Sainavle

Quantum Number

Quantum Number Definition: A set of four numbers that provide complete information about any electron in an atom are known as quantum numbers.

Quantum Number Classification: The four quantum numbers are—

1. Principal Quantum Number (N)
2. Azimuthal Or Subsidiary quantum number (l)
3. Magnetic quantum number (m or m1)
4. Spin quantum number (5 or mg ).
5. To specify an electron in an atom, the following four quantum numbers should be mentioned.
6. Principal quantum number [n]

Quantum Number Origin:

1. From Bohr’s postulates, it is known that each electronic orbit surrounding the nucleus in an atom represents an energy level.
2. The average energy of the electrons revolving in a particular orbit is fixed. So, these orbits are called principal energy levels or principal quantum levels.
3. Depending on their distance from the nucleus, these orbits or principal energy levels are designated by the numbers 1,2,3, 4… etc. These numbers 1,2,3,4… etc. are called principal quantum numbers.

Quantum Number Designation: The principal quantum number is denoted by the letter ‘n ’. For AT-shell n = 1, for L -shell n = 2, for Mshell n = 3 and so on.

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Information obtained:

1. The higher the value of n, the greater the distance of the orbit from the nucleus, and hence, the greater the size ofthe orbit. Thus, r1<r2<r3< r4< …
2. The higher the value of ‘ n,’ the greater will be the electronic energy associated with the orbit.
3. Thus, El<E2<E3<E4<………..
4. A maximum number of electrons that can be accommodated in a principal quantum level n is given by the formula 2n2.
5. Limitations of 2β2 The maximum number of electrons in any orbit can never be more than 32 even if the value of n exceeds 4.
6. The outermost electronic shell does not contain more than 8 electrons.
7. The penultimate shell (i,e., the shell just preceding the outermost shell) does not contain more than 18 electrons.

Azimuthal or subsidiary quantum number

Azimuthal or subsidiary quantum number Origin: A spectrograph with high resolving [/]power has revealed that each bright line in the spectrum of atomic hydrogen consists of some closely spaced finer lines.

This fact suggests that each orbit or energy level in an atom is composed of subshells. Electrons occupying these subshells within the same -shell, exhibit slight differences in energy.

In order to explain the formation of the fine structure of spectral lines, Sommerfeld proposed
the existence of elliptical orbits, besides Bohr’s circular
orbits.

To specify the shape of the elliptical orbit, another supplementary quantum number is necessary.

This supplementary quantum number which indicates the captivity of the electronic orbit is called azimuthal or subsidiary quantum number denoted by the letter.

If the principal quantum number of any orbit is n, then the total number of subshells incorporated in that orbit will also be n.

Magnitude:

1. As per quantum mechanical calculations, the angular momentum of a moving electron in an elliptical path is given by, L = Jl(l+ 1) X.
2. This is often called orbital angular momentum.
3. The value of l determines the shape of the path. So, with the help of the principal quantum number and azimuthal quantum number, a precise idea about the size and shape of the electronic path can be obtained.
4. If n stands for the principal quantum number of an electronic orbit, the values of l will be from to (n- 1) i.e., with respect to the value of principal quantum number n, the azimuthal quantum number / may assume n number of different values including zero, e.g., for n = 4, 1=0, 1, 2 and 3.
5. To indicate the subshells within a shell, spectroscopic symbols are used instead ofthe numbers 0, 1, 2, 3 etc.
6. The symbols s, p, d,f, etc., (spectroscopic coinage) are merely the first letters ofthe words sharp, principal, diffuse, and fundamental, used extensively in spectral analysis.
7. To express the position of an electron in the atom, the principal quantum number should be written first followed by the symbol of the azimuthal quantum number to its right side, e.g., the subshells included in K, L, M, and N-shells are represented as

• The ratio of the major axis to the minor axis of an elliptical path is given by = (/ + 1)/n .
• An elliptical path for which l = (rc- 1), becomes circular e.g., in the case of 4-orbit if 1 = 3, then that orbit becomes circular. The greater the difference between the values of l and n, the larger the ellipticity of that path.
• The penetrating power and screening effect of an elliptical orbit increases on increasing the ellipticity of the orbit.
• So the penetrating and screening powers of different subshells within the same shell follow the sequence: s> p> d>f.
• Due to the difference in the internal energies of the subshells [s, p, d, f, etc.), the electrons moving in those subshells also possess different energies. Energy associated with the subshells in a particular orbit increases in the following order: s <p<d<f.

