Question

Which of the following represent correct order?

• A. Polarity of $M-H$ bond: $P{H}_3<As{H}_3<Sb{H}_3<N{H}_3$
• B. Bond angle: $N{H}_3>As{H}_3>Sb{H}_3>Bi{H}_3$
• C. Bond angle: $N{O}_2^{+}>{NO}_2>N{O}_2^{-}$
• D. Stability: $H_2>H_2^{+}=H_2^{-}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Polarity of $M-H$ bond: $P{H}_3<As{H}_3<Sb{H}_3<N{H}_3$
B Stability: $H_2>H_2^{+}=H_2^{-}$
C Bond angle: $N{H}_3>As{H}_3>Sb{H}_3>Bi{H}_3$
D Bond angle: $N{O}_2^{+}>{NO}_2>N{O}_2^{-}$

All the four statements are correct statements.
(A) The increasing order of polarity is:
$P{H}_3\left(0\right)<As{H}_3\left(0.1\right)<Sb{H}_3\left(0.28\right)<N{H}_3\left(0.9\right)$
This is because lone pair and bond pair polarities almost cancel each other in case of phosphorus, while the bond pair start dominating in Sb and Bi. In nitrogen, lone pair and bond pair support the polarity in the same direction, hence, giving maximum polarity.
(B) The correct order of bond angles $N{H}_3>As{H}_3>Sb{H}_3>Bi{H}_3$
On moving down the group, the electonegativity decreases and the bond angle decreases.
(C) The decreasing order of bond angle is $N{O}_2^{+}>{NO}_2>N{O}_2^{-}$.
$N{O}_2^{+}$ molecule is linear like $C{O}_2$ molecule with bond angle of ${180}^o$. ${NO}_2$ is angular with bond angle of ${134}^o$. In $N{O}_2^{-}$, the bond angle is lower than ${134}^o$ $(i.e.{115}^o)$ due to repulsion between lone pair and bond pair of electrons.
(D) The correct order of stability is $H_2>H_2^{+}=H_2^{-}$
$H_2$ has bond order of 1. $H_2^{+}$ and $H_2^{-}$ have bond orders of 0.5 each. Higher the bond order, higher is the stability.