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Chemical Bonding and Atomic Structure MCQs for NEET — Chemistry Questions with Answers

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The $B F_{3}$ is a planarn molecule where as $N F_{3}$ is pyramidal because:

$B - F$ bond is more polar than $N - F$ bond.

Boron atom is bigger than nitrogen atom.

Nitrogen is more electronegative than boron.

$B F_{3}$ has no lone pair but $N F_{3}$ has a lone pair of electrons.

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Explanation

The BF3 molecule has a planar structure due to the absence of lone pairs on the central boron atom. In contrast, the NF3 molecule adopts a pyramidal structure because the nitrogen atom has one lone pair of electrons, causing repulsion and distorting the molecular geometry.

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Which of the following statement is true for $I O_{2} {F_{2}}^{-}$ according to VSEPR theory?

The lone pair and two I-O double bonds occupy the equatorial positions of trigonal
bipyramid.

It has $s p^{3}$ hybridization and is T-shaped.

Its structure is analogous to $S F_{4}$ .

(A) and (C) both

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Explanation

According to VSEPR theory, the IO2F2- ion has a trigonal bipyramidal structure with the two lone pairs and two I-O double bonds occupying the equatorial positions, which is analogous to the structure of SF4. Therefore, both options (A) and (C) are correct.

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In the structure of $H_{2} C S F_{4}$ , to decide the plane in which C = S is present the following bond angle value are given

Axial FSF angle $(i d e a l i s e d = 180 °)$ $\Rightarrow 170 °$

Equatorial FSF angle $(i d e a l i s e d = 120 °)$ $\Rightarrow 97 °$

After deciding the plane of double bond, which of the following statements is correct?

Two C-H bonds are in the same plane of axial S-F bonds.

Two C-H bonds are in the same plane of equatorial S-F bonds.

Total five atoms are in the same plane.

Equatorial S-F bonds are perpendicular to plane of $π$ bond.

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According to Molecular orbital theory which of the following statement is incorrect?

LUMO level for $C_{2}$ molecule is $σ_{2 p x}$ orbital.

In $C_{2}$ molecules both the bonds $π$ are bonds.

$C_{2}$ is paramagnetic but ${C_{2}}^{2 -}$ is diamagnetic.

In ${C_{2}}^{2 -}$ ion there is one $σ$ and two $π$ bonds.

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Explanation

(A), (B) M.O. for C₂ = ${σ_{1}}^{s^{2}} < {σ_{1}}^{*_{s^{2}}} < {σ_{2}}^{s^{2}} < {σ_{2}}^{*_{s^{2}}} < \frac{π_{2} p^{2} y = π_{2} p^{2} z}{H O M O} < \frac{σ_{2} p_{x}}{L U M O},$ two $π$ molecular orbitals are involved in bonding .

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A diatomic molecule has a dipole moment of 1.2 D. If its bond length is equal to 10 ^-10 m then the fraction of an electronic charge on each atom will be:

42%

52%

37%

25%

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Explanation

(D) μ=q×d

$1.2 \times 10^{- 18} = q \times 10^{- 8} c m$

$q = \frac{1.2 \times 10^{- 18}}{10^{- 8}} = 1.2 \times 10^{- 10^{}} % c h a r g e = \frac{q}{e} \times = \frac{1.2 \times 10^{- 10}}{4.8 \times 10^{- 10}} \times 100 = 25 %$

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Which of the following statements is correct?

${(C H_{3})}_{3}$ COH is less acidic than ${(C H_{3})}_{3}$ SiOH.

Like CO, its analogue of Si is not stable.

In phosgene, C-O bond length is longer than expected while C-Cl bond length is shorter.

All of these

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Explanation

All the given statements are correct. (CH3)3COH is less acidic than (CH3)3SiOH due to the poorer ability of silicon to stabilize the negative charge on the conjugate base. CO is a stable molecule, but its silicon analogue is not stable. In phosgene (COCl2), the C-O bond length is longer than expected due to the presence of resonance, while the C-Cl bond length is shorter.

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The increasing order of the strength of hydrogen bond in the following mentioned linkages
is

(i) O — H — S (ii) O — H — O (iii) F – H – F (iv) F – H – O

(i) < (ii) < (iv) < (iii)

(ii) < (i) < (iv) < (iii)

(i) < (ii) < (iii) < (iv)

(ii) < (i) < (iii) < (iv)

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Explanation

(A) Strength of H-bonds on following factors.
(i) Electronegativity of element covalently bonded to hydrogen atom.
(ii) Size of electronegative element.
(iii) Ease of donation of lone pair of electrons by electronegative element.

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The boiling pot of $C C l_{4}$ is higher than that of $C H C l_{3}$ because :

The dipole moment of $C C l_{4}$ is greater than that of $C H C l_{3}$

$C C l_{4}$ forms hydrogen bonds.

$C C l_{4}$ has higher molecular mass than that of $C H C l_{3}$

$C C l_{4}$ is more ionic than $C H C l_{3}$

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Explanation

(C) Number of polarisable electrons $\propto$ strength of vander waal’s force $\propto$ boiling point.
boiling point $C H C l_{3} = 61 ° C a n d C C l_{4} = 73 ° C$ .

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Assertion : In tetrahedral hybridisation i.e., in ${sp}^{3}$ hybridisation all p-orbitals are involved and no

p-orbital is left for coming with $π$ -bonds.

Reason : Central atom can not form double bonds in the molecules or species having ${sp}^{3}$ hybridisation.

A. If both the assertion and the reason are true and the reason is a correct explanation of the assertion B. If both the assertion and reason are true but the reason is not a correct explanation of the assertion C. If the assertion is true but the reason is false D. If both the assertion and reason are false

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Explanation

The assertion is true because in sp3 hybridization, all four orbitals are involved in sigma (Ïƒ) bonding, and no p-orbital is left for pi (Ï€) bond formation. However, the reason is false because the central atom can form double bonds in molecules or species with sp2 hybridization, where one p-orbital is available for Ï€-bond formation.

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Assertion : $N_{2} {F_{3}}^{+}$ is planar at each nitrogen atom.

Reason : $N_{3} H,$ the bond angle $H ‐ N ‐ N$ is $120 °$ and both the $N ‐ H$ bond lengths are not equal.

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Yes. You can attempt every Chemical Bonding and Atomic Structure question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.