Answer:
Correct option : B
Explanation:
First compare [tex]PF_3[/tex] with [tex]OF_2[/tex]
there are two lone pair in [tex]OF_2[/tex] but only one lone pair presen in the [tex]PF_3[/tex] 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
[tex]bond \, angle \, : OF_2 \textless PF_3[/tex]
[tex]PF_4[/tex] does no exist it exist only in the form of [tex]PF_4^+ \, or\, PF_4^-[/tex] and here assuming [tex]PF_4^+[/tex] because in case of [tex]PF_4^-[/tex] difficult to compare bond angle because here two different type of bond angle is present.
[tex]Compare \, PF_3 \, and \, PF_4^+[/tex]
there is no lone pair in the [tex]PF_4^+[/tex] so here will no be any repulsion and hence bond will be more in [tex]PF_4^+[/tex] as compared to [tex]PF_3[/tex]
[tex]bond \, angle \, : PF_3 \textless PF_4^+[/tex]
[tex]bond \, angle \, : OF_2 \textless PF_3 \textless PF_4^+[/tex]
