyl – 2 – propanamide (s)
  4. 2 – methyl propane – 2 – amine (p)
  5. N – methyl aniline (s)
  6. N – ethyl – N – methyl ethanolamine (t)
  7. 3 – Bromo benzydamine or 3 – Bromoaniline (p)

Question 16. Give one chemical test to distinguish between the following pairs of compounds.

  1. Methylamine and dimethylamine
  2. Secondary and tertiary amines
  3. Ethylamine and aniline
  4. Aniline and benzylamine
  5. Aniline and N – methylaniline

Answer: Methylamine gives an arylamine reaction on heating with CHCl3 and alcoholic KOH. Gives foul smelling (Carbyl amine)

Amine Methylamine And Dimethylamine

whereas Dimethylamine does not give this reaction.

Secondary amines were given Kinsberg’s test. It gives insoluble substance with Ginsberg’s reagent (C6H5SO2Cl) which is not affected by base.

Amine Dialkyl Benzene Sulphonamide

  • Whereas tertiary amine does not react with Hinsberg’s reagent.
  • Aniline gives azo dye test : Dissolve Aniline in cone. HCl and add ice – a cold solution of HNO2 (NaNO2 + dil HCl) at 273 K and then treat it with an alkaline solution of 2-naphthol. The appearance of brilliant orange or red dye indicates aromatic amine (ie Aniline).
  • Whereas aliphatic amine (ie ethylamine) does not form a dye. It will give brisk effervescence due to the evolution of N2 but the solution remains clear.

Amine Ethylamine And Aniline

Nitrous acid test: Benzvlainine reacts with nitrous acid (HNO2) to form a diazonium salt which being unstable even at low temperatures, decomposes with the evolution of N2 gas.

Amine Nitrous Acid Test

Aniline reacts with HNO2 to form benzene diazonium chloride which is stable at 273-278 K and hence does not decompose to evolve N2 gas.

Amine Aniline And Benzenediazonium Chloride

Carbylamine test: Aniline being a primary amine gives a carbylamine test. i.e. when heated with an alcoholic solution of KOH and CHCl3. it gives an offensive smell of phenyl isocyanide. In contrast, N-methyl aniline. being secondary amine does not give this test.

Amine Carbylamine Test

Question 17. Account for the following:

  1. pKb of aniline is more than that of methylamine.
  2. Ethylamine is soluble in water whereas aniline is not.
  3. Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
  4. Although the amino group is o- and p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
  5. Aniline does not undergo Friedel-Crafts reaction.
  6. Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
  7. Gabrial phthalimide synthesis is preferred for synthesizing primary amines,

Answer:

1. In aniline, the lone pair of electrons on the N-atom are delocalized over the benzene ring. Resulting, the electron density of the nitrogen decreases. On the other hand, in CH3NH2 +1 effect of CH3 increases the electron density of the N-atom. Thus, aniline is a weaker base than methylamine and hence its pKb value is higher than that of methylamine.

2. Ethylamine dissolves in water because it forms hydrogen bonds with water molecules. In aniline due to the large, hydrocarbon part, the extent of H-bonding decreases considerably, and hence aniline is insoluble in water.

3. Methylaminc being more basic than water, accepts a proton from water liberating OH ions.

Amine Methylamine

These OHQ ions combine with Fe3+ ions present in H2O to form a brown precipitate of hydrated ferric oxide.

⇒ \(\mathrm{FeCl}_3 \longrightarrow \mathrm{Fe}^{+3}+3 \mathrm{Cl}^{-}\)

Amine Hydrated Ferric Oxide

4. Nitration is usually carried out with a mixture of cones. HNO3, and cone. H2SO4. In the presence of these acids, most of the aniline gets protonated to form anilingus ion. Thus in the presence of acids, the reaction mixture consists of aniline and anilinium ion. Now -NH2 groups in aniline are o, p-directing, and activating while the NH+3 group in anilinium ion is m-directing.

Amine Aniline Ion And M Nitroaniline

5. Aniline being a Lewis base, reacts with Lewis acid AlCl3 to form a salt.

Amine Friedel Crats Reaction

As a result, N of aniline acquires a positive charge and hence it acts as a strong deactivating group for electrophilic substitution reaction, consequently, aniline does not undergo Friedel – Crafts reaction.

6. The diazonium salts of aromatic amines are more stable than those of aliphatic amines due to the dispersal of the positive charge on the benzene ring as shown below.

Amine Dizonium Salts Of Aromatic Amines

7. Gabriel’s phthalimide reaction gives pure primary amines without any contamination of secondary and tertiary amines. Therefore It is preferred for synthesising (aliphatic) primary amines.

Question 18. Arrange the following:

1. In decreasing order of the pKi, values :

⇒ \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2, \mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_3,\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH} \text { and } \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2\)

2. In increasing order of basic strength:

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 \cdot \mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}\left(\mathrm{CH}_3\right)_2 \cdot\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH} \text { and } \mathrm{CH}_3 \mathrm{NH}_2\)

3. In increasing order of basic strength:

  1. Aniline, p-nitroaniline and p-toluidinc
  2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2, \mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_3, \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2\)

4. In decreasing order of basic strength in the gas phase:

⇒ \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2 .\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH} .\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N} \text { and } \mathrm{NH}_3\)

5. In increasing order of boiling point:

⇒ \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} .\left(\mathrm{CH}_3\right)_2 \mathrm{NH} . \mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2\)

6. In increasing order of solubilit> in water:

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 \cdot\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH} . \mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2\)

Answer:

  1. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2>\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}-\mathrm{CH}_3>\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2>\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}\)
  2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2<\mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}\left(\mathrm{CH}_3\right)_2<\mathrm{CH}_3 \mathrm{NH}_2<\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}\)
  3. \(p \text {-nitroaniline }<\text { aniline }<p \text { – toluidine }\)
  4. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2<\mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_3<\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2\)
  5. \(\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N}>\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}>\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2>\mathrm{NH}_3\)
  6. \(\left(\mathrm{CH}_3\right)_2 \mathrm{NH}<\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2<\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\)
  7. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2<\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}<\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2\)

Question 19. How will you convert:

  1. Ethanoic acid into methenamine
  2. Hexamentirile into 1-aminopentane
  3. Methanol to ethanoic acid
  4. Ethanamine into methamine
  5. Ethanoic acid into propionic acid
  6. Methenamine into ethanolamine
  7. Nitromethane into dimethylamine
  8. Propanic acid into ethanoic acid

Answer:

Amine Hoffmann Bromo Amide Reaction

Question 20. Describe a method for the identification of primary, secondary, and tertiary amines. Also, write chemical equations of the reactions involved.
Answer:

Benzene sulphonyl chloride (C6H5SO2Cl), Which is also known as llinsberg’s reagent, reacts with primary and secondary amine to form sulphonamide.

Amine Soluble In Alkali

The hydrogen attached to nitrogen in sulphone amide is strongly acidic due to the presence of a strong electron-withdrawing sulphonyl group. Hence it is soluble in alkali.

Amine N Diethylbenzene Sulphonamide

Tertiary amines do not react with benzene sulphonyl chloride.

This property of amines reacting with benzene sulphonyl chloride differently is used for the distinction of primary, secondary, and tertiary amines.

Question 21. Write short notes on the following:

  1. Carbylamine reaction
  2. Diazotisation
  3. Hafmann’s bromamide reaction
  4. Coupling reaction
  5. Ammonolysis
  6. Acetylation
  7. Gabriel phthalimide synthesis

Answer:

Carbylamine Reaction: Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanide or carbylamines which are foul-smelling substances. Secondary and tertiary amines do not show this reaction, this reaction is known as the carby lamine reaction or idiocy and test and is used as a least for primary amine.

Amine Carbylamine Reacton

Diazotization reaction: When a cold solution of a primary aromatic amine in a dilute mineral acid (MCI or IfS()4) is treated with a cold solution of nitrous acid at 273 – 278 K. arene diazonium salt is formed. This reaction is called the diazotization reaction.

For example: \(\mathrm{NaNO}_2+\mathrm{HCl} \longrightarrow \mathrm{HNO}_2+\mathrm{NaCl}\)

Amine Aniline And Benze Diazonium Chloride

Hoffmann’s Bromamide Reactions: The conversions of a primary amide to a primary amine-containing one carbon atom less than the original amide on heating with a mixture of Br2 in the presence of NaOH or KOH is called Haffmann’s bromamide reaction, for example.

Amine Hoffmanns Bromamide Reaction

This reaction is extremely useful for converting a higher homolog to the next lower ho mo log lie.

Coupling Reaction: The reaction of diazonium salts with phenols and aromatic amines to form a/o compounds of the general formula Ar — N = N — Ar is called a coupling reaction. In this reaction, the nitrogen atoms of the diazo group are retained in the product. The coupling with phenols takes place in a mildly alkaline medium while with amines it occurs under faintly acidic conditions.

Amine Coupling Reaction

Animonolysis: The process of cleavage of the C – X bond by ammonia molecule is known as ammonolysis.

Amine Ammonolysis

Acetylation: The process of introducing an acetyl groupAmine Acetyl Group into a molecule is called acetylation. Common acetylating agents used arc acetyl chloride and acetic anhydride.

Amine Acetylation

Gabriel Phthalimide Synthesis: In this reaction, phthalimide is converted into its potassium salt by treating it with alcoholic potassium hydroxide. Then potassium phthalimide is heated with an alkyl halide to yield an N alkylphthalimide which is hydrolyzed to phthalic acid and primary amine by heating with HCl or KOH solution.

Amine Gabriel Phthalimide Synthesis

aromatic primary amines cannot be prepared by this method.

Question 22. Accomplish the following conversions:

  1. Nirobenzene to benzoic acid
  2. Benzene to m-bromophenol
  3. Benzoic acid to aniline
  4. Aniline to 2.4.6-tribromolluorobenxcne
  5. Benzyl chloride to 2-phenylethanolamine
  6. Chlorobenzene to p-chloroaniJine
  7. Aniline to p-bromoaniline
  8. Benzamide to toluene
  9. Aniline to benzyl alcohol

Answer:

Amine Nitrobenzene To Benzoic To Aniline To Bezyl Alcohol

Amine Nitrobenzene To Benzoic To Aniline To Benzyl Alcohol

Question 23. Give the structure of A, B, and C in the following reaction:

Amine Structure Of The Following Reactions

Answer:

Amine Benzenediazonium Chloride

Question 24. An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br2 and KOH forms a compound C of molecular formula C6H7N. Write the structures and IUPAC names of compounds A, B, and C.
Answer:

Amine Benzoic Acid And Aniline

Question 25. Complete the following reactions:

Amine Phenyl Isocynide.

Answer:

Amine Phenyl Isocynide

Question 26. Win aromatic primary amines can not be prepared by Gabriel phthalimide synthesis.
Answer:

The success of Gabriel’s phthalimide reaction depends upon the nucleophilic attack by the phthalimide anion on the organic halogen compound.

Amine Gabriel Phthalimide Synthesis.

As aryl halides do not undergo nucleophilic substitution reaction easily, aromatic primary amines cannot be prepared by Gabriel phthalimide reaction.

Question 27. Write the reactions of (1) aromatic and (2) aliphatic primary amines with nitrous acid.
Answer:

Amine Aromatic And Primary Amines With Nirous Acid

Question 28. Give a plausible explanation for each of the following:

  1. Why are amines less acidic than alcohols of comparable molecular masses?
  2. Why do primary amines have a higher boiling point than tertiary amines?
  3. Why do aliphatic amines arc stronger bases than aromatic amines?

Answer:

Loss of a proton from an amine gives an amide ion (NH) while loss of a proton from alcohol gives an alkoxide ion (OR) as shown below.

Amine Milecular Mass

As O is more electronegative than N, So negative charge on a more electronegative atom is more stable. As R—O is more stable than RNH. Therefore amines are less acidic than alcohols.

Due to the presence of two H-atoms on N-atoms of 1° amines, they undergo extensive intermolecular H-bonding while at 3° amines due to the absence of H-atoms on N-atom, there is no hydrogen bonding takes place. So primary amines have higher b.p. than tertiary amines of comparable molecular mass.

Amine Higher Boling Point Than Tertiary Amines

Inter Molecular Hydrogen Bonding In 1° Amines.

Amine Hydrogen Bonding

Aliphatic amines are stronger bases than aromatic amines because:

Due to resonance in aromatic amines, the lone pair of electrons on the nitrogen atom gets delocalized over the benzene ring and thus is less available for protonation.

The aryl amine ions have lower stability than the corresponding aliphatic amines i.e. protonation of aromatic amines is not favored.

Important Questions for Class 12 Chemistry Chapter 8 Aldehydes, Ketones and Carboxylic Acids

Aldehyde Ketones And Carboxylic Acid

Question 1. Write the structure of the following reactions:

Aldehydes Ketones And Carboxylic Acid Structure Of Product Of The Following Reactions

Answer:

Aldehydes Ketones And Carboxylic Acid Structure Of Product Of The Following Reactions.

Question 2. Arrange the following compounds in the increasing order of their boiling points. CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3

Answer:

Aldehyde Ketones And Carboxylic Acids Increasing The Order Of Their Boiling Points

Question 3. Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions.

  1. Enthanal, Propanal, Propanone,, Butanone,
  2. Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone.

Hint: Consider the steric effect and electronic effect.

Read and Learn More Class 12 Chemistry with Answers Chapter Wise

Answer: Reactivity of carbonyl compound α (+)ve changeAldehydes Ketones And Carboxylic Acid Steric Hindrance

Aldehydes Ketones And Carboxylic Acid Carbonyl Compound

The +I effect of the alkyl group increases in the order:

Ethanal < Propanal < Propanone < Butanone

If the +I effect increases then (+)ve change on sp2 decreases.

So the reactivity order is: Butanone < Propanone < Propanal < Ethanal

Aldehydes Ketones And Carboxylic Acid Benzaldehyde

-I group on >C = O in reuses (+)ve charge so reactivity order is Acetophenone < p-tolualdchyde < Benzaldehvde < p-Nitrohenzaldehyde

Question 4. Predict the product of the following reactions:

Aldehydes Ketones And Carboxylic Acid Predict The Product Of The Following Reactions

Answer:

Aldehydes Ketones And Carboxylic Acid Predict The Product Of The Following Reactions.

Aldehydes Ketones And Carboxylic Acid Predict The Product Of The Following Reactions..

Question 5. Show how each of the following compounds can be converted to benzoic acid.

  1. Ethylbenzene
  2. Acetophenone
  3. Bromobenzene
  4. Phenylethane (Styrene)

Answer:

Aldehydes Ketones And Carboxylic Acid Ethybenzene To Phenylethene

CBSE Class 12 Chemistry Chapter 8 Aldehydes, Ketones and Carboxylic Acids Important Question And Answers

Question 6. Which of each pair shown here would you expect to be stronger?

  1. \(\mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H}_2 \text { or } \mathrm{CH}_2 \mathrm{FCO}_2 \mathrm{H}\)
  2. \(\mathrm{CH}_2 \mathrm{FCO}_2 \mathrm{H} \text { or } \mathrm{CH}_2 \mathrm{ClCO}_2 \mathrm{H}\)
  3. \(\mathrm{CH}_2 \mathrm{FCH}_2 \mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H} \text { or } \mathrm{CH}_3 \mathrm{CHFCH}_2 \mathrm{CO}_2 \mathrm{H}\)
  4. Aldehydes Ketones And Carboxylic Acid Acid Of Each Should Be Stronger

Answer: Acidic Strength

  1. \(\mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H}_2 \text { < } \mathrm{CH}_2 \mathrm{FCO}_2 \mathrm{H}\)
  2. \(\mathrm{CH}_2 \mathrm{FCO}_2 \mathrm{H} \text { > } \mathrm{CH}_2 \mathrm{ClCO}_2 \mathrm{H}\)
  3. \(\mathrm{CH}_2 \mathrm{FCH}_2 \mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H} \text { < } \mathrm{CH}_3 \mathrm{CHFCH}_2 \mathrm{CO}_2 \mathrm{H}\)
  4. Aldehydes Ketones And Carboxylic Acid Acid Of Each Should Be Stronger

Question 7. Give names of the reagents to bring about the following transformations:

  1. Hexan-1-ol to hexanal
  2. Cyclohexanol to cyclohexanone
  3. p-FluorotoIuene to p-fluorobenzaldchydc
  4. Ethanenitrile to ethanal
  5. Allyl alcohol to propenal
  6. But-2-ene to ethanal

Answer:

  1. C5H5NH+CrO3Cl(PCC)
  2. K2Cr2O7 in an acidic medium
  3. CrO in the presence of acetic anhydride 1. CrO2Cl2 2. HOH
  4. (Diisobutyl) aluminium hydride (DIBAL-H)
  5. PCC
  6. O3/H2O-Zn dust

Question 8. Arrange the following compounds in the increasing order of their boiling points:

⇒\(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CHO}, \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}, \mathrm{H}_5 \mathrm{C}_2-\mathrm{O}-\mathrm{C}_2 \mathrm{H}_3, \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3\)

Answer:

  • The molecular masses of these compounds are in the range of 72 to 74. Since only butane-1-ol molecules are associated due to extensive intermolecular hydrogen bonding, therefore, the boiling point of butan-1-ol would be the highest.
  • Butanal is more polar than ethoxyethane. Therefore, the intermolecular dipole-dipole attraction is stronger in the former, n-Pentane molecules have only weak van der Waals forces. Hence increasing order of boiling points of the given compounds is as follows :

⇒\(\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3<\mathrm{H}_3 \mathrm{C}_2-\mathrm{O}-\mathrm{C}_2 \mathrm{H}_5\)<\(\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CHO}<\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}\)

Question 9. Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reactions than propanal? Explain your answer.
Answer:

The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than the carbon atom of the carbonyl group present in propanal. The polarity of the carbonyl group is reduced in benzaldehyde due to resonance as shown below and hence it is less reactive than propanal.

