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 \(PH _{4}{ }^{+}\).

Correct answer is (A) : Lone pair - bond pair repulsion exists in $PH_3$