Magnetic quantum number (m or mt)

Origin:

1. Zeeman in 1896 observed that each fine line in atomic spectra splits further into finer lines in the presence of the highly powerful magnetic field.
2. In the absence of a magnetic field, such finer splitting i.e., hyperfine splitting disappears. This phenomenon is called the Zeeman effect. To explain the Zeeman effect, a third type of quantum number, known as a magnetic quantum number was introduced.

Discussion:

1. Due to the angular motion of electrons around the nucleus, a magnetic field is produced, which interacts with the external magnetic field.
2. As a result subshells of definite energy split into three-dimensional spatial regions called orbitals.
3. Magnetic quantum number (MI) signifies the orientation of the orbitals in space in which the electron exists.
4. The value of m depends on the azimuthal quantum number l.
5. For a certain value of l, m has an o total of (2Z +1 ) different values. These values may be any whole number starting from -Z to +1 (including zero).
6. For s- subshell, l = 1 and m – 1. This subshell, l = 0 and m – 0. This orbital (i.e., s-orbital). Z = 1 denotes p -subshell consisting of three orbitals which are directed along three axes.
7. These are marked as px, py, and pz orbitals which have the respective values of m = -1, 0, and + 1 . Similarly, d and /-subshells contain 5 and 7 orbitals respectively.
8. The negative values of the magnetic quantum number signify that these orbitals are inclined in the direction opposite to the magnetic field and the positive values indicate that these orbitals are inclined in the direction ofthe magnetic field.
9. shows the different directions of the d -d-subshell (Z = 2) in the magnetic field.

Orientation of different orbitals of ZV-shell (n = 4] under the influence of magnetic field.

Values of magnetic quantum number (m] for different values of azimuthal quantum number [l]

Spin quantum number [s or ms]

Uhlenbeck and Goudsmit introduced a fourth quantum number called the spin quantum number.

This is because the other three quantum numbers were not able to give sufficient explanation to the hyperfine structure of the atomic spectra.

% Just like the earth, an electron while moving around the nucleus also spins about its own axis either in a clockwise or in an anti-clockwise direction,

Each type of spin can give rise to characteristic spectral lines with the formation of a hyperfine spectrum in the spectral series.

The spin quantum number denoted by the symbol ‘s’ expresses two opposite types of spinning motions of each electron.

The spin quantum number ‘s’ can have only two values, \(+\frac{1}{2} \text { and }-\frac{1}{2}\) The positive and negative signs represent two opposite directions of spinning motion of any spinning motion of any spinning motion of electron are very often represented by two arrows pointing in opposite directions,| and.

Q A spinning electron behaves like a tiny magnet with a definite magnetic moment. The angular mentum associated with the spinning electron is given by the mathematical expression.

Spin Quantum number (s) signifies the mode of Electron Spin (Clockwise or Anti-clockwise).

Heisenberg’s Uncertainty Principle

Werner Heisenberg introduced his uncertainty principle in 1927 which is a direct consequence of the dual nature of electrons.

Heisenberg’s uncertainly principle: it is impossible to measure simultaneously both the position and the momentum of a sub-atomic particle like an electron, accurately, at any instant of time.

Explanation: If at a particular moment, the uncertainty in position and the uncertainty in the momentum of a sub-atomic particle be Ax and Ap respectively, then it can be
shown that the product of these two uncertainties must be at least equal to or greater than \(\frac{h}{4 \pi}.\)

Mathematically it can be expressed as \(\Delta x \times \Delta p \frac{h}{4 \pi}\) [where h = Planck’s constant]

Heisenberg’s uncertainty principle

Werner Heisenberg introduced his uncertainty principle in 1927 which is a direct consequence of the dual nature of electrons.

Heisenberg’s uncertainly principle: it is impossible to measure simultaneously both the position and the momentum of a sub-atomic particle like an electron, accurately, at any instant of time.

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Explanation: If at a particular moment, the uncertainty in position and the uncertainty in the momentum of a sub-atomic particle are Ax and Ap respectively, then it can be shown that the product ofthese two uncertainties must be at least equal to or greater than \(\frac{h}{4 \pi}.\)

Mathematically it can be expressed as, \(\Delta x \times \Delta p \frac{h}{4 \pi}\) [where h = Plancks constant ….[1]

⇒ \(\text { or, } \Delta x \times \Delta(m v) \frac{h}{4 \pi} \quad \text { or, } \Delta x \times m \Delta v \frac{h}{4 \pi}\)

⇒\(\text { or, } \Delta x \times \Delta v \frac{h}{4 \pi m}\)

[since m is constant]

Suppose we are going to measure simultaneously both the position and momentum of an electron in an atom. If an attempt is made to measure the position of the electron with high accuracy, then the measured value of the momentum will be less accurate and vice versa.

