The correct increasing order of adjacent bond | vedclass

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(C) In $BF_3$,the central atom $B$ is $sp^2$ hybridized with a trigonal planar geometry,resulting in a bond angle of $120^{\circ}$.In $PF_3$,the centr

Vedclass · Accepted Answer

$ClF_3 < PF_3 < BF_3$

Vedclass · Answer

(C) In $BF_3$,the central atom $B$ is $sp^2$ hybridized with a trigonal planar geometry,resulting in a bond angle of $120^{\circ}$.In $PF_3$,the central atom $P$ is $sp^3$ hybridized with a trigonal pyramidal geometry. Due to the presence of one lone pair,the bond angle is compressed to approximately $96^{\circ}$.In $ClF_3$,the central atom $Cl$ is $sp^3d$ hybridized with a $T$-shaped geometry. Due to the presence of two lone pairs in the equatorial positions,the bond angle is further compressed to approximately $87^{\circ}$.Thus,the increasing order of bond angles is $ClF_3 (87^{\circ}) < PF_3 (96^{\circ}) < BF_3 (120^{\circ})$.

The correct increasing order of adjacent bond angle among $BF_3$,$PF_3$,and $ClF_3$ is:

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