Question 10. An organic compound (A) with molecular formula CsHsO forms an orange-red precipitate with 2,4-DNP reagent and gives a yellow precipitate on heating with iodine in the presence of sodium hydroxide. It neither reduces Tollens’ or Foldings’ reagent nor does it decolourise bromine water or Baeyer’s reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid (B) having molecular formula C7H6O2. Identify the compounds (A) and (B) and explain the reactions involved.
Answer:

(A)forms 2,4-DNP derivative. Therefore, it is an aldehyde or a ketone. Since it does not reduce Tollens’ or Fehling reagent, (A) must be a ketone. (A) responds to the iodoform test. Therefore, it should be a methyl ketone.

  • The molecular formula of (A) indicates a high degree of unsaturation, yet it does not decolourise bromine water or Baeyer’s reagent.
  • This indicates the presence of unsaturation due to an aromatic ring. Compound (B), being an oxidation product of a ketone should be a carboxylic acid.
  • The molecular formula of (B) indicates that it should be benzoic acid and compound (A) should, therefore, be a monosubstituted aromatic methyl ketone. The molecular formula of (A) indicates that it should be phenyl methyl ketone (acetophenone). Reactions are as follows:

Aldehydes Ketones And Carboxylic Acid Dinitrophenylhydrazine

Question 11. Write chemical reactions to affect the following transformations:

  1. Butan-1-ol to butanoic acid
  2. Benzyl alcohol to phenylethanoid acid
  3. 3-Nilrobromobenzene to 3-nitrobenzoic acid
  4. 4-Methylacetophenone to benzene-1,4-dicarboxylic acid
  5. Cyelohexene to hexane-1,6-dioic acid
  6. Butanal to butanoic acid.

Answer:

Aldehyde Ketones And Carboxylic Acids Hydroxyphenylacetic Acid

Question 12. What is meant by the following terms? Give an example of the reaction in each case.

  1. Cyanohydrin
  2. Acetal
  3. Semicarbazone
  4. Aldol
  5. Hemiacetal
  6. Oxime
  7. Ketal
  8. Inline
  9. 2,4- DNP-derivative
  10. Schiffs base

Answer: Cyanohydrin: The organic compounds have -CN and -OH groups in the structure.

Aldehydes Ketones And Carboxylic Acid Cyanohydrin

Acetal: Acetals are gem-dial koxy alkanes in which two alkoxy groups are present on the terminal carbon atom.

Aldehydes Ketones And Carboxylic Acid Acetal

Semicarbazone: Semicarbazone,s are derivatives of aldehydes and ketones produced by the condensation reaction between a ketone or aldehyde and semicarbazide.

Aldehydes Ketones And Carboxylic Acid Semicarbazone

Aldol: A β-hydroxy aldehyde or ketone is known as an aldol. It is produced by the condensation reaction of two molecules of the same or one molecule each of two different aldehydes or ketones with aH in the presence of a base.

Aldehydes Ketones And Carboxylic Acid Aldol

Hemiacetal: Hemiacetal are a-alkoxyalcohols Aldehyde reacts with one molecule of a monohydric alcohol in the presence of dry HCl gas.

Aldehydes Ketones And Carboxylic Acid Hemiacetal

Oxime: Oximes are a class of organic compounds having a general formula.

Aldehydes Ketones And Carboxylic Acid Oxime

On treatment with hydroxyl amine in a weakly acidic medium, aldehydes or ketones form oximes.

Aldehydes Ketones And Carboxylic Acid Aldehydes Or Ketone Form Oximes

Ketal: Ketals are gem-dialkoxyalkanes in which two alkoxy groups are present on the same carbon atom within the chain.

Aldehydes Ketones And Carboxylic Acid Ketal

Imine:

Aldehydes Ketones And Carboxylic Acid ImineIminces Imines are produced when aldehydes and ketones react with R- NH2.

Aldehydes Ketones And Carboxylic Imine.

2, 4-DNP-derivative: 2, 4-dinitrophenyIhydrazones are 2, 4-DNP-derivatives, which are produced when aldehydes or ketones react with 2. 4-dinitrophenylhydrazine in a weakly acidic medium.

Aldehydes Ketones And Carboxylic Acid DNA Derivative

To identify and characterize aldehydes and ketones, 2, 4-DNP derivatives are used.

Sehiffs Base: Is >c = NR (R is not Hydrogen) (R may be -Ph)

Aldehydes and ketones on treatment with primary aliphatic or aromatic amines in the presence of a trace of an acid yield a Schiffs base.

Aldehydes Ketones And Carboxylic Acid Schiff Base

Question 13. Draw structures of the following derivatives.

  1. The 2, 4-dinitrophenylhydrazone of benzaldehyde
  2. Cyclopropane oxime
  3. Acctaldehydedimcthylacctal
  4. The semicarbazone of cyclobutanone
  5. The ethylene ketal of hexane-3-one
  6. The methyl hemiacetal of formaldehyde

Answer:

Aldehydes Ketones And Carboxylic Structure Of The Following Derivatives

Question 14. Predict the products formed when cyclohexanecarbaldehyde reacts with the following reagents.

  1. PhMgBr and then H3O+
  2. Tollen’s reagent
  3. Semicarbazide and weak acid
  4. Excess ethanol and acid
  5. Zinc amalgam and dilute hydrochloric acid

Answer:

Aldehydes Ketones And Carboxylic Acid Cyclohexanecarbaldehyde

Question 15. Which of the following compounds would undergo aldol condensation, which is the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.

  1. Methanal
  2. 2-Methylpentanal
  3. Benzaldehyde
  4. Benzophenone
  5. Cyclohexanone
  6. 1-Phenylpropanone
  7. Phenylacetaldehyde
  8. Bulan-1-ol
  9. 2,2-Dimethylbutanal

Answer: Aldehydes and ketones having at least one a-hydrogen undergo aldol condensation (2), (5), (6), and (7) give aldol condensation.

  • Aldehydes (only) having no a-hydrogen undergo Cannizzaro reactions, (1), (3), (9) give Cannizzaro reaction.
  • Compound (4) is a ketone having no a-hydrogen atom and compound (8) Bulan-l-ol is an alcohol. Hence, these compounds do not undergo either aldol condensation or Cannizzaro reactions.

Aldol condensation:

Aldehydes Ketones And Carboxylic Acid Aldol Condensation

Cannizaro Reaction:

Aldehydes Ketones And Carboxylic Acid Cannizzaro Reaction

Question 16. How will you convert cthanal into the following compounds?

  1. Butane-1,3-diol
  2. But-2-enal
  3. But-2-enoic acid

Answer:

Aldehydes Ketones And Carboxylic Acid 3 Hydroxybutanal

When treated with Tollen’s reagent, But-2-enal produced in the above reaction produces but-2-enoic acid.

Question 17. Write structural formulas and names of four possible aids from propanal and butanal. In each case, indicate which aldehyde acts as a nucleophile and which is an electrophile.
Answer:

Take two molecules of propanal. one which acts as a nucleophile and the other as an electrophile.

Aldehydes Ketones And Carboxylic Acid Molecules Of Propanal

Taking two molecules of butanal. one which acts as a nucleophile and the other as an electrophile.

Aldehydes Ketones And Carboxylic Acid Molecule Of Butanal

Take one molecule each of propanal and butanal in which propanal acts as a nucleophile and butanal act as an electrophile.

Aldehydes Ketones And Carboxylic Acid One Molecule Each Of Propanal And Butanal

Taking one molecule each of propanal and butanal in which propanal acts as an electrophile and butanal acts as a nucleophile.

Aldehydes Ketones And Carboxylic Acid One Molecule Each Of Propanal And Butanal

Question 18. An organic compound (A) (molecular formula (C8H16O2) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-cue. Write equations for the reactions involved.
Answer: Hydrolysis of ester gives acid and alcohol

Aldehydes Ketones And Carboxylic Acid Hydrolysis Of Ester Gives Acid And Alcohol

Aldehydes Ketones And Carboxylic Acid Alkanol Gives Acid On Oxidation

Question 19. Arrange the following compounds in increasing order of their property as indicated:

Acetaldehyde, Acetone, Di-tert-butyl ketone. Methyl tert-butyl ketone

⇒ \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}(\mathrm{Br})\mathrm{COOH}\), \(\mathrm{CH}_3\mathrm{CH}(\mathrm{Br})\mathrm{CH}_2\mathrm{COOH},\left(\mathrm{CH}_3\right)_2\mathrm{CHCOOH},\mathrm{CH}_3 \mathrm{CH}_2\mathrm{CH}_2\mathrm{COOH}\)

Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid

⇒ \(\mathrm{CH}_3 \mathrm{COCl}, \mathrm{CH}_3 \mathrm{CONH}_2, \mathrm{CH}_3 \mathrm{COOCH}_3,\left(\mathrm{CH}_3 \mathrm{CO}\right)_2 \mathrm{O}\)

Answer: Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde

Aldehyde Ketones And Carboxylic Acids 4 Methoxybenzo

4-Methoxybezo

Question 20. Give simple chemical tests to distinguish between the following pairs of compounds.

  1. Propanal and Propanonc
  2. Acetophenone and Bcnzophenonc
  3. Phenol and Benzoic acid
  4. Benzoic acid and Ethyl benzoate
  5. Pentan-2-one and Pentan-3-one
  6. Benzaldehyde and Acetophenone
  7. Ethanal And Propanal

Answer: 1. By Tollen’s test

Propanal reduces Tollen’s reagent, Fehling solution, and Benedict solution but propanonc (ketone) does not. reduces. Iodoform test: Propanone gives an iodoform test but propanal (CH3-CH2-CHO) does not.

2. Acetophenone and Benzophenone can be distinguished using the iodoform test. Acetophenone gives an iodoform test but benzophenone does not.

3. Phenol and benzoic acid can be distinguished by the ferric chloride test and NaHCO3, test.

  1. Phenol gives a violet colour with FeCl3,
  2. Benzoic acid gives CO2 with NaHCO3 while phenol does not.

4. Benzoic acid and Ethyl benzoate can be distinguished by the sodium bicarbonate test. Sodium bicarbonate test: Benzoic acid reacts with NaHCO3 to produce brisk effervescence due to the evolution of CO2 gas but ethylbenzoate does not.

5. Pentan-2-one and pentan-3-one can be distinguished by iodoform lest. Pentan-2-one gives an iodoform test but pentan-3-one does not.

6. Benzaldehyde and acetophenone can be distinguished by (1) Tollen’s test and (2) the Iodoform test. Benzaldehyde gives Tollen’s test whereas acetophenone gives an iodoform test.

7. Ethanal and propanal can be distinguished by the iodoform test. Ethanal gives an iodoform test but propanal does not.

Question 21. How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic regent having not more than one carbon atom

  1. Methoxybenzoic
  2. M-Nitrobenzoic acid
  3. P-Nitrobenzoic acid
  4. Phenylacetic acid
  5. P-Nitrobenzaldehyde

Answer:

Aldehydes Ketones And Carboxylic Acid Inorganic And Organic Reagent

Question 22. How will you about the following conversions in more than two steps?

  1. Propanone to Propene
  2. Benzoic acid to Benzaldehyde
  3. Ethanol to 3-Hydroxybutyanal
  4. Benzene to m- Nitroacetophenone
  5. Benzaldehyde to Benzophenone
  6. Bromobenze to 1-Phenylethanol
  7. Benzaldehyde to 3-Phenylpropan-1-ol
  8. Benzaldehyde to α-Hydroxyphenylacetic acid

Answer:

Aldehyde Ketones And Carboxylic Acids Hydroxyphenylacetic Acid.

Question 23. Describe the following:

  1. Acetylation
  2. Carnnizzaro reaction
  3. Cross aldol condensation
  4. Decarboxylation

Answer: Acetylation: To introduce acetyl groupAldehydes Ketones And Carboxylic Acid Acetyl Groupin an organic compound.

Aldehydes Ketones And Carboxylic Acid Acetylation

Carnnizzaro reaction: The self-oxidation-reduction (disproportionation) reaction of aldehydes having no ex-hydrogens on treatment with concentrated alkalis is known as the Cannizzaro reaction. In this reaction, two molecules of aldehydes participate where one is reduced to alcohol and the other is oxidized to carboxylic acid.

Aldehydes Ketones And Carboxylic Acid Cannizzaro Reaction..

Cross-aldol condensation: The reaction between two different aldehydes, or two different ketones, or an aldehyde and a ketone (having α-H) in the presence of a base is called a cross-aldol condensation.

Aldehydes Ketones And Carboxylic Acid Cross Aldol Condensation

Decarboxylation: Decarboxylation refers to the reaction in which carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with soda-lime.

Aldehydes Ketones And Carboxylic Acid Decarboxylation

Question 24. Complete each synthesis by giving missing starting material, reagent or products.

Aldehydes Ketones And Carboxylic Acid Synthesis By Starting Material Reagent Or Product

Answer:

Aldehyde Ketones And Carboxylic Acids Synthesis Reagents Or Product

Aldehydes Ketones And Carboxylic Acid Synthesis By Starting Material Reagent Or Product 2

Aldehydes Ketones And Carboxylic Acid Synthesis By Starting Material Reagent Or Product 3

Question 25. Give a plausible explanation for each of the following:

  1. Cyclohexanone forms cyanohydrin in good yield but 2, 2, 6-trimethyl cyclohexanone does not.
  2. There are two, groups in semicarbazide. However, only one is involved in the formation of semicarbazone.
  3. During the preparation of esters, a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.

Answer: Cyclohexanones form cyanohydrins according to the following equation.

Aldehydes Ketones And Carboxylic Acid Cyclohexane And Cynohydrin

In 2, 2. 6 – trimethyl cyclohexanone, methyl groups at α-positions offer steric hindrance and as a result.

CN cannot attack effectively.

Aldehydes Ketones And Carboxylic Acid Trimethyl Cyclohexanone

2, 2, 6 – trimethyl cyclohexanone,

For this reason, it does not form a cyanohydrin.

Aldehydes Ketones And Carboxylic Acid Formation Of Semicarbazone

The electron density on the -NH2, group involved in the resonance also decreases. As a result, it cannot act as a nucleophile. Since the other -NH2, the group is not involved in resonance: it can act as a nucleophile and can attack carbonyl – carbon atoms of aldehydes and ketones to produce semicarbazone.

Aldehydes Ketones And Carboxylic Acid Carboxylic Acid To Water

If either water or ester is not removed as soon as it is formed, then it reacts to give back the reactants as the reaction is reversible.

Question 26. An organic compound contains 69.77% carbon. 11.63% hydrogen and the rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollen’s reagent but forms an additional compound with sodium hydrogen sulphite and gives a positive iodoform test. On vigorous oxidation, it gives ethanoic and propanoic acid. Write the possible structure of the compound.
Answer:

% of carbon = 69.77%

% of hydrogen = 11.63%

% of oxygen = [100-(69.77 + 11.63)]% = 18.6%

Thus, the ratio of the number of carbon, hydrogen and oxygen atoms in the organic compound can be given as:

⇒ \(\mathrm{C}: \mathrm{H}: \mathrm{O}=\frac{69.77}{12}: \frac{11.63}{1}: \frac{18.6}{16}=5.81: 11.63: 1.16=5: 10: 1\)

Therefore, the empirical formula of the compound is C5H10O. Now, the empirical formula mass of the compound can be given as:

5 × 12 +10 × 1 + 1 × 16 = 86

Molecular mass of the compound = 86 (given)

Therefore, the molecular formula of the compound is given by C5H10O.

Since the given compound does not reduce Tollen’s reagent, it is not an aldehyde. Again the compound forms sodium hydrogen sulphate addition products and gives a positive iodoform test. Since the compound is not an aldehyde, it must be a methyl ketone.

The given compound also gives a mixture of ethanoic acid and propanoic acid.

Hence, the given compound is plan-2-one.

Aldehyde Ketones And Carboxylic Acids Pentan Two One

The given reactions can be explained by the following equations:

Aldehyde Ketones And Carboxylic Acids Sodium Butanoate

Question 27. Although phenoxide ion has more number of resonating structures than carboxylate ion. carboxylic acid is a stronger acid than phenol. Why?
Answer: Resonance structures of phenoxide ion are:

Aldehyde Ketones And Carboxylic Acids Phenoxic Ion

Unequivalent resonating structures

Aldehyde Ketones And Carboxylic Acids Equivalent Resonating Structure

Equivalent resonating structures are more stable.

Important Questions for Class 12 Chemistry Chapter 7 Alcohols, Phenols and Ethers

Alcohol Phenol And Ether

Question 1. Classify the following as primary, secondary, and tertiary alcohols:

Alcohol Phenol And Ether Primary Secondary And Teritary

Answer:

  1. Primary alcohol > (1), (2), (3)
  2. Secondary alcohol > (4), (5)
  3. Tertiary alcohol > (6)

Question 2. Identify allylic alcohols in the above examples.
Answer:

  • Allylic that means C = C – C – OH
  • The alcohols given in (2) and (6) are allylic.

Question 3. Name the following compounds according to the IUPAC system.

Alcohol Phenol And Ether According To IUPAC System

Answer:

  1. 3-Chloromethyl-2-isopropylpentan-1-ol
  2. 2, 5 – Dimelhylhcxane – 1,3- cliol
  3. 3-Bromocyclohexanol
  4. Hex-1 -en-3-ol
  5. 2-Bromo-3-methylbut-2-en-1 -ol

Question 4. Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal.

Alcohol Phenol And Ether Grignard Reagent

Answer:

Alcohol Phenol And Ether Grignard Reagent.

Read and Learn More Class 12 Chemistry with Answers Chapter Wise

Question 5. Write structures of the products of the following reactions:

Alcohol Phenol And Ether Structure Of The Product Of The Reactions

Answer:

Alcohol Phenol And Ether Structure Of The Product Of The Following Reactions

CBSE Class 12 Chemistry Chapter 7 Alcohol Phenol And Ether Important Question And Answers

Question 6. Give structures of the products you would expect when each of the following alcohols reacts with

  1. HCl-ZnCl2
  2. HBr and
  3. SOCl2
  4. Bulan-1-ol
  5. 2-Methylbutan-2-ol

Answer:

Alcohol Phenol And Ether Alcohol Reacts With HCl

Tertiary alcohols react immediately with Lucas’ reagent.

