Unit 3: Chemical Bonding and Molecular Structure — Online MCQ Test
CHEMISTRY · NEET (All) · NEET
Practice Unit 3: Chemical Bonding and Molecular Structure with a free chapter-wise online MCQ test.
This chapter covers: Kossel - Lewis approach to chemical bond formation the concept of ionic and covalent bonds. Ionic Bonding: Formation of ionic bonds factors affecting the formation of ionic bonds;....
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Unit 3: Chemical Bonding and Molecular Structure — Important Questions & Answers
Given below are two statements: Statement I: A hypothetical diatomic molecule with bond order zero is quite stable. Statement II: As bond order increases, the bond length increases.
- A. Both statement I and Statement II are true
- B. Both statement I and Statement II are false
- C. Statement I is true but Statement II is false
- D. Statement I is true but Statement II is true
Answer: B. Both statement I and Statement II are false
Bond order zero implies the molecule is unstable. As bond order increases, bond length decreases. Both statements are false.
Bond order zero implies the molecule is unstable. As bond order increases, bond length decreases. Both statements are false.
Given below are two statements: Statement I: A hypothetical diatomic molecule with bond order zero is quite stable. Statement II: As bond order increases, the bond length increases. In the light of the above statements, choose the most appropriate answer from the options given below:
- A. Both statement I and Statement II are true
- B. Both statement I and Statement II are false
- C. Statement I is true but Statement II is false
- D. Statement I is true but Statement II is true
Answer: B. Both statement I and Statement II are false
Bond order 0 implies the molecule is unstable. As bond order increases, bond length decreases (inversely proportional). Both statements are false.
Bond order 0 implies the molecule is unstable. As bond order increases, bond length decreases (inversely proportional). Both statements are false.
Match the List -I with List-II. List-I: (A) XeO₃, (B) XeF₂, (C) XeOF₄, (D) XeF₆. List-II: (I) sp³d; linear, (II) sp³; pyramidal, (III) sp³d³; distorted octahedral, (IV) sp³d²; square pyramidal.
- A. A-II, B-I, C-IV, D-III
- B. A-II, B-I, C-III, D-IV
- C. A-IV, B-II, C-III, D-I
- D. A-IV, B-II, C-I, D-III
Answer: A. A-II, B-I, C-IV, D-III
XeO₃ is sp³ pyramidal (A-II); XeF₂ is sp³d linear (B-I); XeOF₄ is sp³d² square pyramidal (C-IV); XeF₆ is sp³d³ distorted octahedral (D-III).
XeO₃ is sp³ pyramidal (A-II); XeF₂ is sp³d linear (B-I); XeOF₄ is sp³d² square pyramidal (C-IV); XeF₆ is sp³d³ distorted octahedral (D-III).
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 Choose the correct answer from the options given below:
- A. A, D only
- B. B, D only
- C. A, C only
- D. B, C only
Answer: A. A, D only
A is correct (Dipole moments: 1.85D, 1.47D, 1.04D). D is correct (Bond enthalpies of N2 are much higher due to triple bond).
A is correct (Dipole moments: 1.85D, 1.47D, 1.04D). D is correct (Bond enthalpies of N2 are much higher due to triple bond).
Predict the correct order among the following VSEPR repulsion strengths:
- A. lone pair- lone pair > lone pair - bond pair > bond pair - bond pair
- B. lone pair - lone pair > bond pair - bond pair > lone pair - bond pair
- C. bond pair - bond pair > lone pair - bond pair > lone pair - lone pair
- D. lone pair - bond pair > bond pair - bond pair > lone pair - lone pair
Answer: A. lone pair- lone pair > lone pair - bond pair > bond pair - bond pair
According to VSEPR theory, the order of repulsion is: lp-lp > lp-bp > bp-bp.
According to VSEPR theory, the order of repulsion is: lp-lp > lp-bp > bp-bp.