The decreasing values of bond angles from $$NH_3$$ $$(106^\circ)$$ to $$SbH_3$$ $$(101^\circ)$$ down group-15 of the periodic table is due to

In all the hydrides $$EH_3$$ of group-15 ($$E = N,\,P,\,As,\,Sb$$) the central atom is $$sp^3$$ hybridised and carries one lone pair (lp) and three bond pairs (bp). For an ideal $$sp^3$$ arrangement the bond angle is $$109.5^\circ$$, but lp-bp repulsion contracts it. Experimentally the bond angles are

$$NH_3:106^\circ \; \gt \; PH_3:94^\circ \; \gt \; AsH_3:92^\circ \; \gt \; SbH_3:101^\circ$$ (average $$\approx 101^\circ$$ for $$SbH_3$$).

The decisive factor is the electronegativity of the central atom. Down the group the electronegativity decreases: $$N(3.0) \gt P(2.1) \gt As(2.0) \gt Sb(1.9)$$.

• A more electronegative central atom (as in $$NH_3$$) attracts the bonding electrons towards itself; the bp electron clouds are pulled closer to the nucleus. This increases bp-bp and lp-bp repulsions, keeping the hydrogen atoms further apart and producing a larger $$\angle H\!-\!E\!-\!H$$.

• As electronegativity falls (from $$NH_3$$ to $$SbH_3$$) the bonding electrons are held less tightly. The bp electron clouds spread out away from the central atom towards the hydrogens. Repulsion between the bp clouds therefore diminishes, allowing the three $$E-H$$ bonds to come closer together, and the H-E-H bond angle decreases.

Hence the progressive decrease of bond angle in $$EH_3$$ is primarily due to the decreasing electronegativity of the central atom.

Option D which is: decreasing electronegativity