The correct order of bond angle is

P F_{3} = P C l_{3} = P B r_{3} = P I_{3}

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P F_{3} < P B r_{3} < P C l_{3} < P I_{3}

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P I_{3} < P B r_{3} < P C l_{3} < P F_{3}

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P F_{3} < P C l_{3} < P B r_{3} < P I_{3}

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Solution

The correct option is D $P F_{3} < P C l_{3} < P B r_{3} < P I_{3}$
The increasing order of the bond angles is $P F_{3} < P C l_{3} < P B r_{3} < P I_{3}$
As the electronegativity of the halogens decreases, the bond pair of electrons are more towards $P$ atom. This increases the electron-electron repulsion between lone pair and bond pairs. Hence, the bond angle increases.

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