Trend in bond angle down group 15

The hydrides are: $NH_3,\, PH_3,\, AsH_3,\, SbH_3$

Their bond angles decrease approximately as: $106^\circ > 93.5^\circ > 92^\circ > 91^\circ$

Shape and electron pair arrangement

Each molecule has the configuration: $AX_3E$ i.e. one lone pair and three bond pairs, so the shape is trigonal pyramidal.

Key factor controlling bond angle

In $NH_3$, the central atom is relatively small and hybridisation is close to $sp^3$. The lone pair is more concentrated and exerts significant lone pair-bond pair repulsion, keeping the bond angle larger.

Down the group, the central atom becomes larger and hybridisation becomes less effective. The lone pair becomes more diffuse/inert, so its repulsion on bond pairs decreases.

Hence, the lp-bp repulsion decreases, causing the $H-M-H$ bond angle to decrease toward $90^\circ$.

Check the options

• A: increasing bp-bp repulsion
  If bond pair-bond pair repulsion increased, bond angle would not decrease in this manner. Incorrect.

• B: increasing p-orbital character in $sp^3$
  Down the group, bonding does gain more pure $p$ character, and pure $p$ orbitals tend to give angles near $90^\circ$. This is a contributing description often used, but among the given options for the observed decrease in these pyramidal hydrides, the standard explanation emphasizes reduced lone pair-bond pair repulsion. For this question, this is not the best choice.

• C: decreasing lp-bp repulsion
  This directly explains the decrease in bond angle from $NH_3$ to $SbH_3$. Correct.

• D: decreasing electronegativity
  This is not the primary reason for the observed bond angle trend here. Incorrect.

Final answer

$\boxed{\text{C: decreasing lp-bp repulsion}}$