Chemical Bonding and Molecular Structure

Question 1

Match the species in List-I with their geometry in List-II. Choose the correct answer from the options given below: A. PCl₅ B. BrF₅ C. BF₄⁻ D. [Ni(CN)₄]²⁻ I. Tetrahedral II. Square Planar III. Trigonal bipyramidal IV. Square pyramidal

Accepted Answer

A-III, B-IV, C-I, D-II. PCl₅ is sp³d with five bond pairs and no lone pair, giving a trigonal bipyramid. BrF₅ is sp³d² but one position is a lone pair, so the shape is square pyramidal. BF₄⁻ is sp³ and tetrahedral. [Ni(CN)₄]²⁻ is dsp² with cyanide forcing pairing, so it is square planar.

Answer

A-IV, B-III, C-I, D-II

Answer

A-III, B-I, C-II, D-IV

Answer

A-III, B-II, C-I, D-IV

Question 2

Given below are two statements. In the light of the above statements, choose the most appropriate answer from the options given below. Statement I: A hypothetical diatomic molecule with bond order zero is quite stable. Statement II: As bond order increases, the bond length increases.

Accepted Answer

Both Statement I and Statement II are false. Both statements are false. A bond order of zero means the number of bonding and antibonding electrons is equal, so there is no net bonding and the species does not exist as a stable molecule — this is why He₂ is not found. And bond order and bond length are inversely related: a higher bond order pulls the nuclei closer, which is why the bond length falls along the series N₂ < O₂ < F₂ as bond order falls from 3 to 1.

Answer

Both Statement I and Statement II are true

Answer

Statement I is true but Statement II is false

Answer

Statement I is false but Statement II is true

Question 3

Identify the correct orders against the property mentioned. A. H₂O > NH₃ > CHCl₃ – dipole moment B. XeF₄ > XeO₃ > XeF₂ – number of lone pairs on central atom C. O–H > C–H > N–O – bond length D. N₂ > O₂ > H₂ – bond enthalpy

Accepted Answer

A, D only. A is correct: the dipole moments are 1.85 D for H₂O, 1.47 D for NH₃ and 1.04 D for CHCl₃. D is correct because bond enthalpy tracks bond order, and N₂ has a triple bond, O₂ a double bond and H₂ a single bond. B is wrong — the lone pairs on Xe are two in XeF₄, one in XeO₃ and three in XeF₂, so XeF₂ leads. C is wrong because bond length increases with atomic size, giving N–O > C–H > O–H, the reverse of what is stated.

Answer

B, D only

Answer

A, C only

Answer

B, C only

Q1.Match the species in List-I with their geometry in List-II. Choose the correct answer from the options given below:

Q2.Given below are two statements. In the light of the above statements, choose the most appropriate answer from the options given below.

Q3.Identify the correct orders against the property mentioned.