It should be realized that the uncertainty principle is not due to any limitation of the measuring instrument but is the consequence of the dual nature of moving particles and electromagnetic radiation (light).

Heisenberg’s uncertainty principle rules out the concept of a fixed circular path with definite position and momentum electrons in an atom as proposed by Bohr.

It should be remembered that the uncertainty principle is applicable to the position and momentum of a particle along the same axis.

Thus, if Ax represents the uncertainty in position along the x-axis then Ap must be the uncertainty in momentum along the x-axis (let it be represented as Ap ).

Thus \(\Delta x \times \Delta p_x \frac{h}{4 \pi}\)

Similarly, \(\Delta y \times \Delta p_y \frac{h}{4 \pi} \text { and } \Delta z \times \Delta p_z \frac{h}{4 \pi}\)

The uncertainty principle can also be applied to the conjugate pair, energy, and time. If Af represents the uncertainty in measuring the lifetime of a state and AE represents the uncertainty in measuring its energy in that state, then according to uncertainty principle \(\Delta E \times \Delta t \frac{n}{4 \pi}.\)

Uncertainty Principle For Macrocophic Objects: Theoretically, the uncertainty principle holds good for objects of all sizes, but in reality, it has no significance for macroscopic objects (big objects).

To realize this, let us consider a particle of mass of 1 mg, and for this, the approximate value ofthe product of Ax and Aw is given by, \(\begin{aligned}
\Delta x \cdot \Delta v \approx \frac{h}{4 \pi m} & =\frac{6.626 \times 10^{-34} \mathrm{~kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-1}}{4 \times 3.14 \times 10^{-6} \mathrm{~kg}} \\
& =0.53 \times 10^{-28} \mathrm{~m}^2 \cdot \mathrm{s}^{-1}
\end{aligned}\)

Thus the product of Ax and Av is extremely small. In other words, for objects of ordinary size, the uncertainties in position and momentum are very small as compared to the size of the object and the momentum of the object respectively.

Hence from the practical point of view, the values of these uncertainties may be taken as zero.

This means that the position and velocity of large objects can be measured almost accurately at any instant in time.

since in everyday life, we come across big objects only, it can be concluded that Heisenberg’s uncertainty principle has no significance in everyday life.

For a subatomic particle such as an electron, we have

This value is quite large so neither the uncertainty in position nor the uncertainty in velocity of the sub-atomic particle can be neglected.

For example, if the uncertainty in the position of the electron is 10-4m, uncertainty in its velocity will be ~ 0.58 m.s-1, which is quite significant.

It is for this reason, that Bohr’s concept of a fixed circular path with a definite velocity needs modification.

Electrons cannot rollers In Ihn nucleus: The diameter of the Nucleus is of the order 10-15m. For the electron to reside within the nucleus, the maximum uncertainty in
its position should be 10-15m,i.e., Ax = 10-15m.

⇒ \(\begin{aligned}
\Delta x \times \Delta p &\frac{h}{4 \pi} \quad \text { or, } \Delta x \times m \Delta v \frac{h}{4 \pi} \text { or, } \Delta v \frac{h}{4 \pi m \times \Delta x} \\
\text { or, } \Delta v &\frac{6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}{4 \times 3.14 \times 9.1 \times 10^{-31} \mathrm{~kg} \times 10^{-15} \mathrm{~m}} \\
& =5.7 \times 10^{10} \mathrm{~m} \cdot \mathrm{s}^{-1}
\end{aligned}\)

This value is much higher than the velocity of light (3 x 108m-s_1) which is not possible. Thus, an electron cannot reside within the nucleus.

Reasons for the failure of Bohr’s atomic model:

In Bohr’s model of an atom, the electron is regarded as a charged particle moving with definite velocity in a circular orbit of a definite radius.

This is not supported by Heisenberg’s uncertainty principle according to which it is impossible to determine both the velocity and the position of an electron simultaneously with certainty.

