Question

Bond angle in $P{H}_4^{+}$ is more than that of $P{H}_3$. This is because

• A. Lone pair - bond pair repulsion exists in $P{H}_3$
• B. $P{H}_4^{+}$ has square planar structure
• C. $P{H}_3$ has planar trigonal structure
• D. Hybridisation of P changes when $P{H}_3$ is converted to $P{H}_4^{+}$

Answer 1

Answer: (A)

The hydrides of group $15,16$ below the $3^{rd}$ period, follows Drago's rule. The rule states that due to a large energy difference between the atomic orbitals, these compounds do not exhibit hybridization. Thus, $P{H}_3$ will not exhibit hybridization and here the bond formation takes place due to the overlap of pure $p$-orbitals and $s$-orbitals. $P{H}_3$ has a lone pair on the central $P$ atom, which is absent in $P{H}_4{}^{+}$. Thus in $P{H}_3$, there will be bond pair - lone pair repulsion and this is the reason why the bond angle in $P{H}_3$ is less than that of $P{H}_4{}^{+}$.