Question

The correct order of bond angle of $HgCl_2 (A)$, $NH_3 (B)$, $H_2O (C)$ is

• A. $A>C>B$
• B. $B>A>C$
• C. $B>C>A$
• D. $A>B>C$

Hint:

{VSEPR Rule:} Repulsion order is Lone Pair-Lone Pair > Lone Pair-Bond Pair > Bond Pair-Bond Pair. More lone pairs on the central atom result in smaller bond angles.

Answer 1

The Correct Option is D

Step 1: Determine Molecular Geometry and Hybridization:

• $HgCl_2$ (Molecule A):
  • Central atom: Hg (Group 12, 2 valence $e^-$).
  • Bonding: 2 bond pairs, 0 lone pairs.
  • Hybridization: $sp$.
  • Geometry: Linear.
  • Bond Angle: $\mathbf{180^\circ}$.
• $NH_3$ (Molecule B):
  • Central atom: N (Group 15, 5 valence $e^-$).
  • Bonding: 3 bond pairs, 1 lone pair.
  • Hybridization: $sp^3$.
  • Geometry: Trigonal Pyramidal.
  • Bond Angle: Reduced from $109.5^\circ$ to $\mathbf{107^\circ}$ due to lp-bp repulsion.
• $H_2O$ (Molecule C):
  • Central atom: O (Group 16, 6 valence $e^-$).
  • Bonding: 2 bond pairs, 2 lone pairs.
  • Hybridization: $sp^3$.
  • Geometry: Bent / V-shape.
  • Bond Angle: Reduced further to $\mathbf{104.5^\circ}$ due to greater lp-lp repulsion.

Step 2: Order Comparison:
$180^\circ (A)>107^\circ (B)>104.5^\circ (C)$.
Step 3: Final Answer:
Order is $A>B>C$.