Furthermore, Bohr’s model does not take into consideration the concept of the dual nature of an electron (which is a sub-atomic particle). Due to such inherent weakness, Bohr’s atomic model lost its significance

Class 11 Chemistry First Law Of Thermodynamics

Question 1. Calculate die change in internal energy when heat released by the die system Is 200 J and work done on the die system is 120J heat absorbed and work done by the system arc 200J and 120J respectively 0heat released and work done by the system arc 200 J and 120 J respectively.
Answer: As per given data, q = -200 J and w = +120 J

∴ Change in internal energy,

AU = q + w =(- 200 + 120) J =-80 J

As per given data, q = +200 J and w = -120

∴ Change in internal energy,

AU = q+w=(200-120) J = + 80 J

As per given data, q = -200 J and w = -120 J

∴ Change in internal energy,

AU = q + w=(- 200- 120) J = -320 J

Question 2. In a process, the system performs 142 J of work and the internal energy of the system Increases by 879 J. Predict the direction of heat flow and also calculate the quantity of heat transferred.
Answer: Given, w = -142 J and AU = + 879 J

∴ AU = q+w or, 879 = q- 142

∴ q = + 1021 J

The positive value of q indicates that the system absorbs heat from the surroundings i.e., heat flows from the surroundings to the system. The amount of heat transferred in the process is 1 021 J.

Question 3. A system undergoes n process in which It gives up 300J of heat and Its internal energy decreases by 300J. Of the system and its surroundings, which one docs work in this process?
Answer: Given, A U = -300J and q = -900 J

∴ AU = q + w or, -300 = – 900 + w or, w = + 600 J

w is (+)ve. This means work is done on the system by the surroundings and the amount of work done is 600 J

Question 4. The initial volume of a gas, confined in a cylinder fitted with a piston is U.2L. The final volume of the gas becomes 33.6L after expansion against a constant external pressure of 2atm. During expansion if the gas absorbs 1kg of heat from the surroundings then what will be the change in the internal energy of that gas?
Answer: Work done, w = -Pex(V2-V t)

Given: Pgx = 2 atm; Vj = 11.2 L and V2 = 33.6 L

∴ w =-2(33.6- 11.2) = -44.8 L-atm = -44.8 X 101.3
= -4538.24 J [v 1 L-atm = 101.3 J]

So, the amount of work done by the gas is 4538.24 J.

Again, the heat (q) absorbed by the gas = 1 of = 1000J Therefore, the change in internal energy of the gas, AU = q+w =(1000-4538.24) J = -3538.24 J

Therefore, the decrease in internal energy is 3538.24 J

Question 5. An ideal gas is expanded from 1L to 6L in a closed vessel at 2 atm pressure by applying heat at a fixed temperature. Calculate the work done and heat absorbed by the gas. [Given: lLatm = 24.22 cal]
Answer: Work done, w = -PexAV = -Pex(V2- V1)

Given: Pgx = 2 atm; Vy = 1 L and V2 = 6 L

w = -2(6-1) = -10 L-atm =-242.2 cal

From the first law of thermodynamics, AU = q + w.

Since the expansion is carried out isothermally and the system is an ideal gas, the change in internal energy in the process will be equal to zero. Thus, AU = 0.

0 = q + w or, q = -w = + 242.2cal

Therefore, the amount of heat absorbed bythe gas is 242.2 cal.

Question 6. A cylinder fitted with a piston contains an ideal gas with a volume of 21L. The gas is compressed isothermally to l/3rd of its initial volume under a constant external pressure of 3 atm. Calculate q, w, and AU.
Answer: Work done, w = -Pex(V2- Vx) [Vy > V2]
Given: Pgx = 3 atm; Vy = 21L and V2 ,\(=21 \times \frac{1}{3}=7 \mathrm{~L}\)

∴ w = -3(7-21) = 42 L. atm

= 42 X 101.3 J = 4254.6 J p[since 1L.atm =101.3J]

In an isothermal process of an ideal gas, the internal energy ofthe gas does not change i.e., AU = 0

From the first law of thermodynamics, AU = q + w

∴ 0 = q + 4254.6 J or, q = -4254.6 J

Thus, in the process g=- 4254.6 J, w=+ 4254.6 J, AU=0.