Alcohol Phenol And Ether Teritary Alcohol Reacts Immediately With Lucas Reagent

Alcohol Phenol And Ether Alcohol Reacts With SOCl2

Question 7. Predict the major product of acid-catalyzed dehydration of

  1. 1-methylcyclohexane and
  2. Butan-1-ol

Answer:

Alcohol Phenol And Ether Major Product Of Acid Catalysed Dehydration

Alcohol Phenol And Ether Major Product

Question 8. Ortho para nitrophenols are more than phenol. DFrawn the resonance structure of the corresponding phenoxide ions.
Answer:

Alcohol Phenol And Ether Resonance Structure Of The Phenoxide Ion

(Resonance structure of the phenoxide ion).

Alcohol Phenol And Ether Resonance Structure Of P Nitrophenoxide Ion

(Resonance structure of p-nitro phenoxide ion)

Alcohol Phenol And Ether Resonance Structure Of O Nitrophenoxide Ion

(Resonance structure of o-nitro phenoxide ion)

It can be observed that the presence of the nitro group increases the stability of phenoxide ions.

Question 9. Write the equations involved in the following reactions:

  1. Reimer – Tiemann reaction
  2. Kolbe – Schmidt reaction

Answer:

Alcohol Phenol And Ether Tiemann Reaction And Schmidt Reaction

Question 10. Write the reactions of Williamson synthesis of 2-ethoxy-3-inethylpentane starting from ethanol and 3-methylpentan-2-ol.
Answer:

Alcohol Phenol And Ether The Reaction Of Williamson Synthesis

Question 11. Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4- nitrobenzene and why?

Alcohol Phenol And Ether One Methoxy Four Nitrobenzene

Answer: Set (2) is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzcne.

Alcohol Phenol And Ether One Methoxy Four Nitrobenzene.

In set (1), the Bromine atom is stable by resonance.

Question 12. Predict the products of the following reactions:

Alcohol Phenol And Ether Predict The Product Of The Following Reaction

Answer:

Alcohol Phenol And Ether Predict The Product Of The Following Reaction.

Question 13. Give IUPAC names of the following compounds:

Alcohol Phenol And Ether IUPAC Name Of The Following Compound

Answer:

  1. 4-ChIoro-2,3-dimethylpentan-1-oI
  2. 2-Ethoxy propane
  3. 2,6-Diinethylphenol
  4. 1-Elhoxy-2-nilrocyclohexane

Question 14. You will notice that the reaction produces a primary alcohol with methanol, a secondary alcohol with other aldehydes, and a tertiary alcohol with ketones. Give the structures and IUPAC names of the products expected from the following reactions:

  1. Catalytic reduction of butane.
  2. Hydration of propene in the presence of dilute sulphuric acid.
  3. Reaction of propanone with methyl magnesium bromide followed by hydrolysis.

Answer:

Alcohol Phenol And Ether Primary Alcohol Secondary Alcohol And Teritary Alcohol With Ketones

Question 15. Arrange the following sets of compounds in order of their increasing boiling points:

  1. Pentan-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol.
  2. Pentan-1-ol, n-butane, pentanal, ethoxyethane.

Answer:

  1. Methanol, ethanol, propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol.
  2. n-Butane, ethoxyethane, pentanal, and pentan-1-ol.

Question 16. Arrange the following compounds in increasing order of their acid strength: Propan-1-ol, 2, 4, 6-trinitrophenol, 3-nitrophenol, 3,5-dinitrophenol, phenol, 4-methylphenol.
Answer:
Propan-1-ol, 4-methylphenol, phenol, 3-nitrophenol, 3, 5-dinitrophenol, 2, 4, 6-trinitrophenol.

Question 17. Write the structures of the major products expected from the following reactions:

  1. Mononitration of 3-methyl phenol
  2. Denitration of 3-methyl phenol
  3. Mononitration of phenyl methanoate.

Answer: The combined influence of -OH and -CH3 groups determines the position of the incoming group.

Alcohol Phenol And Ether Structure Of Major Products

Question 18. The following is not an appropriate reaction for the preparation of t-butyl ethyl ether.

Alcohol Phenol And Ether T Butyl Ethyl Ether

  1. What would be the major product of this reaction?
  2. Write a suitable reaction for the preparation of t-butyl ethyl ether.

Answer:

1. The major product of the given reaction is 2-methyl prop-1-one.

This is because sodium ethoxide is a strong nucleophile as well as a strong base. Thus elimination reaction predominates over substitution.

Alcohol Phenol And Ether T Butyl Ethyl Ether.

2. Phenols are also converted to ethers by this method. In this, phenol is used as the phenoxide moiety.

Alcohol Phenol And Ether Phenoxide Moiety

Question 19.

Alcohol Phenol And Ether Question 19

Answer:

Alcohol Phenol And Ether Question 19 Answer

Question 20. Write IUPAC names of the following compounds:

Alcohol Phenol And Ether IUPAC Names Of The Following Compound

Answer:

  1. 2, 2, 4-Trimethylpcntan-3-ol
  2. 5-Ethylheplane-2, 4-diol
  3. Butane-2, 3-diol
  4. Propane-1, 2, 3-triol
  5. 2-Methylphenol
  6. 4-Methylphenol
  7. 2. 5-Dimcthylphenol
  8. 2, 6-Dimethylphenol
  9. 1-Methoxy-2-methylpropane
  10. Ethoxyben/ene
  11. 1-Phenoxyheptane
  12. 2-Elhoxvbutane

Question 21. Write structures of the compounds whose IUPAC names are as follows:
Answer:

  1. 2-Methylbutan-2-ol
  2. 1-Phenyl-2-ol
  3. 3,5- Dimethylhexene-1,3,5-triol
  4. 2,3-Diethylphenol
  5. 1-Ethoxypropane
  6. 2-Ethoxy-3-methyl pentane
  7. Cyclohexylmethanol
  8. 3-Cyclohexylpentan-3-ol
  9. Cyclopent-3-en-1-ol
  10. 3-Chloromethylpentan-1-ol

Answer:

Alcohol Phenol And Ether Structure Of The Compound IUPAC Name As Followes

Question 22. Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names. Classify them as primary, secondary, and tertiary alcohols.
Answer:

Alcohol Phenol And Ether Structure Of All Isomeric Alcohols

  1. Primary alcohols – (1), (2)(3), and (4)
  2. Secondary alcohols – (5),(6) and (7)
  3. Tertiary alcohol – (8)

Question 23. Explain why propanol has a higher boiling point than that of the hydrocarbon, butane.
Answer:

Propanol undergoes intermolecular H-bonding because of the presence of the -OH group. On the other hand, butane does not

Alcohol Phenol And Ether Inter Molecular H Bonding

Question 24. Explain why are alcohols comparatively more soluble in water than the corresponding hydrocarbons.
Answer:

Hydrogen bonding exists between alcohol and water molecules while hydrocarbons are nonpolar, they can not form an H-bond with water molecules.

Question 25. What is meant by hydroboration oxidation reaction? Illustrate it with an example.
Answer:

Alcohol Phenol And Ether Hydroboration Oxidation Reaction

Question 26. Give the structures and IUPAC names of monohydric phenols of molecular formula C7H8O.
Answer:

Alcohol Phenol And Ether Structure Of IUPAC Names Of Monohydric Phenols

Question 27. While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which is steam volatile. Give reason.
Answer: O-nitrophenol is steam volatile due to intramolecular hydrogen bonding.

Question 28. Give the equation of reaction of preparation of phenol from cumene.
Answer:

Alcohol Phenol And Ether Reaction Of Phenol From Cumene

Question 29. Write a chemical reaction for the preparation of phenol from chlorobenzene.
Answer:

Chlorobenzene is fused with NaOH (at 623 K and 320 atm pressure) to produce sodium phenoxide, which gives phenol on acidification.

Alcohol Phenol And Ether Phenol From Chlorobenzene

Question 30. You are given a benzene, cone. H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents.
Answer:

Alcohol Phenol And Ether Preparation Of Phenol Using These Reagents

Question 31. Show how will you synthesize:

  1. 1-phenylethanol from a suitable alkane.
  2. CyelohexyI methanol using an alkyl halide by an SN2 reaction
  3. Pentan-1-ol using a suitable alkyl halide?

Answer:

Alcohol Phenol And Ether Chloromethylcyclohexane

Question 32. Give two reactions that show the acidic nature of phenol. Compare the acidity of phenol with that of ethanol.
Answer:

The acidic nature of phenol can be represented by the following two reactions.

1. Phenol reacts with sodium to give sodium phenoxide, liberating H2

Alcohol Phenol And Ether Phenol And Sodium Phenoxide

2. Phenol reacts with sodium hydroxide to give sodium phenoxide and water as by-products.

Alcohol Phenol And Ether Sodium Phenoxide

The acidity of phenol is more than that of ethanol. This is because after losing a proton, the phenoxide ion undergoes resonance and gets stabilized whereas the ethoxide ion does not.

Question 33. Explain why is ortho-nitro phenol more acidic than ortho-methoxy phenol.
Answer: Electron withdrawing effect of nitro group and electron releasing effect of methoxy group.

Question 34. Explain how the -OH group attached to a carbon of benzene ring activates it towards electrophilic substitution.
Answer: Electron density at the o- and p-position increases and thus the electrophile attacks easily.

Alcohol Phenol And Ether Benzene Ring

Question 35. Give equations of the following reactions:

  1. Oxidation of propane-1 -ol with KMnO4 solution.
  2. Bromine is CS2 with phenol.
  3. Dilute HNO3 with phenol.
  4. Treating phenol with chloroform in the presence of aqueous NaOH.

Answer:

Alcohol Phenol And Ether Salicyladehyde

Question 36. Explain the following with an example.

  1. Williamson ether synthesis.
  2. Unsymmetrical ether.

Answer:

1. Williamson’s ether synthesis:

Williamson ether synthesis is a laboratory method to prepare symmetrical and unsymmetrical ethers by allowing alkyl halides to react with sodium alkoxide.

Alcohol Phenol And Ether Williamsons Ether Synthesis

This reaction involves an SN2 attack of the alkoxide ion on the alkyl halide. Better results are obtained in the case of primary alkyl halides.

Alcohol Phenol And Ether Alkoxide Ion On The Alkyl Halide

If the alkyl halide is secondary or tertiary, then elimination competes over substitution.

2. Unsymmetrieal ether:

An unsymmetrical ether is an ether where two alkyl groups on the two sides of an oxygen atom differ. For example: ethyl methyl ether (CH3 -O-CH2CH3).

Question 37. How are the following conversions carried out?

  1. Propene → Propan-2-ol
  2. Benzyl chloride → Benzyl alcohol
  3. Ethyl magnesium chloride → Propan-1-ol.
  4. Methyl magnesium bromide → 2-Methylpropan-2-ol,

Answer:

Alcohol Phenol And Ether Two Methylpropan Two Ol

Question 38. Give a reason for the higher boiling point of ethanol in comparison to methoxyrnethane.
Answer:

  • Ethanol undergoes intermolecular H-bonding due to the presence of the -OH group, resulting in the association of molecules.
  • On the other hand, methoxyrnethane does not undergo H-bonding. Hence, the boiling point of ethanol is higher than that of methoxyrnethane.

Question 39. Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
Answer: The reaction of Williamson synthesis involves an SN2 attack of an alkoxide ion on a primary alkyl
halide.

Alcohol Phenol And Ether The Reaction Of Williamsons Synthesis Involves SN2

  • But if secondary or tertiary alkyl halides are taken in place of primary alkyl halides, then elimination would compete over substitution. As a result, alkenes would be produced.
  • This is because alkoxides are nucleophiles as well as strong bases. Hence, they react with alkyl halides, which results in an elimination reaction.

Alcohol Phenol And Ether Tertiary Alkyl Halide And Alkene

Question 40. How is 1-propoxypropane synthesized from propan- 1-ol? Write the mechanism of this reaction.

Alcohol Phenol And Ether Propane One Ol And One Propoxypropane

Answer:

The mechanism of this reaction involves the following three steps:

Step 1: Protonation

Alcohol Phenol And Ether Protonation

Step 2: Nucleophilic attack

Alcohol Phenol And Ether Nucleophilic Attack

Step 3: Deprotonation

Alcohol Phenol And Ether Deprotonation

Question 41. The preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Answer:

  • The formation of ethers by dehydration of alcohol is a bimolecular reaction (SN2) involving the attack of an alcohol molecule on a protonated alcohol molecule.
  • In this method, the alkyl group should be unhindered. In the case of secondary or tertiary alcohols, the alkyl group is hindered. As a result, elimination dominates substitution. Hence, in place of others, alkenes are formed.

Question 42. Write the equation of the reaction of hydrogen iodide with:

  1. 1-propoxypropane
  2. Methoxybcnezcnc and
  3. Benzyl ethyl ether

Answer:

Alcohol Phenol And Ether Hydrogen Iodide

Question 43. Explain the fact that in aryl alkyl ethers

  1. The alkoxy group activates the benzene ring towards electrophilic substitution and
  2. It directs the incoming substituents to ortho and para positions in the benzene ring.

Answer:

Alcohol Phenol And Ether Aryl Alkyl Ether

In arly alkyl ethers, due to the + R effect of the alkoxy group, the electron density in the benzene ring increases as shown in the following resonance structure.

Alcohol Phenol And Ether The Alkoxy Group Resonance Structure

Thus, benzene is activated towards electrophilic substitution by the alkoxy group.

2. It can also be observed from the resonance structures that the electron density increases more at the ortho and para positions than at the meta position. As a result, the incoming substituents are directed to the ortho and para positions in the benzene ring.

Question 44. Write the mechanism of the reaction of III with methoxymethyl.
Answer:

The mechanism of the reaction of III with methoxymethane involves the following steps:

Step 1: Protonation of methoxymelthane:

Alcohol Phenol And Ether Protonation Of Methoxymethane

Step 2: Nucleophilic attack of I

Alcohol Phenol And Ether Nucleophilic

Question 45. Write equations of the following reactions:

  1. Friedel-Crafts reaction-alkylation of anisole.
  2. Nitration of anisole.
  3. Bromination of anisole in ethanoic acid medium.
  4. Friedcl-Craft’s acetylation of anisole.

Answer:

Alcohol Phenol And Ether Ethanoyl Chloride

Question 46. When 3-methyl butane-2-ol is treated with HBr, the following reaction takes place:

Alcohol Phenol And Ether Two Methylpropan Two Ol Is Treated With HBr

Give a mechanism for this reaction.

Answer: The mechanism of the given reaction involves the following steps:

Step 1: Protonation

Alcohol Phenol And Ether Protonation.

Step 2: Formation of 2° carbonation by the elimination of a water molecule

Alcohol Phenol And Ether Formation 2 Degree Carbocation

Step 3: Re-arrangement by the hydride-ion shift

Alcohol Phenol And Ether Re Arrangement By The Hydride By The Hydride Ion Shift

Step 4: Nucleophilic attack

Alcohol Phenol And Ether Nucleophilic Attack.

Important Questions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes

Haloalkanes And Haloarenes

Question 1. Write The structure of the following compounds:

  1. 2-Cliloro-3-methylpentane
  2. 1-Chloro-4-methylcyclohexane
  3. 4-terl. ButyI-3-iodohcplane
  4. 1, 4-DibmmobiU-2- one
  5. 1-Bromo-4-sec, butyl-2-melhylbenzene

Answer:

Haloalkanes And Haloarenes Structure Of The Following Compounds

Question 2. Draw the structure of major monohalo products in each of the following reactions:

Haloalkanes And Haloarenes Structure Of Major Monohalo Products

Answer:

Haloalkanes And Haloarenes Structure Of Major Monohalo Products.

Question 3. Arrange each set of increasing boiling points.

  1. Bromomethane, Bromoform, Chloromethane, Dibromethane.
  2. 1-chloride, Isopropyl chloride, 1-chlorobutane.

Answer:

Haloalkanes And Haloarenes In Order Of Increasing Boiling Point

Read and Learn More Class 12 Chemistry with Answers Chapter Wise

Question 4. Which alkyl halide from the following pairs would you accept to react more rapidly by an SN2 mechanism? Explain your answer.

Haloalkanes And Haloarenes Alkyl Halide From SN2 Mechanism

Answer:

Haloalkanes And Haloarenes Alkyl Halide From SN2 Mechanism.

Question 5. In the following pairs of halogen compounds, which compound undergoes a faster SN1 reaction?

Haloalkanes And Haloarenes The Following Pair Of Halogen Compound Undergoes Faster Reaction

Answer:

Haloalkanes And Haloarenes The Following Pair Of Halogen Compound Undergoes Faster Reaction.

CBSE Class 12 Chemistry Chapter 6 Haloalkanes And Haloarenes Important Question And Answers

Question 6. Identify A, B, C, D, E, R, and R1 in the following:

Haloalkanes And Haloarenes Question 6

Answer:

Haloalkanes And Haloarenes Answer 6.

Question 7. Write the IUPAC name of the following:

Haloalkanes And Haloarenes IUPAC Name Of The Following

Answer:

  1. 4-Bromopent-2-ene
  2. 3-Bromo-2-methylbut-1-ene
  3. 4-Bromo-3-methyl pent-2-ene
  4. 1-Bromo-2-methylbut-2-ene
  5. 1-Bromobut-2-ene
  6. 3-Bromo-2-methylpropene

Question 8. Write the product of the following reactions:

Haloalkanes And Haloarenes Product Of The Following Reactions

Answer:

Haloalkanes And Haloarenes Product Of The Following Reactions.

Question 9. Haloalkanes react with KCN to form alkyl cyanides as the main product while AgCN forms isocyanides as the chief product. Explain.
Answer:

  • KCN is predominantly ionic and provides cyanide ions in solution. Although both carbon and nitrogen atoms are in a position to donate electron pairs, the attack place mainly through the carbon atom and not through the nitrogen atom since the C-C bond is more stable than the C-N bond.
  • However, AgCN is mainly covalent in nature and nitrogen is free to donate electron pair forming isocyanide as the main product.

