Arrange the following hydrides in the order of decreasing bond angle:
(A) SbH$$_3$$
(B) AsH$$_3$$
(C) PH$$_3$$

Question

Arrange the following hydrides in the order of decreasing bond angle:
(A) SbH$$_3$$
(B) AsH$$_3$$
(C) PH$$_3$$
(D) NH$$_3$$

Accepted Answer

1. **Central Atom Trend:** All these hydrides ($MH_3$) involve Group 15 elements (N, P, As, Sb). They all have a pyramidal structure with one lone pair of electrons.
2. **Bond Angle Trend:** The bond angle depends on the electronegativity and size of the central atom. As we move down the group from Nitrogen to Antimony, the electronegativity decreases and atomic size increases.
3. **Repulsion:** In $NH_3$, nitrogen is small and highly electronegative, pulling the bonding electron pairs closer to itself. This creates greater repulsion between the bond pairs, opening up the bond angle to **107.8°** .
4. **Heavier Analogues:** For $PH_3$, $AsH_3$, and $SbH_3$, the bonding pairs are further away from the central atom due to lower electronegativity and larger size. This reduces bond pair-bond pair repulsion. Additionally, Drago's rule suggests that for heavier elements (P, As, Sb), the s-orbital contribution in bonding is minimal (bonds are formed primarily by p-orbitals), resulting in bond angles closer to **90°**. 
   - Specific angles: $PH_3$ (~93.6°) > $AsH_3$ (~91.8°) > $SbH_3$ (~91.3°).
5. **Conclusion:** The correct order of decreasing bond angle is $NH_3 > PH_3 > AsH_3 > SbH_3$, which corresponds to **D > C > B > A**.

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