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 of repulsion among the following:
- A. lone pair-lone pair > bond pair-bond pair > lone pair-bond pair
- B. bond pair-bond pair > lone pair-bond pair > lone pair-lone pair
- C. lone pair-bond pair > bond pair-bond pair > lone pair-lone pair
- D. lone pair-lone pair > lone pair-bond pair > bond pair-bond pair
Answer: D. lone pair-lone pair > lone pair-bond pair > bond pair-bond pair
According to VSEPR theory, lone pair electrons occupy more space around the central atom than bond pair electrons, leading to the order of repulsion: lone pair-lone pair > lone pair-bond pair > bond pair-bond pair.
According to VSEPR theory, lone pair electrons occupy more space around the central atom than bond pair electrons, leading to the order of repulsion: lone pair-lone pair > lone pair-bond pair > bond pair-bond pair.
Consider the molecules CH4, NH3 and H2O. Which of the given statements is false?
- A. The H – O – H bond angle in H2O is larger than the H – C – H bond angle in CH4
- B. The H – O – H bond angle in H2O is smaller than the H – N – H bond angle in NH3
- C. The H – C – H bond angle in CH4 is larger than the H – C – H bond angle in NH3
- D. The H – C – H bond angle in CH4, the H – N – H bond angle in NH3, and the H – O – H bond angle in H2O are all greater than 90º
Answer: A. The H – O – H bond angle in H2O is larger than the H – C – H bond angle in CH4
Due to VSEPR theory and lone pair-bond pair repulsions, the bond angles decrease in the order CH4 (109.5°) > NH3 (107°) > H2O (104.5°). Thus, the H-O-H angle is smaller than the H-C-H angle, making statement A false.
Due to VSEPR theory and lone pair-bond pair repulsions, the bond angles decrease in the order CH4 (109.5°) > NH3 (107°) > H2O (104.5°). Thus, the H-O-H angle is smaller than the H-C-H angle, making statement A false.
Consider the molecules CH4, NH3 and H2O. Which of the given statements is false?
- A. The H – O – H bond angle in H2O is larger than the H – C – H bond angle in CH4
- B. The H – O – H bond angle in H2O is smaller than the H – N – H bond angle in NH3
- C. The H – C – H bond angle in CH4 is larger than the H – C – H bond angle in NH3
- D. The H – C – H bond angle in CH4, the H – N – H bond angle in NH3, and the H – O – H bond angle in H2O are all greater than 90º
Answer: A. The H – O – H bond angle in H2O is larger than the H – C – H bond angle in CH4
According to VSEPR theory, lone pair-lone pair repulsion > lone pair-bond pair repulsion > bond pair-bond pair repulsion. The bond angles are CH4 (109.5°) > NH3 (107°) > H2O (104.5°). Thus, option (1) is false.
According to VSEPR theory, lone pair-lone pair repulsion > lone pair-bond pair repulsion > bond pair-bond pair repulsion. The bond angles are CH4 (109.5°) > NH3 (107°) > H2O (104.5°). Thus, option (1) is false.
Match List-I with List-II. List-I (Compound) A. NH3 B. BrF5 C. XeF4 D. SF6 List-II (Shape/geometry) I. Trigonal pyramidal II. Square planar III. Octahedral IV. Square pyramidal Choose the correct answer from the options given below:
- A. A-III, B-IV, C-I, D-II
- B. A-II, B-III, C-IV, D-I
- C. A-I, B-IV, C-II, D-III
- D. A-II, B-IV, C-III, D-I
Answer: C. A-I, B-IV, C-II, D-III
NH3 is trigonal pyramidal; BrF5 is square pyramidal; XeF4 is square planar; SF6 is octahedral. Hence A-I, B-IV, C-II, D-III.
NH3 is trigonal pyramidal; BrF5 is square pyramidal; XeF4 is square planar; SF6 is octahedral. Hence A-I, B-IV, C-II, D-III.
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.
In the structure of ClF3, the number of lone pairs of electrons on central atom 'Cl' is:
- A. four
- B. one
- C. three
- D. two
Answer: D. two
Chlorine in ClF3 has 7 valence electrons; 3 are used for bonding with F atoms, leaving 4 electrons (2 lone pairs).
Chlorine in ClF3 has 7 valence electrons; 3 are used for bonding with F atoms, leaving 4 electrons (2 lone pairs).