Question 10. In the following pairs of hydrogen compounds, which would undergo SN2 reaction faster?

Haloalkanes And Haloarenes The Hydrogen Compound Undergoes SN2 Reaction Faster

Answer:

Haloalkanes And Haloarenes Primary Halide Undergoes SN2 Reaction Faster

  • It is a primary halide and therefore undergoes SN2 reaction faster.
  • As iodine is a better-leaving group because of its large size, it will be released at a faster rate in the presence of income nucleophiles.

Question 11. Predict the order of reactivity of the following compounds in SN1 and SN2 reactions:

  1. The four isomeric bromobutan.es
  2. C6H5CH2Br, C6H5CH(C6H5)BR, C6H5CH(CH3)Br, C6H5C(CH3)(C6H5)Br

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}<\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}_2 \mathrm{Br}<\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3<\left(\mathrm{CH}_3\right)_3 \mathrm{CBr}\left(\mathrm{SN}_{\mathrm{N}} 1\right)\)

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}>\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}_2 \mathrm{Br}>\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{Br}_2 \mathrm{CH}_3>\left(\mathrm{CH}_3\right)_3 \mathrm{CBr}\left(\mathrm{S}_{\mathrm{N}} 2\right)\right.\)

Question 12. Although chlorine is an electron-withdrawing group, yet it is ortho-, para-directing in electrophilic aromatic substitution reactions. Why?
Answer: Reactivity is thus controlled by the stronger inductive effect and orientation is controlled by resonance effect.

Question 13. Name the following halides according to the IUPAC system and classify them as alkyl, alkyl, benzene(primary, secondary, tertiary), vinyl, or aryl halide:

  1. \(\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}(\mathrm{Cl}) \mathrm{CH}_3\)
  2. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{CH}\left(\mathrm{C}_2 \mathrm{H}_5\right) \mathrm{Cl}\)
  3. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C}\left(\mathrm{CH}_3\right)_2 \mathrm{CH}_2 \mathrm{I}\)
  4. \(\left(\mathrm{CH}_3\right)_3 \mathrm{CCH}_2 \mathrm{CH}(\mathrm{Br}) \mathrm{C}_6 \mathrm{H}_5\)
  5. \(\mathrm{CH}_3 \mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3\)
  6. \(\mathrm{CH}_3 \mathrm{C}\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{CH}_2 \mathrm{Br}\)
  7. \(\mathrm{CH}_3 \mathrm{C}(\mathrm{Cl})\left(\mathrm{C}_2 \mathrm{H}_5\right) \mathrm{CH}_2 \mathrm{CH}_3\)
  8. \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{C}(\mathrm{Cl}) \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{CH}_3\right)_2\)
  9. \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHC}(\mathrm{Br})\left(\mathrm{CH}_3\right)_2\)
  10. \(\mathrm{p}-\mathrm{ClC}_6 \mathrm{H}_4 \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{CH}_3\right)_2\)
  11. \(\mathrm{m}-\mathrm{ClCH}_2 \mathrm{C}_6 \mathrm{H}_4 \mathrm{CH}_2 \mathrm{C}\left(\mathrm{CH}_3\right)_3\)
  12. \(\mathrm{O}-\mathrm{Br}-\mathrm{C}_6 \mathrm{H}_4 \mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{CH}_2 \mathrm{CH}_3\)

Answer:

Haloalkanes And Haloarenes Vinyl Or Aryl Halides

Haloalkanes And Haloarenes Vinyl Or Aryl Halides.

Question 14. Give the IUPAC names of the following compounds

  1. \(\mathrm{CH}_3 \mathrm{CH}(\mathrm{Cl}) \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3\)
  2. \(\mathrm{CHF}_2 \mathrm{CBrClF}\)
  3. \(\mathrm{ClCH}_2 \mathrm{C} \equiv \mathrm{CCH}_2 \mathrm{Br}\)
  4. \(\left(\mathrm{CCl}_3\right)_3 \mathrm{CCl}\)
  5. \(\mathrm{CH}_3 \mathrm{C}\left(\mathrm{p}-\mathrm{ClC}_6 \mathrm{H}_4\right)_2 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3\)
  6. \(\left(\mathrm{CH}_3\right)_3 \mathrm{CCH}=\mathrm{CClC}_6 \mathrm{H}_4 \mathrm{I}-\mathrm{p}\)

Answer:

Haloalkanes And Haloarenes IUPAC Name Of The Following Compounds

Haloalkanes And Haloarenes IUPAC Name Of The Following Compounds.

Question 15. Write the structures of the following organic halogen compounds.

  1. 2-Chloro-3-methyl pentane
  2. p-Bromochlorobenzene
  3. l-Chloro-4-ethylcyelohexane
  4. 2-(2-Chlorophenyl)-1-iodooctane
  5. 2-Bromobutane
  6. 4-tert-butyl-3-iodoheptane
  7. 1-Bromo-4-sec-butyl-2-methyl benzene
  8. 1,4-Dibromobut-2-ene

Answer:

2-Chloro-3-methyl pentaneHaloalkanes And Haloarenes 3 Methyl Pentane

p-BromochlorobenzeneHaloalkanes And Haloarenes P Bromochlorobenzene

1-Chloro-4-ethylcyclohexaneHaloalkanes And Haloarenes 4 Ethylcyclohexane

2-(2-ChlorophenyI)-1-idodooctaneHaloalkanes And Haloarenes One Iodooctane

2-Bromobutane CHHaloalkanes And Haloarenes 2 Bromobutane

4-Tert-butyl-3-iodoheptaneHaloalkanes And Haloarenes 3 Iodoheptane

1-Bromo-4-sec-butyl-2-methyl benzeneHaloalkanes And Haloarenes Two Methyl Benzene

1,4-Dibromobut-2-eneHaloalkanes And Haloarenes Dibromobut Two Ene

Question 16. Which of the following has the highest dipole moment?

  1. CH2Cl2
  2. CHCl3
  3. CCl4

Answer:

Haloalkanes And Haloarenes The Highest Dipole Moment

CCl4 < CHCl3 < CH2Cl2

Question 17. A hydrocarbon C5H10 does not react with chlorine in the dark but gives a single monochrome compound C5H9Cl in bright sunlight. Identify the hydrocarbon.
Answer:

Haloalkanes And Haloarenes Cyclopentane

The reactions involved in the question are:

Haloalkanes And Haloarenes Mono Chlorocylopentane

Question 18. Write the isomers of the compound having the formula C4H9Br.
Answer: There are four isomers of the compound having the formula Chi I- Hr. These isomers are given below.

Haloalkanes And Haloarenes Isomers Of The Compound

7. Write the equations for the preparation of 1-iodobutane from

  1. 1-butanol
  2. 1-chlorobutane
  3. but-1-ene

Answer:

Haloalkanes And Haloarenes One Butanol

Haloalkanes And Haloarenes 1 Chlorobutane And But One Ene

Question 19. What are ambident nucleophiles? Explain with an example.
Answer: Ambident nucleophiles are nucleophiles having two nucleophilic sites. For example, nitrite ion is an ambident nucleophile.

Haloalkanes And Haloarenes Abident Nucleophile

Question 20. Which compound in each of the following pairs will react faster in SN2 reaction with Oil?

  1. CH3Br or CH3I
  2. (CH3)3CCl or CH3Cl

Answer: CH3-I is more reactive because I+ is a better-leaving group than Br+

Haloalkanes And Haloarenes SN2 Reaction With Oh

In the case of (CH3)3CCl. the attack of the nucleophile at the carbon atom is hindered by the presence of bulk)- substituents on that carbon atom bearing the leaving group.

Question 21. Predict all the alkanes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene.

  1. 1-Brorao-1-methylcyclohexane
  2. 2-Chloro-2-methyl butane
  3. 2,2, 3-Trirpethyl-3-bromopentane

Answer:

Haloalkanes And Haloarenes Dehydration Of The Following Halides With Sodium Ethoxide

Question 22. How will you bring about the following conversion?

  1. Ethanol to but-1-yne
  2. Ethane to broiuoethene
  3. Propene to I-nitropropane
  4. Toluene to benzyl alcohol
  5. Propene to propyne
  6. Ethanol to ethyl fluoride
  7. Bromomethane to propanone
  8. But-1-cue to but-2-ene
  9. 1-Chlorobutane to n-octane
  10. Benzene to biphenyl

Answer:

Haloalkanes And Haloarenes Answer 11

Haloalkanes And Haloarenes Answer 11.

Haloalkanes And Haloarenes Answer 11..

Question 23. Explain why

  1. The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
  2. Alkyl halides, though polar, are immiscible with water?
  3. Grignard reagents should be prepared under anhydrous conditions.

Answer:

Haloalkanes And Haloarenes Chlorobenzene And Cyclohexylchloride

+M of -Cl and I of -Cl are in opposite directions so the dipole moment is decreased in chlorobenzene.

  1. No H-bonding of halide with water.
  2. Grignard reagents in the presence of moisture, react with H2O to give alkanes.

Haloalkanes And Haloarenes Grigrard Reagent

Question 24. Give the uses of freon 12. DDT, carbon tetrachloride, and iodoform.
Answer:

  • Uses of Freon-12 Freon-12 (dichlorodifluoromethane. CF2Cl2) is commonly known as CFC. It is used as a refrigerant in refrigerators and air conditioners.
  • The use of DDT (p-p-dichlorodiphenyltrichloroethane) is one of the best-known insecticides.

Uses of carbonletrachloride (CCI4)

  1. It is used for manufacturing refrigerants and propellants for aerosol cans.
  2. It is used as a solvent in the manufacture of pharmaceutical products.
  3. It is used as a fire extinguisher.

Uses of iodoform (CHI3) Iodoform was used earlier as an antiseptic. The antiseptic property of iodoform is due to the liberation of free iodine when it comes in contact with the skin.

Question 25. Write the structure of the major organic product in each of the following reactions.

  1. Haloalkanes And Haloarenes Structure Of Major Organic Product Reaction

Answer:

Haloalkanes And Haloarenes Major Organic Product Of The Reaction

Question 26. Write the mechanism of the following reaction:

Haloalkanes And Haloarenes Mechanism Of The Reaction

Answer: The given reaction is:

Haloalkanes And Haloarenes Mechanism Of The Reaction.

The given reaction is an SN2 reaction. In this reaction. CM acts as the nucleophile and attacks the carbon atom to which Br is attached. CN ion is an ambident nucleophile and can attack through both C and N. In this case, it attacks through the C-atom.

Haloalkanes And Haloarenes Ambident Nucleophile

Question 27. Arrange the compounds of each set in order of reactivity towards SN2 displacement:

  1. 2-Bromo-2-metln Ihutane. 1-Bromopentane, 2-Bromopentane
  2. 1-Bromo-3-methyl butane.2-Bromo-2-methyl butane.3-Bromo-2-inethyIhutane
  3. 1-Bromobutane. l-Bromo-2. 2-dimethyIpropane. 1-Bromo-2-methylbutane. 1-Bromo 3-methyIhutane

Answer:

Haloalkanes And Haloarenes In Order Of Reactivity Towards SN2 Displacement

(more branching at a nearer distance)

Hence, the increasing order of reactivity of the given compounds towards SN2 is:

1-Bromo-2,2-dimethy Ipropane < 1-Bromo-2-methy Ihutane < 1-Bromo-3-methylbutane < 1-Bromobutane.

Question 28. Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolyzed by aqueous KOH?
Answer: C6H5CHClC6H5 (more reactive)

Haloalkanes And Haloarenes More Easily Hydrolysed By Aqueous KOH

(more stable intermediate)

Question 29. p-Dichlorobenzene has higher m.p. and lower solubility than those of o-and m-isomers. Discuss.
Answer:

Haloalkanes And Haloarenes P Dichlorobenzene

p-Dichlorobenzene is more symmetrical than o-and m-isomers. For this reason, molecules of p-dichlorobenzene are more closely than o-and m-isomers in the crystal lattice. Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene.

Question 30. How the following conversions can be carried out?

  1. Propene to propane-1-ol
  2. Ethanol to but-1-yne
  3. 1-Bromopropane to 2-bromopropane
  4. Toluene to benzyl alcohol
  5. Benzene to 4-bronitobenzene
  6. Benzyl alcohol to 2-phenylethanoid acid
  7. Ethanol to propane nitrile
  8. Aniline to chloro benzene
  9. 2-Chlorobutane to 3, 4-dimethyl hexane
  10. 2-Methyl-1-propane to 2-chloro-2-methyl propane
  11. Ethyl chloride to propanoic acid
  12. But-1-ene to n-butyliodide
  13. 2-chloropropane to 1-propanol
  14. Isopropyl alcohol to iodoform
  15. Chlorobenzene to p-nitrophenol
  16. 2-Bromopropane to 1-bromopropane
  17. Chloroethane to butane
  18. Benzene to diphene
  19. tert-Butyl bromide to isobutyl bromide
  20. Aniline to phenyl isocyanide

Answer:

Haloalkanes And Haloarenes Following Conversion Carried Out

Haloalkanes And Haloarenes Following Conversion Carried Out.

Question 31. The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products, Explain.
Answer:

OH ion is highly solvated in an aqueous solution and as a result, the basic character of OH ion decreases. Therefore, it cannot abstract hydrogen from the β-carbon. So

⇒ \(\mathrm{R}-\mathrm{Cl}+\mathrm{Aq} \cdot \mathrm{KOH} \longrightarrow \mathrm{R}-\mathrm{OH}+\mathrm{KCl}\)

On the other hand, an alcoholic solution of KOFI contains an alkoxide (RO) ion. which is a strong base. Thus, it can abstract hydrogen from the β-carbon of the alkyl chloride and form an alkene by eliminating a molecule of HCl.

Haloalkanes And Haloarenes Alkyl Chloride And Alkene

Question 32. Primary alkyl halide C4H9Br(a) reacted with alcoholic KOH to given compound (b). Compound (b) is reacted with lIBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d). C8H18 which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.
Answer:

Two primary alkyl halides have the formula. C4H9Br. They are n-butyl bromide and isobutyl bromide.

Haloalkanes And Haloarenes Isobutyl Bromide

Therefore, compound (a) is either n-butyl bromide or isobutyl bromide.

Now, compound (a) reacts with Na metal to give compound (h) of molecular formula. C8H18. Which is different from the compound formed when n-butyl bromide reacts with Na metal. Hence. compound (a) must be isobutyl bromide.

Haloalkanes And Haloarenes N Butyl Bromide

Thus, compound (d) is 2. 5-Dimethylhexane.

It is given that compound (a) reacts with alcoholic KOH to give compound (b). compound (b) is 2-methyl propene.

Haloalkanes And Haloarenes 2 Methyl Propene

Also, compound (b) reacts with HBr to give compound (c) which is an isomer of (a). Hence compound (e) is -2-Bromo-2-methy propane.

Haloalkanes And Haloarenes 2 Bromo And Two Methylpropane

Question 33. What happens when

  1. n-butyl chloride is treated with alcoholic KOH?
  2. Broinohen/ene is treated with Mg in the presence of dry ether.
  3. Chlorobenzene is subjected to hydrolysis.
  4. Ethyl chloride is treated with aqueous KOH.
  5. Methyl bromide is treated with sodium in the presence of dry ether.
  6. Methyl chloride is treated with KCN.

Answer:

Haloalkanes And Haloarenes N Butyl Chloride

Chlorobenzene does not undergo hydrolysis under normal conditions. However, it undergoes hydrolysis when heated in an aqueous sodium hydroxide solution at a temperature of 623 K and a pressure of 300 atm to form phenol.

Haloalkanes And Haloarenes Chlorobenzene And Phenol

Haloalkanes And Haloarenes Ethyl Chloride And Methyl Bromide And Methyl Chloride

 

Important Questions for Class 12 Chemistry Chapter 5 Coordination Compounds

Coordination Compounds

Question 1. Write the formulas for the following coordination compounds:

  1. Tetraamminediaquacobalt(3) chloride
  2. Potassium tetracyanonickelate (2)
  3. Tris (ethane-1,2-diamine) chromium(3) chloride
  4. Ammincbromideochloridonitrito-N-platinate(2)
  5. Dichloridobis(ethane-1,2-diamine)platinum(4) nitrate
  6. Iron(3) hexacyanoferrate(2)

Answer:

⇒ \(\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_2\left(\mathrm{NH}_3\right)_4\right] \mathrm{Cl}_3\)

⇒ \(\mathrm{K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]\)

⇒ \(\left[\mathrm{Cr}(\text { en })_3\right] \mathrm{Cl}_3\)

⇒ \(\left[\mathrm{Pl}(\mathrm{NH})_3 \mathrm{BrCl}\left(\mathrm{NO}_2\right)\right]^{-}\)

⇒ \(\left[\mathrm{PtCl}_2(\mathrm{en})_2\right]\left(\mathrm{NO}_3\right)_2\)

⇒ \(\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3\)

Question 2. Write the IUPAC name of the following coordination compounds:

⇒ \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3\)

⇒ \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2\)

⇒ \(\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]\)

⇒ \(\mathrm{K}_3\left[\mathrm{Fe}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]\)

⇒ \(\mathrm{K}_2\left[\mathrm{PdCl}_4\right]\)

⇒ \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}\left(\mathrm{NH}_2 \mathrm{CH}_3\right)\right] \mathrm{Cl}\)

Answer:

  1. Hexarmninecobalt(3) chloride
  2. Pentaamminechloridocobalt(3) chloride
  3. Potassium hexacyanoferrate(3)
  4. Potassium trioxalaloferrate(3)
  5. Potassium tetrachloridopalladate(2)
  6. Diamminechiorido(methylamine)platinum(2) chloride

Read and Learn More Class 12 Chemistry with Answers Chapter Wise

Question 3. Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers:

⇒ \(\mathrm{K} \mid \mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_2\left(\mathrm{C}_2 \mathrm{O}_4\right)_2\)

⇒ \(\left[\mathrm{Co}(\mathrm{en})_3\right] \mathrm{Cl}_3\)

⇒\(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_2\right)\right]\left(\mathrm{NO}_3\right)_2\)

⇒ \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)\left(\mathrm{H}_2 \mathrm{O}\right) \mathrm{Cl}_2\right]\)

Answer: 1. Both geometrical (cis-, trans-) isomers for K[Cr(H2O),(C2O4)2] can exist. Also, optical isomers for cis-isomers exist.