Question 1. The number of isomers possible for disubstituted borazine, B₃N₃H₄X₂ is-

1. 3
2. 4
3. 5
4. 6

Answer: 2. 4

Question 2. Pentaborane-9 (B₅H₉) is an example of—

1. Arachno-borane
2. Pseudo-borane
3. Nido-borane
4. Closo-borane

Answer: 1. Arachno-borane

Question 3. Si when reacts with A forms B. A & B respectively are

1. HF, H₂SiF₄
2. HF, H₂SiF₆
3. HCl, H₂SiCl₆
4. HI, H₂SiI₆

Answer: 2. HF, H₂SiF₆

Question 4. Boric acid is a

1. Monobasic and weak Lewis acid
2. Monobasic and weak Bronsted acid
3. Monobasic and strong Lewis acid
4. Tribasic and weak Bronsted acid

Answer: 1. Monobasic and weak Lewis acid

Question 5. Which ofthe following does not exist in a free state

1. BF₃
2. BCl₃
3. BBr₃
4. BH₃

Answer: 4. BH₃

Question 6. The correct order of decreasing Lewis acid character is

1. BCl₃ > AlCl₃ > GaCl₃ > InCl₃
2. AlCl₃ > BCl₃ > InCl₃ > GaCl₃
3. AlCl₃ > GaCl₃ > BCl₃ > InCl₃
4. InCl₃ > GaCl₃ > AlCl₃ > BCl₃

Answer: 1. BCl₃ > AlCl₃ > GaCl₃ > InCl₃

Question 7. Which of the following is present in the chain structure of silicate

1. (Si₃O^2-₅)_n
2. (Si₃O^2-₃)_n
3. (SiO^4-₄)
4. Si₂O^6-₇

Answer: 2. (Si₃O^2-₃)_n

Question 8. A metal, M forms chlorides in +2 and +4 oxidation states. Which ofthe following statements about these chlorides is correct

1. MCl₂ is more volatile than MCl₄
2. Ml₂ is more ionic than MCl₄
3. MCl₂ is more soluble in any. ethanol than MCl₄
4. MCl₂ is more easily hydrolysed than MCl₄

Answer: 2. Ml₂ is more ionic than MCl₄

Question 9. The number of O-atoms that are shared per Si04 tetrahedra in silicate anion of beryl is

1. 4
2. 3
3. 2
4. 1

Answer: 3. 2

Question 10. Which of the following on hydrolysis produces crosslinked silicone polymer

1. R₄Si
2. RSiCl₃
3. R₂SiCl₂
4. R₃SiCl

Answer: 2. RSiCl₃

Question 11. The antidote of poisoning caused by CO is

1. Carborundum
2. Carbogen
3. Carbonic acid
4. Pure oxygen

Answer: 2. Carbogen

Question 12. Carbon suboxide in reaction with water produces

1. Oxalic acid
2. Formic acid
3. Lactic acid
4. Malonic acid

Answer: 4.  Malonic acid

Question 13. The volume of which liquid metal increases on solidification

1. Ga
2. Al
3. Zn
4. Cu

Answer: 1. Ga

Question 14. Which ofthe following reacts only with alkali

1. B₂O₃
2. Al₂O₃
3. Ga₂O₃
4. ln₂O₃

Answer: 1. B₂O₃

Question 15. Which is the strongest Lewis acid

1. BF₃
2. BCl₃
3. BBr₃
4. BI₃

Answer: 4. BI₃

Question 16. The atomic radius of Ga is slightly less than that of Al. The reason is

1. Weaker shielding effect of -electrons of Ga
2. Stronger shielding effect of s -electrons of Ga
3. Weaker shielding effect of d -electrons of Ga
4. Stronger shielding effect of d -electrons of Ga

Answer: 3. Weaker shielding effect of d -electrons of Ga

Question 17. Carbon does not form complexes, because

1. Vacant d – orbitals are absent in it
2. It is not a metal
3. Its atomic radius is small
4. It is neutral

Answer: 1. Vacant d -d-orbitals are absent in it

Question 18. Supercritical CO₂ is used as

1. Dry ice
2. Fire extinguisher
3. A solvent for the extraction of organic compounds from natural sources
4. The inert solvent in various reactions

Answer: 3. Asolvent for the extraction of organic compounds from natural sources

Question 19. The stability of the +1 oxidation state increases in the sequence—

1. Al < Ga < In < Tl
2. Tl<In<Ga<Al
3. In < Tl < Ga < Al
4. Ga<In<Al<Tl

Answer: 1. Al < Ga < In < Tl

Question 20. Which ofthe following is acidic—

1. B₂O₃
2. Al₂O₃
3. Ga₂O₃
4. ln₂O₃

Answer: 1.B₂O₃

Question 21. The correct order of first ionisation enthalpy for Gr-13 elements is

1. B > Al > Ga > In > Tl
2. B < Al < Ga < In < Tl
3. B < Al > Ga < In > Tl
4. B > Al < Ga > In < Tl

Answer: 4.  B > Al < Ga > In < Tl

Question 22. Which of the following elements is not likely to be the central atom in MF^3-₆

