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Chemical Bonding and Atomic Structure NEET Questions — Free MCQ Practice

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A yellow metallic powder is burnt in a stream of fluorine to obtain a colourless gas X which is thermally stable and chemically inert. Its molecule has octahedral geometry. Another colourless gas Y with same constituent atoms as that of X is obtained when sulphur dichloride is heated with sodium fluoride. Its molecule has triangular pyramidal structure X and Y are

A.

SF₄ and S₂F₂

B.

SF₆ and SF₄

C.

NaF and NaCl

D.

SF₄ and SF₆

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Explanation

$S + 3 F_{2} \to S F_{6} (X) c o l o u r l e s s g a s S C l_{2} + 4 N a F \to S F_{4} (Y) + S_{2} C l_{2} + 4 N a C l$

SF₄ has trigonal pyramidal shape due to sp³d hybridisation. SF₆ is octahedral due to sp³d² hybridisation. Sulphur has S₈ molecular geometry

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In nitrogen family the H-M-H bond angle in the hybrides MH₃ gradually becomes closer to 90^o on going from N to Sb. This shows that gradually

A.

The basic strength of hybrides increases

B.

Almost pure p-orbitals are used for M-H bonding

C.

The bond energies of M-H bond increases

D.

The bond pairs of electrons becomes nearer to the central atom

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Explanation

The hybrides have a pyramidal or tetrahedral shape with a lone pair of electrons in on of the orbitals.The H-M-H bond angle is less than the original 109^o28' tetrahedral bond angle(H-N-H in NH₃ is 106^o45') because of greater repulsion between lone pair and a bond pair than between two obnd pairs of electrons.Because electronegativity of M decreases from N to Bi, the bond pair lie farther away from the central atom, and the lone pair causes greater distortion of bond angle.

Thus H-P-H bond in PH₃ is 94^o , while in AsH₃ and SbH₃ it is about 91.8^o and 91.3^o respectively(closer to 90^o). This suggests that orbitals used for bonding are closer to pure p-orbitals.

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Assertion : In trimethyalamine (N(CH₃)₃), the nitrogen has a pyramidal geometry, wheres in trisilylamine (N(SiH₃)₃), it has a planer geometry.

Reason : In trimethylamine, nitrogen shows sp³ hybridization while in trisilyamine it shows sp² hybridization.

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

In trimethylamine, nitrogen shows sp³ hybridization therefore trimethylamine has a pyramidal geometry while in trisilyamine it shows sp² hybridization resulting in a planer square.

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Assertion : ClF₃ exists bul FCl₃ does not.

Reason : FCl₃ is not formesd due to greater steric hinderance by the larger Cl atoms.

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

ClF₃ exists bul FCl₃ does not because FCl₃ is not formed due to greater steric hinderance by the larger Cl atoms.

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Which of the following is not correct?

A.

Liquid helium is used as cryogenic liquid

B.

$X e O_{3}$ has four $σ$ and four $π$ - bonds

C.

The hybridisation of Xe in $X e F_{4}$ is $s p^{3} d^{2}$

D.

Among noble gases, the occurrence of argon in highest in air

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The molecular shape of $S F_{4}, C F_{4} a n d X e F_{4}$ are:

A.

the same with 2, 0 and 1 lone pairs of electrons on the central atoms respectively

B.

the same with 1, 1 and 1 lone pairs of electrons on the central atoms respectively

C.

different with 0, 1 and 2 lone pairs of electrons on the central atoms, respectively

D.

different with 1, 0 and 2 lone pairs of electrons on the central atoms, respectively

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The correct order of the thermal stability of hydrogen halide (H-X) is:

A.

HI > HBr > HCl > HF

B.

HF > HCl > HBr > HI

C.

HCl < HF > HBr < HI

D.

HI > HCl < HF > HBr

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Explanation

(2)

As the size of the halogen atom increases from F to I, H – X bond length in HX molecules also increases from H – F to H – I (H – F < H – Cl < H – Br < H – I).

The increase in H – X bond length decreases the strength of H – X bond from H – F to H – I (H – F > H – Cl > H – Br > H – I). The decrease in the strength of H – X bond is evident from the fact that H – X bond dissociation energies decrease from H – F to H – I.

Due to successive decrease in the strength of H – X bond from H – F to H – I, thermal stability of HX molecules also decreases from HF to HI (HF > HCl > HBr > HI).

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Which of the following interhalogen anions is linear?

A.

$I_{3}^{-}$

B.

$I C l_{2}^{-}$

C.

Both (a) and (b)

D.

none of these

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The number of lone pairs of electrons present in central atom of $C I F_{3}$ are:

A.

0

B.

1

C.

2

D.

3

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Explanation

Cl has 7 valance electron and 3 is bonded ,so 4 electron is left. hence its 2 pair.

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The geometry of $H_{2} S$ and its dipole moment are:

A.

angular and non-zero

B.

angular and zero

C.

linear and zero

D.

linear and non-zero

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Explanation

The geometry of H2S is angular or bent, with an angle of approximately 92.1Â°. It has a non-zero dipole moment due to the difference in electronegativity between the hydrogen and sulfur atoms, resulting in a polarized molecule.

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

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