Geometrical isomers

Coordination Compound Geometrical Isomers

Truns-iso me r is optically inactive. On the other hand, cis-isomer is optically active.

Coordination Compound Trails Isomer Is Optically Inactive

2. Two-optical isomers for [CO(en)3]Cl3 exist.

Coordination Compound TwoOptical Isomers

Two optical isomers arc possible Tor this structure

Coordination Compound Two Optical Isomers Are Possible For This Structure

3. [CO(NH3)5(NO2)](NO3)2

A pair of optical isomers

Coordination Compound A Pair Of Optical Isomers

It can also show linkage isomerism [CO( NH3 )5(NO2)] ( NO3)2 and [CO(NH )5 (ONO)]( NO3)2

It can also show ionization isomerism [Co(NH3)5 (NO2)](NO3)2 and [CO(NH3)5 (NO3)] (NO3) (NO2)

Geometrical ( cis-trails-) isomers of [Pt( NH3)( H2O)Cl2] can exist.

Coordination Compound Geometrical Isomers.

Question 4. Give evidence that [Co(NH3)5Cl]SO4 and [Co(NH3)5SO4]CI are ionization isomers.
Answer:

When ionization isomers are dissolved in water, they ionize to give different ions, l liese ions then react differently

⇒ \(\left[\mathrm {CO} (\mathrm {NH}_{3})_{5}\mathrm {Cl} \left]\mathrm{SO}_4+\mathrm{Ba}^{2+} \longrightarrow \mathrm{BaSO}_4 \downarrow\right.\right.\) White precipitate

⇒ [CO(NH3)5Cl]SO4 + Ag+ →  No reaction

⇒ [CO(NH3)5SO4]Cl + Ba2+ →  No reaction

⇒ \(\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_5 \mathrm{SO}_4 \mathrm]{Cl}+\mathrm{Ag}^{+} \longrightarrow \mathrm{AgCl} \downarrow\right.\)

Question 5. Explain based on valence bond theory that [Ni(CN)4]2- ion with square planar structure is diamagnetic and the [Ni(Cl4]2- ion with tetrahedral geometry Is paramagnetic.
Answer:

Ni is in the +2 oxidation state i.e., in the d8 configuration.

d8 configuration:

Coordination Compound Tetrahedral Geometry Is Paramagnetic

There are 4 CN ions. Thus, it can either have a tetrahedral geometry or square planar geometry. Since CN ion is a strong field ligand, it causes the pairing of unpaired 3d electrons.

Coordination Compound CN Ion Is A Strong Field Ligand It Causes The Pairing Of Unpaired 3d Electrons

It now undergoes dsp2 hybridization. Since all electrons are paired, it is diamagnetic. in. the case of [NiCl4]2-, Cl ion is a weak field ligand. Therefore, it does not lead to the pairing of impaired 3d electrons. Therefore, it undergoes sp3 hybridization.

Coordination Compound There Are 2 Unpaired Electrons In This Case It Is Paramagnetic In Nature

Since there are 2 unpaired electrons in this case, it is paramagnetic.

CBSE Class 12 Chemistry Chapter 5 Coordination Compounds Important Question And Answers

Question 6. [NiC4]2- is paramagnetic while [NitCO)4] is diamagnetic though both are tetrahedral. Why?
Answer:

Though both [NiCl4]2- and [Ni(CO)4] are tetrahedral, their magnetic characters are different. This is due to a difference like ligands. Cl is a weak field ligand and it does not cause the pairing of unpaired 3d electrons. Hence. [NiCl4]2- is paramagnetic.

Coordination Compound Tetrahedral Their Magnetic Characters Are Different Paramagnetic

In Ni(CO)4, Ni is in the zero oxidation state Le., it has a configuration of 3d8 4s2.

Coordination Compound Zero Oxidation State

But CO is a strong field ligand. Therefore, it causes the pairing of unpaired 3d electrons. Also. it causes the 4s electrons to shift to the 3d orbital, thereby giving rise to sp3 hybridization. Since no unpaired electrons are present in this case, [Ni(CO)4] is diamagnetic.

Question 7. [Fe(H2O)6]3+ is strongly paramagnetic whereas ji’e(CN)6J’ is weakly paramagnetic. Explain.
Answer:

In both [Fe(H2O)6]3+ and [Fe(CN)6]3-, Fe exists in the +3 oxidation state i.e., in d5 configuration.

Coordination Compound Strong Field Ligand

Since CN is a strong field ligand, it causes the pairing of impaired electrons. Therefore, there is only one unpaired electron left in the d-orbital.

Coordination Compound Weak Field Ligand

Therefore, \(\mu=\sqrt{n(n+2)}=\sqrt{1(1+2)}=\sqrt{3}=1.732 B M\)

On the other hand. (H2O is a weak field ligand. Therefore, it cannot cause the pairing of electrons. This means that the number of unpaired electrons is 5.

Therefore, \(\mu=\sqrt{n(n+2)}=\sqrt{5(5+2)}=\sqrt{35} \approx 6 B M\)

Thus, it is evident that [Fe(H2O)6]3+ is strongly paramagnetic, while [Fe(CN)6]3- is weakly paramagnetic.

Question 8. Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex.
Answer:

Coordination Compound Inner Orbital Complex And Outer Orbital Complex

Question 9. Predict the number of unpaired electrons in the square planar [Pt(CN)6]2- ion.
Answer: [Pt(CN)6]2-

In this complex. Pt is in the +2 state. It forms a square planar structure. This means that it undergoes dsp2 hybridization. Now. the electronic configuration of Pd(+2) is 5d8.

Coordination Compound Square Planar Structure

CN being a strong field ligand causes the pairing of unpaired electrons. Hence, there are no unpaired electrons in [Pt(CN)4]2-.

Question 10. The hexaquo manganese(ll) ion contains live unpaired electrons, while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory.
Answer:

Coordination Compound Crystal Field Theory

Question 11. Calculate the overall complex dissociation equilibrium constant for the Cu(NH3)2+4‘ ion. given that the β4 for this complex is 2.1 × 1013,
Answer:

⇒ \(\beta^4=2.1 \times 10^{13}\)

The overall complex dissociation equilibrium constant is the reciprocal of the overall stability constant (f.

⇒ \(\frac{1}{\beta_4}=\frac{1}{2.1 \times 10^{13}}=4.7 \times 10^{-14}\)

Question 12. Explain the bonding in coordination compounds in terms of Werner’s postulates.
Answer:

  1. Werner’s postulates explain the bonding in coordination compounds as follows: A metal exhibits two types of valencies namely, primary and secondary valencies. Primary valencies are satisfied by negative ions while secondary valencies are satisfied by both negative and neutral ions.
  2. A metal ion has a definite number of secondary valencies around the central atom. Also, these valencies project in a specific direction in the space assigned to the definite geometry of the coordination compound.
  3. Primary valencies are usually ionizable. while secondary valencies are non-ionizable.

Question 13. FeSO4 solution mixed with (NH4)2SO4 solutions in a 1:1 molar ratio gives the least of Fe2+ ion but CuSO4 solution mixed with aqueous ammonia in a 1:4 molar ratio does not give the least of Cu2+ ion. Explain why?
Answer:

⇒ \(\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4+\mathrm{FeSO}_4+6 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text { Mohr’s Salt }}{\mathrm{FeSO}_4 \cdot\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \cdot 6 \mathrm{H}_2 \mathrm{O}}\)

⇒ \(\mathrm{CuSO}_4+4 \mathrm{NH}_3+5 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text { Tetraam min copper(2)sulphate }}{\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right] \mathrm{SO}_4 .5 \mathrm{H}_2 \mathrm{O}}\)

Both the compounds i.e., FeSO4. (NH4), SO4. 6H2O and [Cu(NH3)4] SO4. 5H2O falls under the category of addition compounds with only one major difference i.e., the former is an example of a double salt, while the latter is a coordination compound.

Question 14. Explain with two examples of each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.
Answer:

Coordination entity: A coordination entity is an electrically charged radical or species carrying a positive or negative charge. In a coordination entity, the central atom or ion surrounded by a suitable number of neutra 1 molecules or negative ions (called ligands For example:

⇒\(\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+},\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_6\right]^{3+} \text { etc. }\) ⇒ cationic complex

⇒ \(\left[\mathrm{PtCl}_4\right]^{2-},\left[\mathrm{Ag}(\mathrm{CN})_2\right]^{-} \quad \text { etc. }\) ⇒ anionic complex

⇒ \(\left[\mathrm{Ni}(\mathrm{CO})_4\right] \cdot\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \text { etc. }\) ⇒ neutral complex

Ligands: The neutral molecules or negatively charged ions that surround the metal atom in a coordination entity or a coordinate complex are known as ligands. For example Cl, FLO. H, N CFLCFLNFL etc.

Coordination number: The total number of ligands (either neutral molecules or negative ions) that get attached to the central metal atom in the coordination sphere is called the coordination number of the central metal atom. It is also referred to as its legacy.

For example:

  1. In the complex, K2[PtCI]6 there as six chloride ions attached to Pt in the coordinate sphere. Therefore, the coordination number of Pt is 6.
  2. Similarly, in the complex [Ni(NH3)4 ]Cl2, the coordination number of the central atom (Ni) is 4.

Coordination polyhedron: Coordination polyhedrons about the central atom can be
defined as the spatial arrangement of the ligands that are directly attached to the central metal ion in the coordination sphere. For example:

Coordination Compound Square Planar And Tetrahedral

Homoleptic complexes: These are those complexes in which the metal ion is bound to only one kind of a donor group.

For example., [Co(NH3)6]3+ .[PtCl4]2- etc.

Heteroleptic complexes: Heteroleptic complexes are those complexes where the central metal ion is bound to more than one type of donor group.

For example., [Co(NH3)4 Cl2]+ ,[Co(NH3)5Cl]2+

Question 16. What is meant by unidentate? dentate and ambidentate ligands? Give two examples for each.
Answer:

A ligand may contain one or more unshared pail’s of electrons which are called the donor sites of ligands. Now. depending on the number of these donor sites, ligands can be classified as follows:

Unidenatate ligands: Ligands with only one donor site arc are called unidentate ligands.

⇒ \(\ddot{\mathrm{N}} \mathrm{H}_3, \mathrm{Cl}^{-2} \text { etc. }\)

Didentate ligands: Ligands that have two donor sites arc called bidentate ligands. For example.,

Ethane- 1, 2-diamine

Coordination Compound Didentate Ligands

Ambidentate ligands: Ligands that can attach themselves to the central metal atom through two different atoms are called ambidentate ligands. For example:

Coordination Compound Ambidentate Ligands

Question 17. Specify the oxidation numbers of the metals in the following coordination entities:

  1. \(\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)(\mathrm{CN})(\mathrm{en})_2\right]^{2+}\)
  2. \(\left[\mathrm{CoBr}_2(\mathrm{en})_2\right]^{+}\)
  3. \(\left[\mathrm{PtCl}_4\right]^{2-}\)
  4. \(\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]\)
  5. \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]\)

Answer:

1.  \(\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)(\mathrm{CN})(\mathrm{en})_2\right]^{2+}\)

Let the oxidation number of Co be x.

The charge on the complex is +2

Coordination Compound Oxidation Numbers Of The Metals 1

x- 1 = +2

x = +3

2. [PtCl4]

Let the oxidation number of Pt be x.

The charge on the complex is -2,

Coordination Compound Oxidation Numbers Of The Metals 2

x = +2

3. Coordination Compound Oxidation Numbers Of The Metals 3

x- 2= +1

x =+3

4. \(\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]\)

Coordination Compound Oxidation Numbers Of The Metals 4

x – 2 = +1

x = +3

5. K3[Fe(CN)6]

x = +3

Coordination Compound Oxidation Numbers Of The Metals 5

x – 3 = 0

x = +3

Question 18. Using IUPAC norms write the formulas for the following:

  1. Tetrahydroxozincatc (2)
  2. Potassium tetrachloridopalladate (2)
  3. Diamminedichloridoplatinum (2)
  4. Potassium tetracyanonickelate (2)
  5. Pentaamminenitrito-O-cobalt (3)
  6. Hexaamminecobalt (3) sulphate
  7. Potassium tri(oxalate)chromate (3)
  8. Hexaammineplatinum (4)
  9. Tetrabromidocuprate (2)
  10. Pentaamminenitrito-N-cobalt (3)

Answer:

  1. \(\left[\mathrm{Zn}(\mathrm{OH})_4\right]^{2-}\)
  2.  \(\mathrm{K}_2\left[\mathrm{PdCl}_4\right]\)
  3. \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]\)
  4. \(\mathrm{K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]\)
  5. \(\left[\mathrm{Co}(\mathrm{ONO})\left(\mathrm{NH}_3\right)_5\right]^{2+}\)
  6. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]_2\left(\mathrm{SO}_4\right)_3\)
  7. \(\mathrm{K}_3\left[\mathrm{Cr}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]\)
  8. \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_6\right]^{4+}\)
  9. \(\left[\mathrm{Cu}(\mathrm{Br})_4\right]{^2-}\)
  10. \(\left[\mathrm{Co}\left[\mathrm{NO}_2\right)\left(\mathrm{NH}_3\right)_5\right]^{2+}\)

Question 19. Using IUPAC norms write the systematic names of the following:

  1. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3\)
  2. \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}\left(\mathrm{NH}_2 \mathrm{CH}_3\right)\right] \mathrm{Cl}\)
  3. \(\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}\)
  4. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}\left(\mathrm{NO}_2\right)\right] \mathrm{Cl}\)
  5. \(\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)
  6. \(\left[\mathrm{NiCl}_4\right]^{2-}\)
  7. \(\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_2\)
  8. \(\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}\)
  9. \(\left[\mathrm{Ni}(\mathrm{CO})_4\right]\)

Answer:

  1. Hexaamminecobalt (3) chloride
  2. Diamminechlorido (methylamine) platinum (2) chloride
  3. Hexaaquatitanium (3) ion
  4. Tetraamminechloridonitrito-N-Cobalt (3) chloride
  5. Hexaaquamangancse (2) ion
  6. Tetrachloridonickelate (2) ion
  7. Hexaamminenickcl (2) chloride
  8. Tris (ethane- 1, 2-diamine) cobalt (3) ion
  9. Tetracarbonylnickel (0)

Question 20. List various types of isomerism possible for coordination compounds, giving an example of each.
Answer:

Coordination Compound Isomerism In Coordination Compounds

1. Geometrical Isomerism: This type of isomerism is common in heteroleptic complexes. It arises due to the different possible geometric arrangements of the ligands. For example:

Coordination Compound Cis Isomer And Trans Isomer

2. Optical Isomerism: This type of isomerism arises in chiral molecules. Isomers are mirror images of each other and are non-superim possible.

Coordination Compound Optical Isomerism

3. Linkage Isomerism: This type of isomerism is found in complexes that contain ambidentate ligands.

Coordination Compound Linkage Isomerism

4. Coordination Isomerism: This type of isomerism arises when the ligands are interchanged between cationic and anionic entities of different metal ions present in the complex.

For example \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]\left[\mathrm{Cr}(\mathrm{CN})_6\right] \text { and }\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]\left[\mathrm{Co}(\mathrm{CN})_6\right]\)

5. Ionization Isomerism: This type of isomerism arises when a counter ion replaces a ligand within the coordination sphere. For example

⇒ \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{SO}_4\right] \mathrm{Br} \text { and }\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Br}\right] \mathrm{SO}_4\)

6. Solvate Isomerism: Solvate isomers differ by whether or not the solvent molecule is directly bonded to the metal ion or merely present as a free solvent molecule in the crystal lattice.

Coordination Compound Solvate Isomerism

Question 21. How many geometrical isomers are possible in the following coordination entities?

  1. [Cr(C2O4)3]3+
  2. [Co(NH3)3Cl3]

Answer:

For [Cr(C2O4)3]3+ no geometrical isomer is possible as it is a bidentate ligand.

Coordination Compound Isomer Is Possible As It Is A Bidentate Ligand

[Co(NH3)3Cl3]  Two geometrical isomers are possible.

Coordination Compound Facial And Meridional

Question 22. Draw the structure of optical isomers of:

  1. \(\left[\mathrm{Cr}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}\)
  2. \(\left[\mathrm{PtCl}_2(\mathrm{en})_2\right]^{2+}\)
  3. \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2(\mathrm{en})\right]^{+}\)

Answer: 1. \(\left[\mathrm{Cr}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}\)

Coordination Compound Structure Of Optical Isomers 1

2. \(\left[\mathrm{PtCl}_2(\mathrm{en})_2\right]^{2+}\)

Coordination Compound Structure Of Optical Isomers 2

3. \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2(\mathrm{en})\right]^{+}\)

Coordination Compound Structure Of Optical Isomers 3

Question 23. Draw all the isomers (geometrical and optical) of:

  1. \(\left[\mathrm{CoCl}_2(\mathrm{en})_2\right]^{+}\)
  2. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right) \mathrm{Cl}(\mathrm{en})_2\right]^{2+}\)
  3. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2(\mathrm{en})\right]^{+}\)

Answer: 1. \(\left[\mathrm{CoCl}_2(\mathrm{en})_2\right]^{+}\)

Coordination Compound The Geometrical Isomers

In total, three isomers are possible.

2. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right) \mathrm{Cl}(\mathrm{en})_2\right]^{2+}\)

Coordination Compound The Geometrical Isomers 2

Trans-isomers are optically inactive.

Cis-isomers are optically active.