1. B
2. Al
3. Ga
4. In

Answer: 1. B

Question 23. The tendency of catenation in Gr-14 elements follows the order

1. C > Si > Ge > Sn
2. C>>Si > Ge s: Sn
3. Si > C > Sn > Ge
4. Ge > Sn > Si > C

Answer: 2. C>>Si > Ge s: Sn

Question 24. The repeating structural unit in silicone is—

Answer: 2

Question 25. Which of the following allotropic forms of carbon is isomorphous with crystalline silicon

1. Graphite
2. Coal
3. Coke
4. Diamond

Answer: 4.  Diamond

Question 26. The shape and hybridisation of the B-atom of BH₄^– is

1. Pyramidal, sp³
2. Octahedral, sp³d²
3. Tetrahedral, sp³
4. None of these

Answer: 3.  Tetrahedral, sp³

Question 27. Germanium is transparent in

1. Visible light
2. Infrared region
3. Ultraviolet region
4. Ultraviolet region

Answer: 2. Infrared region

Question 28. The chain length of silicone polymer can be controlled by adding

1. MeSiCl₃
2. Me₂SiCl₂
3. Me₃SiCl
4. Me₄Si

Answer: 3. Me₃SiCl

Question 29. Higher B—F (in BF₃) bond dissociation energy as compared to that of C— F (in CF₄) is due to

1. Stronger σ-bond between B and F in BF₃ as compared to that between C and F in CF₄
2. Significant pn-pn interaction between B and F in
3. BF₃ whereas there is no possibility of such interaction between C and F in CF₄
4. A lower degree of pπ-pπ interaction between B and F in BF₃ than that between C and F in CF₄ smaller size of B -atom as compared to that of C -atom

Answer: 2. Significant pn-pn interaction between B and F in

Question 30. The reaction of diborane with ammonia initially gives

1. B₂H₆.NH₃
2. Borazol
3. B₂H₆-3NH₃
4. [BH₂(NH₃)₂]⁺[BH₄]^–

Answer: 4.  [BH₂(NH₃)₂]⁺[BH₄]^–

Question 31.    X, y, Z and their respective colours are

1. X = Cu(BO₂)₂ (blue), Y = Cu(BO₂) (colourless),Z = Cu (red)
2. X = CuBOz (blue) , Y = Cu(BO₂)₂ (colourless), Z = Cu (Black)
3. X = CU(BO₂)₂ (red) , Y = CuBO₂ (blue) Z = (red)
4. X = Cu (red) , Y = Cu(BO₂)₂ (blue) , Z = CuBO₂ (colourless)

Answer: 1. X = Cu(BO₂)₂ (blue) , Y = Cu(BO₂) (colourless) ,Z = Cu (red)

Question 32. The correct formula for borax is

1. Na₂[B₄O₄(OH)₃].9H₂O
2. Na₂[B₄O₄(OH)₄].8H₂O
3. Na₂[B₄O₆(OH)₅].7H₂O
4. Na₂[B₄O₇(OH)₆]-6H₂O

Answer: 2. Na₂[B₄O₄(OH)₄].8H₂O

Question 33. Which of the following statements is correct—

1. Sn (II) and Pb (IV) salts are used as oxidants
2. Sn (II) and Pb (IV) salts are used as reductants
3. Sn (II) salts are used as oxidants and Pb (IV) salts are used as reductants
4. Sn (II) salts are used as reductants and Pb (IV) salts are used as oxidants

Answer: 4. Sn (II) salts are used as reductants and Pb (IV) salts are used as oxidants

Question 34. SiCl₄ gets readily hydrolysed but CCl₄ does not, because

1. Si can expand its octet but C does not
2. The ionisation enthalpy of C is greater than that of Si
3. C forms both double and triple bonds
4. The electronegativity of C is greater than that of Si

Answer: 1. Si can expand its octet but C does not

Question 35. PbCl₄ exists but PbBr₄ and Pbl₄ do not, because

1. Chlorine is a most electronegative element
2. Bromine and iodine are larger
3. Bromine and iodine cannot oxidise Pb to Pb⁴⁺
4. Bromine & iodine are stronger oxidants than chlorine

Answer: 3. Bromine and iodine cannot oxidise Pb to Pb⁴⁺

Question 36. Which of the following resembles CO in terms of physical properties

1. O₂
2. Cl₂
3. N₂
4. F₂

Answer: 3.N₂

Question 37. Which ofthe following statements is incorrect

1. Most of the silicones are water repellents
2. Silicones get dissociated at high temperature
3. Silicones do not get oxidised in air at high temperature
4. Silicones are good thermal and electrical insulators

