Question

Which of the following orders of bond angle is correct?

• A. B(OH)₃ < B(OMe)₃ (BOX, where X = H, Me)
• B. NH₄⁺ < NH₃ (HNH)
• C. BF₃ < BCl₃ (XBX)
• D. SiCl₄ < CCl₄ (ClACl where A = Si, C)

Answer 1

Answer: (D)

SiCl₄ < CCl₄ (ClACl where A = Si, C)

Answer 2

The question asks to identify the correct order of bond angles among the given options. We will analyze each option based on VSEPR theory, hybridization, and factors influencing bond angles (lone pair repulsion, electronegativity of central/terminal atoms, steric hindrance).

Key Concepts:

1. VSEPR Theory: Predicts molecular geometry and approximate bond angles based on minimizing electron pair repulsion.
2. Hybridization: Determines the ideal bond angles (e.g., sp³=109.5°, sp²=120°, sp=180°).
3. Lone Pair Repulsion: Lone pair-lone pair > lone pair-bond pair > bond pair-bond pair. Lone pairs compress bond angles.
4. Electronegativity of Central Atom: If the electronegativity of the central atom decreases (moving down a group), the bond pair electrons are less attracted to the central atom and are pushed more towards the terminal atoms. This effectively increases the repulsion between bond pairs, leading to a smaller bond angle. (e.g., H₂O > H₂S, NH₃ > PH₃).
5. Electronegativity of Terminal Atoms: If the electronegativity of the terminal atoms increases, the bond pair electrons are pulled closer to the terminal atoms, reducing the repulsion between bond pairs, leading to a smaller bond angle. (e.g., PCl₃ > PF₃).
6. Steric Hindrance: Bulky groups can increase bond angles due to repulsion.

Let's evaluate each option:

(A) B(OH)₃ < B(OMe)₃ (BOX, where X = H, Me)

• The notation (BOX) typically refers to the angle around the central boron atom, i.e., O-B-O.
• In both B(OH)₃ and B(OMe)₃, the central boron atom is sp² hybridized, and the molecules are trigonal planar.
• Therefore, the O-B-O bond angle in both cases is ideally 120°.
• If the angles are 120° for both, then the inequality B(OH)₃ < B(OMe)₃ is incorrect.
• Alternatively, if (BOX) refers to the B-O-X angle:
  • In B(OH)₃, the B-O-H angle involves an sp³ hybridized oxygen with two lone pairs. The angle is approximately 109.5°.
  • In B(OMe)₃, the B-O-CH₃ angle involves an sp³ hybridized oxygen with two lone pairs. The methyl group (-CH₃) is bulkier than hydrogen (-H). Due to increased steric repulsion between the bulkier methyl group and the boron atom, the B-O-CH₃ angle would be slightly larger than the B-O-H angle.
  • So, B-O-H < B-O-CH₃ would be correct.
  • However, the most common interpretation of "bond angle" in such a comparison is the angle around the primary central atom (Boron). Given the common interpretation, this option is incorrect.

(B) NH₄⁺ < NH₃ (HNH)

• NH₃ (Ammonia): The central nitrogen atom has 3 bond pairs and 1 lone pair. It is sp³ hybridized. The lone pair-bond pair repulsion compresses the H-N-H bond angle from the ideal 109.5° to approximately 107°.
• NH₄⁺ (Ammonium ion): The central nitrogen atom has 4 bond pairs and 0 lone pairs. It is sp³ hybridized. The geometry is perfectly tetrahedral, and the H-N-H bond angle is the ideal 109.5°.
• Therefore, NH₄⁺ (109.5°) > NH₃ (107°).
• The given order NH₄⁺ < NH₃ is incorrect.

(C) BF₃ < BCl₃ (XBX)

• In both BF₃ and BCl₃, the central boron atom is sp² hybridized, and the molecules are trigonal planar.
• The X-B-X (F-B-F or Cl-B-Cl) bond angle in both cases is ideally 120°.
• While back-bonding (pπ-pπ overlap from halogen to boron) is more significant in BF₃ than in BCl₃, this primarily affects bond length and strength, not the ideal 120° angle of the trigonal planar geometry which is maintained due to symmetry and lack of lone pairs.
• Therefore, BF₃ = BCl₃ = 120°.
• The given order BF₃ < BCl₃ is incorrect.

(D) SiCl₄ < CCl₄ (ClACl where A = Si, C)

• Both CCl₄ and SiCl₄ have a central atom (C or Si) with 4 bond pairs and 0 lone pairs. Both are sp³ hybridized and have a tetrahedral geometry.
• The ideal Cl-A-Cl bond angle for a tetrahedral molecule is 109.5°.
• However, subtle deviations can occur based on the electronegativity of the central atom.
• Rule: When comparing molecules with the same number of bond pairs and lone pairs, if the central atom belongs to a higher period (i.e., less electronegative, moving down a group), the bond angle generally decreases. This is because the bond pair electrons are less attracted to the central atom and are pushed more towards the terminal atoms. This effectively increases the repulsion between bond pairs (as they are further from the central atom), leading to a smaller bond angle.
• Electronegativity of C (2.55) > Electronegativity of Si (1.90).
• Applying the rule: Since C is more electronegative than Si, CCl₄ should have a larger bond angle than SiCl₄.
• Thus, SiCl₄ < CCl₄ is the correct order.
• (Actual values are very close to 109.5°, but this general trend holds for subtle differences).