3. \(\left[\mathrm{CoCl}_2(\mathrm{en})_2\right]^{+}\)

Coordination Compound The Geometrical Isomers 3

Question 24. Write all the geometrical isomers of [Pt(NH3)(Br)(G)(py)j and how many of these will exhibit optical isomers?
Answer: [Pt(NH3)(Br)(Cl)(py)]

Coordination Compound Tetrahedral Complexes Rarely Show Optical Isomerization

From the above prisoners, none will exhibit optical isomers. Tetrahedral complexes rarely show optical isomerization. They do so only in the presence of unsymmetrical chelating agents.

Question 25. Aqueous copper sulphate solution (blue) gives:

  1. A green precipitate with aqueous potassium fluoride, and
  2. A bright green solution with aqueous potassium chloride Explain these experimental results.

Answer:

Aqueous CuSO4 exists as [Cu(H2O)4]SO4 It is blue due to the presence of [Cu(H2O)4]2+ ions.

When KF is added.

Coordination Compound Aqueous Potassium Fluoride

When KCI is added:

⇒ \(\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}_4\right)\right]^{2+}+4 \mathrm{Cl}^{-} \longrightarrow \underset{\text { (bright green) }}{\left[\mathrm{CuCl}_4\right]^{2-}}+4 \mathrm{H}_2 \mathrm{O}\)

In both these cases, the weak field ligand water is replaced by the F and Cl ions.

Question 26. What is the coordination entity formed when an excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H2S (g) is passed through this solution?
Answer:

⇒ \(\mathrm{CuSO}_{4(\mathrm{aq})}+4 \mathrm{KCN}_{(\mathrm{aq})} \longrightarrow \mathrm{K}_2\left[\mathrm{Cu}(\mathrm{CN})_4\right]_{(\mathrm{aq})}+\mathrm{K}_2 \mathrm{SO}_{4(\mathrm{aq})}\)

⇒ \(\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_4\right]^{2+}+4 \mathrm{CN}^{-} \longrightarrow\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{2-}+4 \mathrm{H}_2 \mathrm{O}\)

Thus, the coordination entity formed in the process is K2[Cu(CN)4]. It is a very stable complex, which does not ionize to give Cu2+ ions when added to water. Hence, Cu2+ ions are not precipitated when H2S, i is passed through the solution.

Question 27. Discuss the nature of bonding in the following coordination entities based on valence bond theory:

  1. [Fe(CN)6]4-
  2. [FeF6]3-
  3. [Co(C2CO4)3]3-
  4. [CoF6]3-

Answer: [Fe(CN)6]4-

In the above coordination complex, iron exists in the +2 oxidation state.

Fe2+: Electronic configuration is 3d6

Orbitals of Fe2+ ion:

Coordination Compound CN Minous ion Is A Strong Field Ligand

As CN ion is a strong field ligand, it causes the pairing of the unpaired 3d electrons.

Coordination Compound Feasible Hybridization

Since there are six ligands around the central metal ion, the most feasible hybridization is d2sp3.

d2sp3 hybridized orbitals of Fe2+ are:

Coordination Compound 6 Electron Pairs From CN Ions Occupy The Six Hybrid Orbitals

6 electron pairs from CN ions occupy the six hybrid dfsp3 orbitals.

Then,

Coordination Compound 6 Pairs Of Electrons From 6 CN Ions

Hence, the geometry of the complex is octahedral and the complex is diamagnetic (as there are no unpaired electrons).

2. [FeF6]3-

In this complex, the oxidation state of Fe is +3.

Orbitals of Fe-3 ion:

Coordination Compound The Pairing Of The Electrons In The 3d Orbitals

There are 6 F ions. Thus, it will undergo d2sp3 or sp2d3 hybridization. As F is a weak field ligand, it does not cause the pairing of the electrons in the 3d orbitals. Hence, the most feasible hybridization is sp3d2.

sp3d2 hybridized orbitals of Fe3+ it are:

Coordination Compound The Geometry Of The Complex Is Found To Be Octahedral

Hence, the geometry of the complex is found to be octahedral.

3. [Co(C2CO4)3]3-

Cobalt exists in the +3 oxidation state in the given complex.

Orbitals of Co3+ ion:

Coordination Compound Fluoride Ion Is A Weak Field Ligand

Oxalate ion is a weak field ligand. Therefore, it cannot cause the pairing of the 3d orbital electrons. As there are 6 ligands, hybridization has to be sp3d2.

sp3d2 hybridized orbitals of Co3+

Coordination Compound Hybridized Orbitals Of Co 3 Plus

The 6 electron pairs from the 3 oxalate ions (oxalate anion is a bidentate ligand) occupy these sp3d2 orbitals.

Coordination Compound Geometry Of The Complex Is Found To Be Octahedral

Hence, the geometry of the complex is found to be octahedral.

4. [CoF2]3-

Cobalt exists in the +3 oxidation state.

Orbitals of Co3+ ion:

Coordination Compound Oxalate Ion Is A Weak Field Ligand

Again, fluoride ion is a weak field ligand. It cannot cause the pairing of the 3d electrons. As a result, the Co!+ ion will undergo sp’d2 hybridization. sp3d2 hybridized orbitals of Co3+ ion are:

Coordination Compound Geometry Of The Complex Is Octahedral And Paramagnetic

Hence, the geometry of the complex is octahedral and paramagnetic.

Question 28. Draw a figure to show the splitting of d orbitals in an octahedral crystal field.
Answer:

Coordination Compound Octahedral Crystal Field

Question 29. What is a spectrochemical series? Explain the difference between a weak field and a strong field ligand.
Answer:

A spectrochemical series is the arrangement of common ligands in the increasing order of their crystal-field splitting energy (CFSE) values. The ligands present on the R.H.S of the series are strong field ligands while that on the L.H.S are weak field ligands. Also, strong field ligands cause higher splitting in the d orbitals than weak field ligands.

I-< Br- <S2 <SCN <CI <N3- <OH <C2O 2-4 <~H2O < NCS <edta4- <NH3<en <CN < CO

Question 30. What is crystal field splitting energy? How does the magnitude of Δ0 decide the actual configuration of d-orbitals in a coordination entity?
Answer:

  • The degenerate d-orbitals (in a spherical field environment) split into two levels i.e., eg and t2g in the presence of ligands. The splitting of the degenerate levels due to the presence of ligands is called crystal-field splitting while the energy difference between the two levels (eg and t2g) is called the crystal-field splitting energy.
  • It is denoted by Δ0. After the orbitals have split, the filling of the electrons takes place. After 1 electron (each) has been filled in the three t2g orbitals, the filling of the fourth electron takes place in two ways.
  • It can enter the e orbitals (giving rise to t32g e1g like electron configuration) or the pairing of the electrons can take place in the t2g orbitals (giving rise to t42g e0g like electronic configuration).
  • If the A0 value of a ligand is less than the pairing energy (P), then the electrons enter the e orbital.  On the other hand, if the Δ0 value of a ligand is more than the pairing energy (P), then the electrons enter the t2g orbital.

Question 31. [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2- is diamagnetic. Explain why?
Answer:

In [Cr(NH3)6]3+ Cr is in the +3 oxidation state i.e., d3 configuration. Since there are three unpaired electrons in 3d orbitals.

Coordination Compound Unparied Electrons In 3d Orbitals

Therefore, it undergoes d2sp3 hybridization and the electrons in the 3d orbitals remain unpaired. Hence, it is paramagnetic.

In [Ni(CN)4]2- Ni exists in the +2 oxidation state i.e., d8 configuration.

Coordination Compound Paramagnetic In Nature

CN is a strong field ligand. It causes the pairing of the 3d orbital electrons. Then, Ni2+ undergoes dsp2 hybridization.

Coordination Compound It Cause The Pairing Of The 3d Orbital Electrons

As there are no unpaired electrons, it is diamagnetic.

Question 32. A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2- is colourless. Explain.
Answer:

  • In [Ni(H2O)6]2+, H2O is a weak field ligand. Therefore, there are unpaired electrons in Ni-‘. In this complex, the 3d electrons from the lower energy level can be excited to the higher energy level i.e., the possibility of d-d transition is present.
  • Hence,[Ni(H2O)6]2+  is coloured. In [Ni(CN)4]2- the electrons are all paired as CN is a strong field ligand. Therefore, d-d transition is not possible in [Ni(CN)4]2-. Hence, it is colourless.

Question 33. [Fe(CN)6]4- and [Fe(H2O)6]2+ are of different colours in dilute solutions. Why?
Answer:

  • The colour of a particular coordination compound depends on the magnitude of the crystal-field splitting energy, Δ. This CFSE in turn depends on the nature of the ligand.
  • In the case of [Fe(CN)6]4- and [Fe(H2O)6]2+, the colour differs because there is a difference in the CFSE. Now CN is a strong field ligand having a higher CFSE value of compound to the CFSE value of water. This means that the absorption of energy for the infra d-d transition also differs. Hence the transmitted colour also differs.

Question 34. Discuss the nature of bonding in metal carbonyls.
Answer:

  • The metal-carbon bonds in metal carbonyls have both σ and π characters. An σ bond is formed when the carbonyl carbon donates a lone pair of electrons to the vacant orbital of the metal.
  • A π bond is formed by the donation of a pair of electrons from the filled metal d orbital into the valent anti-bonding π* orbital (also known as back bonding of the carbonyl group).
  • The σ bond strengthens the π bond and vice versa. Thus, a synergic effect is created due to this metal-ligand bonding. This synergic effect strengthens the bond between CO and the metal.

Coordination Compound Synergic Bonding In Metal Carbonyls

Question 35. Give the oxidation state, d-orbital occupation and coordination number of the central metal ion in the following complexes:
Answer:

  1. \(\mathrm{K}_3\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]\)
  2. \(\text { cis- }\left[\mathrm{Cr}(\mathrm{en})_2 \mathrm{Cl}_2\right] \mathrm{Cl}\)
  3. \(\left(\mathrm{NH}_4\right)_2\left[\mathrm{CoF}_4\right]\)
  4. \(\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{SO}_4\)

Answer:

1. \(\mathrm{K}_3\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]\)

The central metal ion is Co3+

Its coordination number is 6.

The oxidation state can be given as:

x – 6 = -3

x = +3

The d orbitals occupation for Co3+ is t62g e0g.

2. \(\text { cis- }\left[\mathrm{Cr}(\mathrm{en})_2 \mathrm{Cl}_2\right] \mathrm{Cl}\)

The central metal ion is Cr3+

The coordination number is 6.

The oxidation state can be given as:

x + 2(0) + 2(-1) = +1

x- 2 = +1

x = +3

The d orbitals occupation for Cr3+ is t32g.

3. \(\left(\mathrm{NH}_4\right)_2\left[\mathrm{CoF}_4\right]\)

The central metal ion is Co2+.

The coordination number is 4.

The oxidation state can be given as:

x- 4 = -2

x = +2

The d orbitals occupation for Co2+ is e 4g t32g

4. \(\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{SO}_4\)

The central metal ion is Mn2+.

The coordination number is 6.

The oxidation state can be given as:

x + 0 = +2

x = +2

The d orbitals occupation for Mn2+ is t32g e2g.

Question 36. Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also, give the stereochemistry and magnetic moment of the complex:

  1. \(\mathrm{K}\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_2\left(\mathrm{C}_2 \mathrm{O}_4\right)_2\right] \cdot 3 \mathrm{H}_2 \mathrm{O}\)
  2. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2\)
  3. \(\mathrm{CrCl}_3(\mathrm{py})_3\)
  4. \(\mathrm{Cs}\left[\mathrm{FeCl}_4\right]\)
  5. \(\mathrm{K}_4\left[\mathrm{Mn}(\mathrm{CN})_6\right]\)

Answer:

1. \(\mathrm{K}\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_2\left(\mathrm{C}_2 \mathrm{O}_4\right)_2\right] \cdot 3 \mathrm{H}_2 \mathrm{O}\)

IUPAC Name: Potassium diaquadioxalatochromate (TIT) trihydrate.

The oxidation state of chromium = +3

Electronic configuration: 3d3: t32g

Coordination number = 6

Shape: octahedral

Stereochemistry:

Coordination Compound Potassium Diaquadioxalatochromate

Coordination Compound Trans Optically Inactive.

Coordination Compound Cis Optically Active

Magnetic moment, \(\mu=\sqrt{n(n+2)}\)

\(\mu=\sqrt{3(3+2)}=\sqrt{5} \sim 4 \mathrm{BM}\)

2. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2\)

TUPAC name: Pcntaamminechloridocoball (3) chloride

The oxidation state of Co = +3

Coordination number = 6

Shape: octahedral.

Electronic configuration : d6: t62g.

Stereochemistry:

Coordination Compound Cis Optically Active 3

Magnetic Moment = 0

3. CrCl3(py)3

TUPAC name: Trichloridotripyridincchromiuixi (3)

The oxidation state of chromium = +3

Electronic configuration for d3: t32g

Coordination number = 6

Shape: octahedral.

Stereochemistry:

Coordination Compound Facial Isomer And Meridional Isomers

Both isomers are optically active. Therefore, a total of 4 isomers exist.

Magnetic moment, \(\mu=\sqrt{n(n+2)}=\sqrt{3(3+2)}=\sqrt{15} \sim 4 B M\)

4. Cs[FeCl4]

IUPAC name: Caesium tetrachloroferrate (3)

The oxidation state of Fe = +3

Electronic configuration d5: e2g t32g

Coordination number = 4

Shape: tetrahedral

Stereochemistry: optically inactive

Magnetic moment:

⇒ \(\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{5(5+2)}=\sqrt{35} \sim 6 \mathrm{BM}\)

5. K4[Mn(CN)6]

IUPAC Name: Potassium hexacyanomanganate (2)

The oxidation state of manganese = +2

Electronic configuration: d5: t52g

Coordination number = 6

Shape: octahedral

S(geochemistry:optically inactive

Magnetic moment,

⇒ \(\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{1(1+2)}=\sqrt{3}=1.732 \mathrm{BM}\)

Question 37. What is meant by the stability of a coordination compound in solution? State the factors which govern the stability of complexes.
Answer:

The stability of a complex in a solution refers to the degree of association between the two species involved in a state of equilibrium. Stability can be expressed quantitatively in terms of stability constant or formation constant.

⇒ \(\mathrm{M}+3 \mathrm{~L} \rightleftharpoons \mathrm{ML}_3\)

Stability constant,

⇒ \(\beta=\frac{\left[\mathrm{ML}_3\right]}{[\mathrm{M}][\mathrm{L}]^3}\)

For this reaction, the greater the value of the stability constant, the greater is the proportion of ML3 in the solution.

Stability can be of two types:

  1. Thermodynamic stability: The extent to which the complex will be formed or will be transformed into another species at the point of equilibrium is determined by thermodynamic stability.
  2. Kinetic stability: This helps determine the speed with which the transformation will occur to attain the state of equilibrium.

Question 38. What is meant by the chelate effect? Give an example.
Answer:

When a ligand attaches to the metal ion in a manner that forms a ring, then the metal-ligand association is found to be more stable. In other words, we can say that complexes containing chelate rings are more stable than complexes without rings. This is known as the chelate effect.

For example: ⇒ \(\mathrm{Ni}_{(\mathrm{aq})}^{2+}+6 \mathrm{NH}_{3 \mathrm({aq})} \longleftrightarrow\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]_{(\mathrm{aq})}^{2+} \log \beta=7.99\)

⇒ \(\mathrm{Ni}_{(\mathrm{aq})}^{2+}+3 \mathrm{en}_{(\mathrm{aq})} \longleftrightarrow\left[\mathrm{Ni}(\mathrm{en})_3\right]_{(\mathrm{aq})}^{2+} \log \beta=18.1 \text { (more stable) }\)

Coordination Compound Chelate Effect

Question 39. Discuss briefly giving an example in each case the role of coordination compounds in:

  1. Biological system
  2. Medicinal chemistry
  3. Analytical Chemistry
  4. Extraction/metallurgy of metals

Answer:

  1. Role of coordination compounds in biological system: We know that photosynthesis is made possible by the presence of the chlorophyll pigment. This pigment is a coordination compound of magnesium.
  2. Role of coordination compounds in medicinal chemistry: Certain coordination compounds of platinum (for example, cis-plutin) are used for inhibiting the growth of tumours.
  3. Role of coordination compounds in analytical chemistry: During salt analysis, several basic radicals are detected with the help of the colour changes they exhibit with different reagents.
  4. Role of coordination compounds in extraction or metallurgy of metals: From [Au(CN)2]+ solution, gold is extracted by the addition of zinc metal.

Question 40. How many ions are produced from the complex Co(NH3)6Cl2 in solution?

  1. 6
  2. 4
  3. 3
  4. 2

Answer: 3

The given complex can be written as Co(NH3)6Cl2 Thus, [Co(NH3)6]2+ along with two Cl ions are produced.

Question 41. Amongst the following ions which one has the highest magnetic moment value?

  1. [Cr(H2O6)]3+
  2. [Fe(H2O)6]2+
  3. [Zn(H2O)6]2+

Answer:

No. of unpaired electrons in [Cr(H2O6)]3+ = 3

⇒ \(\mu=\sqrt{n(n+2)}=\sqrt{3(3+2)}=\sqrt 15 {\sim 4 B M}\)

No. of unpaired electrons in [Fe(H2O)6]2+ = 4

⇒ \(\mu=\sqrt{4(4+2)}=\sqrt{24} \sim 5 \mathrm{BM}\)

No. of unpaired electrons in [Zn(H2O)6]2+ = 0

Hence, [Fe(H2O)6]2+ has the highest magnetic moment value.

Question 42. The oxidation number of cobalt in K[Co(CO)4] is

  1. +1
  2. +3
  3. -1
  4. -3

Answer:

We know that CO is a neutral ligand and K carries a charge of + 1. Therefore, the complex can be written as K+[Co(CO)4]. Therefore, the oxidation number of Co in the given complex is -1. Hence, option (3) is correct.

