Question

Place the following in order of increasing F-A-F bond angle, where A represents the central atom in each molecule. PF3 OF2 PF4

Answer 1

Answer:

Correct option : B

Explanation:

First compare $PF_3$ with $OF_2$

there are two lone pair in $OF_2$ but only one lone pair presen in the $PF_3$ so more lone pair means more repulsion and hence less bond bond angle beacuse of lone pair repulsion bonded atom come closer and hence bond angle decreased .

bond angle inversely proporional to repulsion so more repulsion means less bond angle

$bond \, angle \, : OF_2 \textless PF_3$

$PF_4$ does no exist it exist only in the form of $PF_4^+ \, or\, PF_4^-$ and here assuming $PF_4^+$ because in case of $PF_4^-$ difficult to compare bond angle because here two different type of bond angle is present.

$Compare \, PF_3 \, and \, PF_4^+$

there is no lone pair in the $PF_4^+$ so here will no be any repulsion and hence bond will be more in $PF_4^+$ as compared to $PF_3$

$bond \, angle \, : PF_3 \textless PF_4^+$

$bond \, angle \, : OF_2 \textless PF_3 \textless PF_4^+$