Question

Which one of the following molecules has higher bond angle?

• A. \(H_2O\)
• B. \(NH_3\)
• C. \(CH_4\)
• D. \(BF_3\)

Answer 1

The Correct Option is D

To determine which molecule has the highest bond angle, we need to analyze the molecular geometry and electronic structure of each molecule using the Valence Shell Electron Pair Repulsion (VSEPR) theory.

1. Understanding VSEPR Theory:
The VSEPR theory predicts the shape of a molecule based on the repulsion between electron pairs (both bonding and lone pairs) around the central atom. The bond angle is influenced by the following factors:
- Number of electron pairs around the central atom: More electron pairs lead to greater repulsion, resulting in smaller bond angles.
- Presence of lone pairs: Lone pairs exert more repulsive force than bonding pairs, leading to further distortion of bond angles.

2. Analyzing Each Molecule:

(1) $\text{H}_2\text{O}$ (Water):
- Central atom: Oxygen ($\text{O}$)
- Valence electrons: 6 (from oxygen) + 1 from each hydrogen = 8 electrons.
- Electron pair arrangement: Tetrahedral (4 electron pairs: 2 bonding pairs and 2 lone pairs).
- Molecular geometry: Bent due to the presence of two lone pairs.
- Bond angle: Approximately $104.5^\circ$.

(2) $\text{NH}_3$ (Ammonia):
- Central atom: Nitrogen ($\text{N}$)
- Valence electrons: 5 (from nitrogen) + 1 from each hydrogen = 8 electrons.
- Electron pair arrangement: Tetrahedral (4 electron pairs: 3 bonding pairs and 1 lone pair).
- Molecular geometry: Trigonal pyramidal due to the presence of one lone pair.
- Bond angle: Approximately $107^\circ$.

(3) $\text{CH}_4$ (Methane):
- Central atom: Carbon ($\text{C}$)
- Valence electrons: 4 (from carbon) + 1 from each hydrogen = 8 electrons.
- Electron pair arrangement: Tetrahedral (4 electron pairs: all bonding pairs).
- Molecular geometry: Tetrahedral.
- Bond angle: Exactly $109.5^\circ$.

(4) $\text{BF}_3$ (Boron Trifluoride):
- Central atom: Boron ($\text{B}$)
- Valence electrons: 3 (from boron) + 7 from each fluorine = 24 electrons.
- Electron pair arrangement: Trigonal planar (3 electron pairs: all bonding pairs).
- Molecular geometry: Trigonal planar.
- Bond angle: Exactly $120^\circ$.

3. Comparing Bond Angles:
- $\text{H}_2\text{O}$: $104.5^\circ$
- $\text{NH}_3$: $107^\circ$
- $\text{CH}_4$: $109.5^\circ$
- $\text{BF}_3$: $120^\circ$

4. Conclusion:
Among the given molecules, $\text{BF}_3$ has the highest bond angle of $120^\circ$.

Final Answer: ${D} $