Answer: 2.  Silicones get dissociated at high temperature

Question 38. Wollastonite is a

1. Three-dimensional silicate
2. Chain silicate
3. Sheet silicate
4. Cyclic silicate

Answer: 4. Cyclic silicate

Question 39. B(OH)₃ + NaOH ⇌ NaBO₂ + Na[B(OH)₄] + H₂O; The above reaction be made to proceed in the forward direction by

1. Addition of diol
2. Addition of borax
3. Addition of KHF₂
4. Addition of NaHPO₄

Answer: 1. Addition of diol

Question 40. Which ofthe following is correct

1. Al(OH)₃ is more acidic than B(OH)₃
2. B(OH)₃ is basic but Al(OH)₃ is amphoteric in nature
3. B(OH)₃ is acidic but Al(OH)₃ is amphoteric in nature
4. Both B(OH)₃ and Al(OH)₃ are amphoteric

Answer: 3. B(OH)₃ is acidic but Al(OH)₃ is amphoteric in nature

Question 41. Which of the following is correct

1. B₂H₆-2NH₃ is known as inorganic benzene
2. Boric acid is a protonic acid
3. Be exhibits coordination number = 6
4. BeCl₃ and AlCl₃ have bridged chlorine structures in the solid phase

Answer: 4. BeCl₃ and AlCl₃ have bridged chlorine structures in the solid phase

Question 42. B cannot form B³⁺ ion, because

1. Formation of B³⁺ ion requires a greater amount of energy and this cannot be obtained from lattice energy or hydration energy
2. B is a non-metal
3. B do not possess any vacant d -orbitals
4. B possess the highest melting point among its group members

Answer: 1.  Formation of B³⁺ ion requires a greater amount of energy and this cannot be obtained from lattice energy or hydration energy

Question 43. Which of the following has the minimum heat of dissociation

1. (CH₃)₃N : → BF₃
2. (CH₃)₃N :→B(CH₃)₂F
3. (CH₃)₃N :→ B(CH₃)₃
4. (CH₃)₃N :→B(CH₃)F₂

Answer: 3. (CH₃)₃N :→ B(CH₃)₃

Question 44. The correct statement concerning CO is

1. It combines with H₂O to give carbonic acid
2. It reacts with haemoglobin
3. It acts only as a reducing agent
4. It cannot form adducts

Answer: 2.  It reacts with haemoglobin

Question 45. Foamite mixture consists of

1. Al₂ (SO₄)₃ + NaHCO₃
2. Al₂(SO₄)₃ + Na₂CO₃
3. Fe₂(SO₄)₃ + Na₂CO₃
4. CuSO₄ + NaHCO₃

Answer: 1. Al₂ (SO₄)₃ + NaHCO₃

Question 46. In which of the following compounds, the 3c-2e bond is present

1. AI₂(CH₃)₆
2. In (C₆H₅)₃
3. B₂H₆
4. Al₂Cl₆

Answer: 1 and 2

Question 47. Which ofthe following oxides do not get reduced by CO

1. ZnO
2. Fe₂O₃
3. CaO
4. Na₂O

Answer: 1 and 3

Question 48. Which of the following is not isostructural with CO₂

1. SnCl₂
2. HgCl₂
3. SCl₂
4. Znl₂

Answer: 2 and 4

Question 49. C(OH₄) is unstable but Si(OH)₄ is stable. Possible reasons are

1. C — O bond energy is low
2. C — O  bond energy is high
3. Si — O bond energy is low
4. Si — O bond energy is high

Answer: 1 and 4

Question 50. Which ofthe following statements are correct

1. Fullerenes have dangling bonds
2. Fullerenes are cage-like molecules
3. Graphite is thermodynamically the most stable allotrope of carbon
4. Graphite is the purest allotrope of carbon

Answer: 2 and 3

Question 51. Boron trifluoride (BF₃) is

1. An electron-deficient compound
2. A Lewis acid
3. An ionic compound
4. Used as rocket fuel

Answer: 1 and 2

Question 52. Compounds which readily undergo hydrolysis are

1. AlCl₃
2. CCl₄
3. SiCl₄
4. PbCl₄

Answer: 1, 3 and 4

Question 53. Which of the following compounds undergo disproportionation in aqueous solution

