Question

Bond angle in $PH^+_4$ is more than that of $PH_3$. This is because

• A. Lone pair - bond pair repulsion exists in $PH_3$
• B. $PH^+_4$ has square planar structure
• C. $PH_3$ has planar trigonal structure
• D. Hybridisation of P changes when $PH_3$ is converted to $PH^+_4$

Answer 1

The Correct Option is 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, $PH _{3}$ will not exhibit hybridization and here the bond formation takes place due to the overlap of pure $p$-orbitals and $s$-orbitals. $PH _{3}$ has a lone pair on the central $P$ atom, which is absent in $PH _{4}{ }^{+}$. Thus in $PH _{3}$, there will be bond pair - lone pair repulsion and this is the reason why the bond angle in $PH _{3}$ is less than that of $PH _{4}{ }^{+}$.