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 $PH_3$

Answer 2

In PH$_3$, phosphorus is sp$^3$ hybridized with a lone pair of electrons, resulting in a trigonal pyramidal structure. The bond angle in PH$_3$ is less than 109.5° due to the lone pair-bond pair repulsion, which pushes the bonded atoms closer together.
When PH$_3$ is converted to PH$_4^+$, the lone pair on phosphorus is removed, leading to sp$^3$ hybridization of phosphorus and a tetrahedral structure. In a tetrahedral structure, the bond angle is 109.5°, which is greater than that in PH$_3$.

Therefore, the bond angle in PH$_4^+$ is more than that of PH$_3$ because the hybridization of phosphorus changes when PH$_3$ is converted to PH$_4^+$.