1. TlCl₃
2. GaCl
3. InCl
4. TlCl

Answer: 2 and 3

Question 54. Me₃SiCl is used during the polymerisation of organosilicon because

1. The chain length of organosilicon polymers can be controlled by adding Me₃SiCl
2. Me₃SiCl blocks the end terminal of the silicone polymer
3. Me₃SiCl improves the quality and yield of the polymer
4. Me₃SiCl acts as a catalyst during polymerisation

Answer: 1 and 2

Question 55. Which of the following acids, on dehydration, produce oxides of carbon

1. Succinic acid
2. Propanoic acid
3. Mlonicacid
4. Formic acid

Answer: 3 and 4

Question 56. Which of the following are basic nature

1. B₂O₃
2. Tl₂O
3. ln₂O₃
4. Al₂O₃

Answer: 2 and 3

Question 57. The linear shape of CO₂ is due to

1. sp³ -hybridisation of C
2. sp -hybridisation of C
3. pπ-pπ bonding between C and 0
4. sp² -hybridisation of C

Answer: 2 and 3

Question 58. Which metallic salts exhibit the same colouration both in oxidising and reducing flame in the borax-bead test

1. Fe
2. Mn
3. Co
4. Cr

Answer: 3 and 4

Question 59. Which of the following two acidic substances react to give an alkaline solution

1. H₂B₄O₇
2. H₃BO₃
3. HF
4. KHF₂

Answer: 2 and 4

Question 60. Which of the following are the ingredients of baking powder

1. NaOH
2. Tartaric acid
3. Formic acid
4. Potassium hydrogen tartrate

Answer: 2 and 4

Question 61. Which ofthe following are sheet silicates

1. Diopside
2. Kaolinite
3. Talc
4. Beryl

Answer: 2 and 3

Question 62. Identify the correct resonating structures

1. O -C ≡ O
2. O = C = O
3. O ≡ C –  O⁺
4. ^–O – C ≡  O⁺

Answer: 2 and 4

Question 63. Which of the following species are not known

1. [SiCl₆]^2-
2. [CF₆]^2-
3. [PbCl₆]^2-
4. [SiF₆]^2-

Answer: 2 and 3

Question 64. Which of the following is correct concerning Gr-14 elements

1. Stability of dihalides: CX₂ > SiX₂ > GeX₂ > SnX₂
2. The tendency to form pπ-pπ multiple bonds increases down the group
3. The tendency of catenation decreases down the group
4. Each of them forms oxide ofthe type MO₂

Answer: 2, 3 and 4

Question 65. Which ofthe following has a bridge bond

1. Water
2. Inorganic benzene
3. Phenol
4. Diborane

Answer: 4. Diborane

Question 66. An aqueous solution of borax is

1. Neutral
2. Amphoteric
3. Basic
4. Acidic

Answer: 3. Basic

An aqueous solution of borax is basic because it is a salt of strong base (NaOH) and weak acid (H₃BO₃)

Question 67. Boric acid is polymeric due to

1. Its acidic nature
2. The presence of hydrogen bonds
3. Its monobasic nature
4. Its geometry

Answer: 2. The presence of hydrogen bonds

Boric acid is polymeric due to the presence of; hydrogen bonds

Question 68. The type of hybridisation of boron In dlborane

1. sp
2. sp²
3. sp³
4. dsp²

Answer: 2.sp²

In B₂HO, the hybridisation state of B is sp²

Question 69. Thermo dynamically the most stable form of carbon is

1. Diamond
2. Graphite
3. Fullerenes
4. Coal

Answer: 2. Graphite

Question 70. Elements of Gr-14

1. Exhibit oxidation state of +4 only
2. They exhibit oxidation state of +2 and +4
3. Form M^2-and M²⁺Ions
4. Form M²⁺ and M⁴⁺ ions

Answer: 2. They exhibit oxidation states of +2 and +4

Elements of group 14 exhibit oxidation states of +2 and +4 due to the inert pair effect

• Chapter 1 Solution Important Questions
• Chapter 2 Electrochemistry  Important Questions
• Chapter 3 Chemical Kinetics Important Questions
• Chapter 4 D & F Block Important Questions

• Chapter 5 Coordination Compound Important Questions
• Chapter 6 Haloalkanes And Haloarenes Important Questions
• Chapter 7 Alcohol, Phenol and Ether Important Questions
• Chapter 8 Aldehydes, Ketones and Carboxylic Acid Important Questions
• Chapter 9 Amines  Important Questions
• Chapter 10 Biomolecules Important Questions