Question 43. Amongst the following, the most stable complex is

  1. [Fe(H2O)6]3+
  2. [Fe(NH3)6]3+
  3. [Fe(C2O4)3]3-
  4. [FeCl6]3-

Answer:

We know that the stability of a complex increases by chelation. Therefore, the most stable complex is [Fe(C2O4)3]3-

Coordination Compound Stability Of A Complex Increases By Chlation

Question 44. What will be the correct order for the wavelengths of absorption in the visible region for the following:

⇒\(\left[\mathrm{Ni}\left(\mathrm{NO}_2\right)_6\right]^{4-},\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+},\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)

Answer:

The central metal ion in all three complexes is the same. Therefore, absorption in the visible region depends on the ligands. The order in which the CFSE value of the ligands increases in the spectrochemical series is as follows:

H2O < NH3 < NO2

Thus, the amount of crystal-field splitting observed will be in the following order:

Δ0[H2O] < Δ0[NH3] < Δ0[NO2-]

Hence, the wavelengths of absorption in the visible region will be in the following order:

⇒ \(\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}>\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}>\left[\mathrm{Ni}(\mathrm{NO})_6\right]^{4-}\)

Important Questions for Class 12 Chemistry Chapter 4 The d- and f-Block Elements

D And F Block

Question 1. The silver atom has filled d orbitals (4d ) in its ground state. How can you say that it is a transition element?
Answer: Ag has a filled 4d orbital (4d10 5s1) in its ground state. Now, silver displays two oxidation states (+1 and +2). In the +1 oxidation state, an electron is removed from the s-orbital. However, in the +2 oxidation state, an electron is removed from the d-orbital.

Thus, the d-orbital now becomes incomplete (4d9). Hence, it is a transition element.

Question 2. In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomization of zinc is the lowest, i.e., 126 kJ mol-1Why?
Answer: The extent of metallic bonding a clement undergoes decides the enthalpy of atomization.

  • The more extensive the metallic bonding of an element, the more will be its enthalpy of atomization. In all transition metals (except Zn, electronic configuration: 3d10 4s2), some unpaired electrons account for their stronger metallic bonding.
  • Due to the absence of these unpaired electrons, the inter-atomic electronic bonding is the weakest in Zn and as a result, it has the least enthalpy of atomization.

Question 3. Which of the 3d scries of the transition metals exhibits the largest number of oxidation states and why?
Answer: Mn (Z = 25) = 3d5 4s2‘ Mm has the maximum number of unpaired electrons present in the d-subshell (5 electrons). Hence. Mn exhibits the largest number of oxidation states, ranging from +2 to +7.

Question 4. The E0(M2+/M) value for copper is positive (+0.34V). What is possibly the reason for this? (Hint: consider its high energy and low energy)
Answer: The E0(M2+/M) value of a metal depends on the energy changes involved in the following:

1. Sublimation: The energy required for converting one mole of an atom from the solid state to the gaseous state.

M(s) → M(g) ΔsH(Sublimation energy)

2. Ionization: The energy required to take out electrons from one mole of atoms in the gaseous state.

M(g) → M2+(g) ΔiH(Ionization energy)

3. Hydration: The energy released when one mole of ions is hydrated.

M2+(g)→ M2+(aq) ΔhydH(Hydration energy)

Now. copper has a high energy of atomization and low hydration energy.

Hence, the Eθ(M2+/M) value for copper is positive.

Read and Learn More Class 12 Chemistry with Answers Chapter Wise

Question 5. How would you account for the irregular variation of ionization enthalpies (first and second) in the first series of (the transition elements?
Answer: Ionization enthalpies are found to increase in the given series due to a continuous filling of the inner d-orbitals.

  • The irregular variations of ionization enthalpies can be attributed to the extra stability of configurations such as d0, d5, and d10. Since these states arc exceptionally stable, their ionization enthalpies are very high.
  • In the case of first ionization energy. Cr has low ionization energy. This is because after losing one electron, it attains the stable configuration (3d5 ). On the other hand,  Zn has exceptionally high first ionization energy as an electron has to be removed from stable and fully-filled orbitals (3d10 4s2).
  • The second ionization energies are higher than the first since it becomes difficult to remove an electron when an electron has already been taken out.
  • Also, elements like Cr and Cu have exceptionally high second ionization energies as after losing the first electron, they have attained the stable configuration (Cr+:3d5 and Cu+:3d10). Hence, taking out one electron more from this stable configuration will require a lot of energy.

CBSE Class 12 Chemistry Chapter 4 The d And f Block Elements Important Question And Answers

Question 6. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Answer: Both oxide and fluoride ions are highly electronegative and have a very small size. Due to these properties, they can oxidize the metal to its highest oxidation state.

Question 7. Which is a stronger reducing agent Cr2+ or Fe2+ and why?
Answer: The following reactions are involved when Cr2+ and Fe2+ act as reducing agents.

→ Cr Cr2+ 3+ Fe2+ → Fe3+

The value is -0.41 V and is +0.77 V. This means that Cr2+ can be easily oxidized to Cr3+, but Fe does not get oxidized to Fe3+ easily. Therefore. Cr2+ is a better-reducing agent than Fe3+.

Question 8. Calculate the ‘spin only’ magnetic moment of M2+(aq), ion (Z = 27).
Answer:

Z = 27

⇒ [ Ar] 3d7 4s2

M2+ = [Ar] 3d7

D And F Block 3 Unpaired Electrons

i.e., 3 unpaired electrons

.’. n = 3

⇒ \(\sqrt{n(n+2)}=\mu\)

⇒ \(\sqrt{3(3+2)}=\mu\)

⇒ \(\sqrt{15}=\mu\)

⇒ \(\mu \approx 4 B M\)

Question 9. Explain why the Cu+ ion is not stable in aqueous solutions.
Answer: In an aqueous medium, Cu2+ is more stable than Cu+. This is because although energy is required to remove one electron from Cu+ to Cu2+, the high hydration energy of Cu2+ compensates for it. Therefore. Cu+ ions in an aqueous solution are unstable. It is disproportionate to give Cu2+ and Cu.

Question 10. Actinoid contraction is greater from element to element than lanthanide contraction. Why?
Answer: In actinoids, 5f orbitals are filled. These 5f orbitals have a poorer shielding effect than 4f orbitals (in lanthanides).

Thus, the effective nuclear charge experienced by electrons in valence shells in the case of actinides is much more than that experienced by lanthanides. Hence, the size contraction in actinides is greater as compared to that in lanthanides.

Question 11. Write down the electronic configuration of:

  1. Cr3+
  2. Pm3+
  3. Cu+
  4. Ce4+
  5. Co2+
  6. Lu2+
  7. Mn2+
  8. Th4+

Answer:

  1. Cr3+: 1s2 2s2 2p6 3s2 3p6 3d3 Or, [AT]18 3d3
  2. Pm3+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f4 Or, [Xe]54 4f4
  3. Cu+ : 1s2 2s2 2p6 3s2 3p6 3d10 Or, [Ar]18 3d10
  4. Ce4+ : 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4P6 4d10 5s2 5p6 Or, [Xe]54.
  5. Co2+: 1s2 2s2 2p6 3s2 3p6 3d7 Or, [Ar ]18 3d7
  6. Lu2+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f14 5d1 Or, [Xe]54 4f14 5d1
  7. Mn2+: 1s2 2s2 2p6 3s2 3p6 3d5 Or, [Ar]18 3d5
  8. Th4+:  1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d10 6s2 6p6 Or, [Rn]86

Question 12. Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state?
Answer: The electronic configuration of Mn2+ is [Ar]18 3d5. (half-filled stability) The electronic configuration of Fe2+ is [Ar]18 3d6. (After losing one electron gain half-filled stability).

Question 13. Explain briefly how the +2 state becomes more and more stable in the first half of the first-row transition elements with increasing atomic number.
Answer: The oxidation states displayed by the first half of the first row of the transition metals arc are given in the table below.

D And F Block Oxidation State

It can be easily observed that except Sc, all other metals display a +2 oxidation state. Also, on moving from Sc to Mn, the atomic number increases from 21 to 25. This means the number of electrons in the 3d-orbital also increases from 1 to 5.

  1. Sc (+2) = d1
  2. Ti (+2) = d2
  3. V (+2) = d3
  4. Cr (+2) = d4
  5. Mn (+2) = d5

+2 oxidation state is attained by the loss of the two 4s electrons by these metals. Since the number of d electrons in the (+2) state also increases from Tit+2) to Mn(+2), the stability of+2 state increases (as d-orbital is becoming more and more half-filled). Mn(+2) has d electrons (that is a half-filled d shell, which is highly stable).

Question 14. To what extent do the electronic configurations decide the stability of oxidation states in the first series of transition elements? Illustrate your answer with examples.
Answer: The elements in the first – half of the transition series exhibit many oxidation states with Mn exhibiting a maximum number of oxidation states (+2 to +7). The stability of the +2 oxidation state increases with the increase in atomic number. This happens as more electrons are getting filled in the d-orbital.

Question 15. What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: 3d3, 3d5, 3d8 and 3d4?
Answer:

D And F Block Electron Configurations In The Ground State Of Their Atoms

Question 16. Name the oximeter anions of (the first series of transition metals in which the metal exhibits the oxidation state equal to its group number.
Answer:

  1. Vanadate, VO3 → The oxidation state of V is +5.
  2. Chromate. CrO2-4 → The oxidation state of Cr is +6
  3. Permanganate, MnO4 → Oxidation state of Mn is +7.

Question 17. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
Answer: As we move along the lanthanoid series, the atomic number increases gradually by one.

  • This means that the number of electrons and protons present in an atom also increases by one. As electrons are added to the same shell, the effective nuclear charge increases.
  • This happens because the increase in nuclear attraction due to the addition of a proton is more pronounced than the increase in the inter-electronic repulsions due to the addition of an electron.
  • Also, with the increase in atomic number, the number of electrons in the 4f orbital increases. The 4f electrons have a poor shielding effect. Therefore, the effective nuclear charge experienced by the outer electrons increases.
  • Consequently, the attraction of the nucleus for the outermost electrons increases. This results in a steady decrease in the size of lanthanoids with an increase in the atomic number. This is termed as lanthanoid contraction.

Consequences of lanthanoid contraction

  1. There is a similarity in the properties of the second and third transition series.
  2. Separation of lanthanoids is possible due to lanthanide contraction.
  3. It is due to lanthanide contraction that there is variation in the basic strength of lanthanide hydroxides. (Basic strength decreases from La (OH)3 to Lu (OH)3)

Question 18. What are the characteristics of the transition elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?
Answer:

Transition elements are those elements in which the atoms or ions (in a stable oxidation state) contain partially filled d-orbital. These elements lie in the d-block and show a transition of properties between the s-block and the p-block.

Therefore, these are called transition elements. Elements such as Zn, Cd and Hg cannot be classified as transition elements because these have filled the d-subshell.

Question 19. In what way is the electronic configuration of the transition elements different from that of the non-transition elements?
Answer:

Transition metals have a partially filled d-orbital. Therefore, the electronic configuration of transition elements is (n — 1 ) d1-10 ns0-2. The non-transition elements either do not have a d-orbital or have a filled d-orbital. therefore, the electronic configuration of non-transition elements is ns1-2 or ns2 np1-16.

Question 20. What are the different oxidation states exhibited by the lanthanoids?
Answer: In the lanthanide series. +3 oxidation stale is most common i.e., Ln (3) compounds are predominant. However. +2 and +4 oxidation states can also be found in the solution or in solid compounds.

Question 21. Explain giving reasons:

  1. Transition metals and many of their compounds show paramagnetic behaviour.
  2. The enthalpies of atomisation of the transition metals are high.
  3. The transition metals generally form coloured compounds.
  4. Transition metals and their many compounds act as good catalysts.

Answer:

  1. Transition metals show paramagnetic behaviour. Paramagnetism arises due to the presence of unpaired electrons with each electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum.
  2. Transition elements have a high effective nuclear charge and a large number of valence electrons. Therefore, they form very strong metallic bonds. As a result, the enthalpy of atomization of transition metals is high.
  3. Most of the complexes of transition metals are coloured. This is because of the absorption of radiation from the visible light region to promote an electron from one of the d-orbitals to another. In the presence of ligands, the d-orbitals split up into two sets of orbitals having different energies.
    1. Therefore, the transition of electrons can take place from one set to another. The energy required for these transitions is quite small and falls in the visible region of radiation.
    2. The ions of transition metals absorb the radiation of a particular wavelength and the rest is reflected, imparting colour to the solution.
  4. The catalytic activity of the transition elements can be explained by two basic facts.
    1. Owing to their ability to show variable oxidation states and form complexes, transition metals form unstable intermediate compounds. Thus, they provide a new path with lower activation energy, Ea, for the reaction.
    2. Transition metals also provide a suitable surface for the reactions to occur.

Question 22. What are interstitial compounds? Why are such compounds well-known for transition metals?
Answer:

Transition metals are large and contain lots of interstitial sites. Transition elements can trap atoms of other elements (that have small atomic sizes), such as H, C, and N, in the interstitial sites of their crystal lattices. The resulting compounds are called interstitial compounds.

Question 23. How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.
Answer:

In transition elements, the oxidation state can vary from +1 to the highest oxidation state by removing all its valence electrons. Also, in transition elements, the oxidation stales differ by

  1. (Fe2+ and Fe3+: Cu+ and Cu2+). In non-transition elements, the oxidation states differ by
  2. For example. +2 and +4 or + 3 and +5. etc.

Question 24. Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?
Answer:

Potassium dichromate is prepared from chromite ore (FeCr2O4) in the following steps.

Step (1): Preparation of sodium chromate

⇒ \(4 \mathrm{FeCrO}_4+16 \mathrm{NaOH}+7 \mathrm{O}_2 \longrightarrow 8 \mathrm{Na}_2 \mathrm{CrO}_4+2 \mathrm{FeO}_2+8 \mathrm{H}_2 \mathrm{O}\)

Step (2): Conversion of sodium chromate into sodium dichromate

⇒ \(2 \mathrm{NaCrO}_4+\text { conc. } \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{Na}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}\)

Step (3): Conversion of sodium dichromate to potassium dichromate

⇒ \(\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7+2 \mathrm{KCl} \longrightarrow \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+2 \mathrm{NaCl}\)

Potassium dichromate being less soluble than sodium dichromate is obtained in the form of
orange-coloured crystals and can be removed by filtration.

The dichromate ion ( Cr2O2-4 ) exists in equilibrium with the chromate ( Cr2O24 ) ion at pH 4.

However, by changing the pH, they can be interconverted.

⇒ \(\underset{\text { Chromate ion(Yellow) }}{2 \mathrm{CrO}_4^{2-}}+2 \mathrm{H}^{+} \stackrel{\mathrm{PH}^{-4}}{\longrightarrow} \underset{\text { Dichromate ion(Orange) }}{\mathrm{Cr}_2 \mathrm{O}_7^{2-}}+\mathrm{H}_2 \mathrm{O}\)

On increasing pH, \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+2 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{CrO}_4^{2-}+\mathrm{H}_2 \mathrm{O}\)

Question 25. Describe the oxidation action of potassium dichromate and write the ionic equations for its reaction with:

  1. Iodide
  2. Iron(2) solution and
  3. H2S

Answer:

K2Cr2O7 acts as a very strong oxidising agent in the acidic medium.

⇒ \(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+4 \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{K}_2 \mathrm{SO}_4+\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+4 \mathrm{H}_2 \mathrm{O}+3[\mathrm{O}]\)

K2Cr2O7 takes up electrons to gel reduced and acts as an oxidising agent. The reaction of K Cr 0_ with iodide ion, iron (2) solution and H S are given below.

1. K2Cr2O7 oxidizes iodide to iodine.

⇒ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}\)

2I → I2 + 2e × 3

⇒ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+6 \mathrm{I}^{-}+14 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{I}2+7 \mathrm{H}_2 \mathrm{O}\)

K2Cr2O7 oxidizes iron (2) solution to iron (3) solution i.e., ferrous ions to ferric ions.

⇒ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}\)

Fe2+ →  Fe3+ + e × 6

⇒ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{Fe}^{2+} \longrightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{I}_2+7 \mathrm{H}_2 \mathrm{O}\)

K2Cr2O7 oxidizes H2S to sulphur.

⇒ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}\)

H2S →  S + 2H+ + 2e × 3

⇒ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+3 \mathrm{H}_2 \mathrm{~S}+8 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{~S}+7 \mathrm{H}_2 \mathrm{O}\)

Question 26. Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with

  1. Iron(2) ions
  2. SO2
  3. Oxalic acid?

Write the ionic equations for the reactions.

Answer:

Potassium permanganate can be prepared from pyrolusite (MnO2). The ore is fused with KOH in the presence of either atmospheric oxygen or an oxidising agent, such as KNO3 or KCIO4 to give K2MnO4.

D And F Block Atmospheric Oxygen Or An Oxidising Agent

The green mass can be extracted with water and then oxidized either electrolytically or by passing chlorine/ozone into the solution.

Electrolytic oxidation

⇒ \(\mathrm{K}_2 \mathrm{MnO}_4 \longleftrightarrow 2 \mathrm{~K}^{+}+\mathrm{MnO}_4^{2-}\)

⇒ \(\mathrm{H}_2 \mathrm{O} \longleftrightarrow \mathrm{H}^{+}+\mathrm{OH}^{-}\)

At the anode, manganate ions are oxidized to permanganate ions.

⇒ \(\underset{\text { Green }}{\mathrm{MnO}_4^{2-}} \longleftrightarrow \underset{\text { Purple }}{\mathrm{MnO}_4^{-}}+\mathrm{e}^{-}\)

Oxidation by chlorine

⇒ \(2 \mathrm{~K}_2 \mathrm{MnO}_4+\mathrm{Cl}_2 \longrightarrow 2 \mathrm{KMnO}_4+2 \mathrm{KCl}\)

⇒ \(2 \mathrm{MnO}_4^{2-}+\mathrm{Cl}_2 \longrightarrow 2 \mathrm{MnO}_4^{-}+2 \mathrm{Cl}\)

Oxidation by ozone

⇒ \(2 \mathrm{~K}_2 \mathrm{MnO}_4+\mathrm{O}_3+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{KMnO}_4+2 \mathrm{KOH}+\mathrm{O}_2\)

⇒ \(2 \mathrm{MnO}_4^2+\mathrm{O}_3+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{MnO}_4^{2-}+2 \mathrm{OH}+\mathrm{O}_2\)

Acidified KMnO4 solution oxidizes Fe (2) ions to Fe (3) ions i.e., ferrous to ferric ions.

