Q.(a) Bond angle in PH4+ is higher than that in PH3. Why?

Part (a) — Bond angle comparison.

In $PH_3$, phosphorus has 3 bond pairs + 1 lone pair. Lone-pair–bond-pair (LP–BP) repulsion is stronger than bond-pair–bond-pair (BP–BP) repulsion, so the lone pair pushes the three P–H bonds closer together, compressing the H–P–H angle to about 93.5° (well below the tetrahedral 109.5°).

In $PH_4^+$ (formed by protonation of $PH_3$'s lone pair), phosphorus now has 4 bond pairs and no lone pair. With only BP–BP repulsions (weaker, and symmetric), the ion adopts a regular tetrahedral geometry with the ideal bond angle of 109.5°.

Hence $PH_4^+$'s bond angle > $PH_3$'s bond angle, because the lone pair present in $PH_3$ (and absent in $PH_4^+$) compresses the angle.

Part (b) — Reaction with an acid.

$PH_3$'s lone pair accepts a proton from a strong acid such as HI:

$PH_3 + HI \rightarrow PH_4I$ …

For (a), compare the number of lone pairs on P in the two species and apply the VSEPR repulsion order (LP–BP > BP–BP) to explain the angle difference. For (b), recall that $PH_3$'s lone pair lets it act as …

• Don't say $PH_3$'s angle is close to 109.5° "because it's like $NH_3$" — $PH_3$'s angle (93.5°) is actually much more compressed than $NH_3$'s (107°) because P's larger size gives more s-character retention/weaker hybridisation effects; don't just copy the $NH_3$ figure. …