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