⇒ \(2 \mathrm{MnO}_4^2+\mathrm{O}_3+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{MnO}_4^{2-}+2 \mathrm{OH}+\mathrm{O}_2\)

Fe2+ → Fe3+ + e × 5

⇒ \(\mathrm{MnO}_{+}^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}2 \mathrm{O}\)

Acidified potassium permanganate oxidizes SO2 to sulphuric acid.

⇒ \(\left.\mathrm{MnO}_4^{-}+6 \mathrm{H}^{+}+5 \mathrm{e} \longrightarrow \mathrm{Mn}^{2+}+3 \mathrm{H}_2 \mathrm{O}\right] \times 2\)

⇒ \(\left.2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{SO}_2+\mathrm{O}_2 \longrightarrow 4 \mathrm{H}^{+}+2 \mathrm{SO}_4^{2-}+2 \mathrm{e}\right] \times 5\)

⇒ \(2 \mathrm{MnO}_4^{-}+10 \mathrm{SO}_2+5 \mathrm{O}_2+4 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{SO}_4^{2-}+8 \mathrm{H}^{+}\)

Acidified potassium permanganate oxidizes oxalic acid to carbon dioxide.

⇒ \(\mathrm{MnO}_4+8 \mathrm{H}^{+}+5 \mathrm{e} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} \times 2\)

⇒ \(\left.\mathrm{C}_2 \mathrm{O}_4^2- \longrightarrow 2 \mathrm{CO}_2+2 \mathrm{c}\right] \times 5\)

⇒ \(2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_7^{2+}+16 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}\)

Question 27. For M2+/M and M3+/M2++ systems, the E values for some metals are as follows:

  1. Cr2+ /Cr – 0.9 V
  2. Cr3/Cr2+ – 0.4 V
  3. Mn2+/Mn -1.2V
  4. Mn3+/Mn2+ +1.5V
  5. Fe2+/Fe – 0.4 V
  6. Fe3+/Fe2+ + 0.8V

Use this data to comment upon:

  1. The stability of Fe3+ in acid solution as compared to that of Cr3+ and Mn3+ and
  2. The ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

Answer:

  • The E value for Fe3+/Fe2+ is higher than that for Cr3+ /Cr2+ and lower than that for Mn3+/Mn2+ So. the reduction of Fe3+ to Fe2+ is easier than the reduction of Cr3+ to Cr2+ but not as easy as the reduction of Mn3+ to Mn2+.
  • Hence, Fe3+ is more stable than Mn3+, but less stable than Cr3+. These metal ions can be arranged in the increasing order of their stability as Mn3+< Fe3+ < Cr3+

The reduction potentials for the given pairs increase in the following order.

Mu2+ /Mn < Cr2+ /Cr < Fe2+/Fe

So, the oxidation of Fe to Fe2+ is not as easy as the oxidation of Cr to Cr2+ and the oxidation of Mn to Mn2+. Thus these metals can be arranged in the increasing order of their ability to get oxidised: Fe < Cr < Mn

Question 28. Predict which of the following will be coloured in aqueous solution. Ti3+, V3+, Cu+, Sc3+, Mn2+, Fe3+ and Co2+. Give reasons for each.
Answer: Only the ions that have unpaired electrons in the el-orbital will be coloured. The ions in which the d orbital is empty or filled will be colourless.

D And F Block Only The Ions That Have Unpaired Electrons In D Orbital Will Be Coloured

Question 29. Compare the stability of the +2 oxidation state for the elements of the first transition series.
Answer:

D And F Block The Elements Of The First Transition Series

The relative stability of the +2 oxidation state increases on moving from top to bottom. This is because on moving from top to bottom, it becomes more and more difficult to remove the third electron from the d-orbital.

Question 30. Compare the chemistry of actinoids with that of the lanthanoids with special reference to:

  1. Electronic configuration
  2. Atomic and ionic sizes
  3. Oxidation state
  4. Chemical reactivity.

Answer:

  1. Electronic configuration: The general electronic configuration for lanthanoids is [Xe]54 4f1-145d0-1 6s2 and that for actinoids is [Rn]86 5f1-14 6d0-1 7s2
  2. Atomic and Ionic sizes: Similar to lanthanoids, actinoids also exhibit actinoid contraction (overall decrease in atomic and ionic radii). The contraction is greater due to the poor shielding effect of 5f orbitals.
  3. Oxidation states: The principal oxidation state of lanthanoids is (+3). However, sometimes we also encounter oxidation states of +2 and +4. This is because of the extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d, and 7s levels are of comparable energies.
  4. Chemical reactivity: In the lanthanide series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start behaving similarly to Al. Actinoids, on the other hand- are highly reactive metals, especially when they are finely divided.

Question 31. How would you account for the following:

  1. Of the d4 species, Cr2+ is strongly reducing while manganese(3) is strongly oxidising.
  2. Cobalt(2) is stable in an aqueous solution but in the presence of complexing reagents, it is easily oxidised.
  3. The d1 configuration is very unstable in ions.

Answer:

  1. Cr2+ is strongly reducing in nature. It has a d4 configuration. While acting as a reducing agent, it gets oxidized to Cr3+ (electronic configuration, d3 ). This d3 configuration can be written as a t32g configuration, which is a more stable configuration. In the case of Mn3+ (d4), It acts as an oxidizing agent and gets reduced to Mn (d5). This has an exactly half-filled d-orbital and is highly stable.
  2. Co(2) is stable in aqueous solutions. However, in the presence of strong field complexing reagents, it is oxidized to Co(3). Although the 3rd ionization energy for Co is high, the higher amount of crystal field stabilization energy (CFSE) released in the presence of strong-field ligands overcomes this ionization energy.
  3. The ions in the d1 configuration tend to lose one more electron to get into the stable d0 configuration. Also, the hydration or lattice energy is more than sufficient to remove the only electron present in the cl -orbital of these ions. Therefore, they act as reducing agents.

Question 32. What is meant by ‘disproportionation’? Give two examples of disproportionation reactions in an aqueous solution.
Answer:

It is found that sometimes a relatively less stable oxidation state undergoes an oxidation-reduction reaction in which it is simultaneously oxidised and reduced. This is called disproportionation.

For example,

⇒ \(\underset{\mathrm{Cr}(\mathrm{5})}{3 \mathrm{CrO}_4^{3-}}+8 \mathrm{H}^{+} \longrightarrow \underset{\mathrm{Cr}(\mathrm{6})}{2 \mathrm{CrO}_{+}^{2-}}+\underset{\mathrm{Cr}(\mathrm{3})}{\mathrm{Cr}^{3+}}+4 \mathrm{H}_2 \mathrm{O}\)

Cr(5) is oxidized to Cr(6) and reduced to Crd(3)

⇒ \(\underset{\mathrm{Mn}(\mathrm{6})}{3 \mathrm{MnO}_4^{2-}}+4 \mathrm{H}^{+} \longrightarrow \underset{\mathrm{Mn}(\mathrm{7})}{2 \mathrm{MnO}_4^{-}}+\underset{\mathrm{Mn}(\mathrm{4})}{\mathrm{MnO}_2}+2 \mathrm{H}_2 \mathrm{O}\)

Mn (6) is oxidized to Mn (7) and reduced to Mn (4).

Question 33. Which metal in the first series of transition metals exhibits a +1 oxidation state most frequently and why?
Answer: In the first transition series, Cu exhibits a + 1 oxidation state very frequently. It is because Cu (+1) has an electronic configuration of [Ar] 3d10. The filled el-orbital makes it highly stable.

Question 34. Calculate the number of unpaired electrons in the following gaseous ions: Mn3+ \ Cr3+ \ V3+ and Ti3+. Which one of these is the most stable in aqueous solution?
Answer:

D And F Block The Most Stable In Aqueous Solutions Owing To Half Filled Configuraion

Cr3+ is the most stable in aqueous solutions owing to a t32g half-filled configuration.

Question 35. Give examples and suggest reasons for the following features of the transition metal chemistry:

  1. The lowest oxide of transition metal is basic, (and the highest is amphoteric/acidic.
  2. A transition metal exhibits the highest oxidation state in oxides and fluorides.
  3. The highest oxidation state is exhibited in oxoanions of a metal.

Answer:

  1. In the ease of a lower oxide of a transition metal, the metal atom has a low oxidation stale. This means that some of the valence electrons of the metal atom are not involved in bonding. As a result, it can donate electrons and behave as a base.
    1. On the other hand, in the case of a higher oxide of a transition metal, the metal atom has a high oxidation state. This means that the valence electrons are involved in bonding and so. they are unavailable. There is also a highly effective nuclear charge. As a result, it can accept electrons and behave as an acid.
    2. For example. Mn2O is basic and Mm27O7 is acidic.
  2. Oxygen and fluorine act as strong oxidising agents because of their high electronegatives and small sizes. Hence, they bring out the highest oxidation states from the transition metals.
    1. In other words, a transition metal exhibits higher oxidation states in oxides and fluorides. For example, in OsF6 and V2O5 the oxidation states of Os and V are +6 and +5 respectively.
  3. Oxygen is a strong oxidising agent due to its high electronegativity and small size. So. oxo-anions of metal have the highest oxidation state. For example, in MnO4 the oxidation state of Mn is +7.

Question 36. What are alloys? Name an important alloy which contains some of the lanthanoid metals Mention its uses.
Answer:

  • An alloy is a solid of two or more elements in a metallic matrix. It can either be a partial solid solution or a complete solid solution. Alloys are usually found to possess different physical properties than those of the component elements.
  • An important alloy of lanthanoids is Mischmetal. It contains lanthanoids (94-95%), iron (5%), and traces of S, C. Si. Ca and Al.

Uses:

  1. Mischmetal is used in cigarettes and gas lighters.
  2. It is used in flame-throwing tanks.
  3. It is used in tracer bullets and shells.

Question 37. What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements: 29, 59, 74, 95, 102, 104.
Answer:

Inner transition elements are those elements in which the last electron enters the f-orbital. The elements in which the 4f and 5f orbitals are progressively filled are called f-block elements. Among the given atomic numbers, the atomic numbers of the inner transition elements are 59, 95 and 102.

Question 38. The chemistry of the actinoid elements is not as smooth as that of the Lanthanoids. Justify this statement by giving some examples of the oxidation state of these elements.
Answer:

  • Lanthanoids primarily show three oxidation states (+2, +3, +4). Among these oxidation states. +3 state is the most common. Lanthanoids display a limited number of oxidation states because the energy difference between 4f, 5d and 6s orbitals is quite large.
  • On the other hand, the energy difference between 5f, 6d, and 7s orbitals is very low. Hence aclinoids display a large number of oxidation states. For example, uranium and plutonium display +3. +4. +5. and +7. The most common oxidation state in the case of retinoids is also +3.

Question 39. Which are the last elements in the series of actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.
Answer:

The last element in the actinoid series is lawrencium. Lr. Its atomic number is 103 and its electronic configuration is [Rn] 5f14 6d1 7s2. The most common oxidation state displayed by it is +3; because after losing 3 electrons it attains a stable f14 configuration.

Question 40. Use Hund’s rule to derive the electronic configuration of the Ce3+ ion and calculate its magnetic moment based on the ‘spin-only’ formula.
Answer:

Ce3+:1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f1

Magnetic moment can be calculated as:

⇒ \(\mu=\sqrt{n(n+2)}\)

Where, n = number of unpaired electrons.

In Ce3+, n = 1

Therefore, \(\mu=\sqrt{1(1+2)}=\sqrt{3}=1.732 \mathrm{BM}\)

Question 41. Name the members of the lanthanoid series which exhibit a +4 oxidation state and those which exhibit a +2 oxidation state. Try to correlate this type of behaviour with the electronic configuration of these elements.
Answer:

The lanthanoids that exhibit + 2 and +4 states are shown in the given table. The atomic numbers of these elements are given in parentheses.

D And F Block The Atomic Number Of These Elements Are Given In The Parenthesis

  1. Ce after forming Ce4+ attains a stable electronic configuration of [Xe].
  2. Tb after forming Tb4+ attains a stable electronic configuration of [Xe] 4f7
  3. Eu after forming Eu2+ attains a stable electronic configuration of [Xe] 4f7
  4. Yb after forming Yb2+ attains a stable electronic configuration of [Xe] 4f14

Question 42. Write the electronic configuration of the elements with the atomic numbers 61.91. 101. and 109.
Answer:

D And F Block The Electronic Configuration Of The Elements With The Atomic Numbers

Question 43. Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:

  1. Electronic configurations.
  2. Oxidation states,
  3. Ionisation enthalpies, and
  4. Atomic sizes.

Answer:

1. In the 1st, 2nd and 3rd transition series, the 3d, 4d and 5d orbitals are respectively filled.

We know that elements in the same vertical column generally have similar electronic configurations.

In the first transition series, two elements show unusual electronic configurations:

  1. Cr(24) = 3d5 4s1
  2. Cu(29) = 3d10 4s1

Similarly, there are exceptions in the second transition series. These are:

  1. Nb(41) = 4d4 5s1
  2. Mo(42) = 4d5 5s1
  3. Tc (43) = 4d6 5s1
  4. Ru(44) = 4d7 5s1
  5. Rh(45) = 4d8 5 s1
  6. Pd(46) = 4d10 5s0
  7. Ag(47) = 4d10 5s1

There are some exceptions in the third transition series as well. These are:

  1. Pt(78) = 5d9 6s1
  2. Au(79) = 5d10 6s1

As a result of these exceptions, it happens many times that the electronic configurations of the elements present in the same group are dissimilar.

2. In each of the three transition series the number of oxidation states shown by the elements is the maximum in the middle and the minimum at the extreme ends.

  • However, +2 and +3 oxidation states are quite stable for all elements present in the first transition series. All metals present in the first transition series form stable compounds in the +2 and +3 oxidation states.
  • The stability of the +2 and +3 oxidation states decreases in the second and third transition series, wherein higher oxidation states are more important.

For example \(\left[\begin{array}{l}
\mathrm{2} \\
\mathrm{Fe}(\mathrm{CN})_6
\end{array}\right]^{4-} \cdot\left[\begin{array}{l}
\mathrm{3} \\
\mathrm{Co}\left(\mathrm{NH}_3\right)_6
\end{array}\right]^{3+} \cdot\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^3\) are stable complexes,

but no such complexes are known for the second and third transition series such as Mo, W, Rh, and In. They form complexes in which their oxidation states are high. For example WCI6, ReF7, RuO4 etc.

3. In each of the transition series, the first ionisation enthalpy increases from left to right. However, there are some exceptions.

  • The first ionisation enthalpies of the third transition series are higher than those of the first and second transition series. This occurs due to the poor shielding effect of 4f electrons in the third transition series.
  • Certain elements in the second transition series have higher first ionisation enthalpies than elements corresponding to the same vertical column in the first transition series.
  • There are also elements in the 2′ transition series whose first ionisation enthalpies are lower than those of (the elements corresponding to the same vertical column in the 1 transition series.

4. Atomic si/.e generally decreases from left to right across a period. Now. among the three transition series, the atomic sizes of the elements in the second transition series are greater than those of the element corresponding to the same vertical column in the first transition series.

However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series. This is due to lanthanoid contraction.

Question 44. Write down the number of 3d electrons in each of the following ions: Ti2+, V2+, Cr3+, Mn2+, Fe2+, Fe3+, Co2+, Ni 2+ and Cu2+ Indicate how would you expect the live 3d orbitals to be occupied for these hydrated ions (octahedral).
Answer:

D And F Block The Five 3d Orbitals To Be Occupied For These Hydrated Ions

Question 45. What can be inferred from the magnetic moment values of the following complex species?

D And F Block Inferred From The Magnetic Moment Values

Answer:

Magnetic moment (μ) is given as μ = \(\sqrt{n(n+2)} B M\)

  1. For value n = 1, μ = \(\sqrt{1(1+2)}=\sqrt{3}=1.732 \mathrm{BM}\)
  2. For value n = 2, μ = \(\sqrt{2(2+2)}=\sqrt{8}=2.83 \mathrm{BM}\)
  3. For value n = 3, μ = \(\sqrt{3(3+2)}=\sqrt{15}=3.87 \mathrm{BM}\)
  4. For value n = 4, μ = \(\sqrt{4(4+2)}=\sqrt{24}=4.899 \mathrm{BM}\)
  5. For value n = 5, μ = \(\sqrt{5(5+2)}=\sqrt{35}=5.92 \mathrm{BM}\)

1. K4[Mn(CN)6]

For transition metals, the magnetic moment is calculated from the spin-only formula. Therefore.

⇒ \(\mu=\sqrt{n(n+2)}=2.2 \mathrm{BM}\)

We can see from the above calculation that the given value is closest to n = 1. Also, in this complex. Mn in in the 4-2 oxidation state. This means that Mn has 5 electrons in the d-orbital.

Hence, we can say that CN is a strong field ligand that causes the pairing of electrons.

2. [Fe(H2O)6]2+

⇒ \(\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}=5.3 \mathrm{BM}\)

We can see from the above calculation that the given value is closest to n = 4. Also, in this complex. Fe is in the 4-2 oxidation state. This means that Fe has 6 electrons in the d-orbital. Hence, we can say that H2O is a weak field ligand and does not cause the pairing of electrons.

3. K2[MnCI4]

⇒ \(\mu=\sqrt{n(n+2)}=5.9 B M\)

  • We can see from the above calculation that the given value is closest to n = 5. Also, in this complex, Mn is in the 4-2 oxidation state. This means that Mn has 5 electrons in the d-orbital.
  • Hence, we can say that Cl is a weak field ligand and does not cause the pairing of electrons.