The correct increasing order for bond angles | vedclass

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(C) To determine the bond angles,we analyze the hybridisation and molecular geometry of each molecule:$1$. $BF_3$: Boron is $sp^2$ hybridized with a t

Vedclass · Accepted Answer

$ClF_3 < PF_3 < BF_3$

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(C) To determine the bond angles,we analyze the hybridisation and molecular geometry of each molecule:$1$. $BF_3$: Boron is $sp^2$ hybridized with a trigonal planar geometry. The bond angle is $120^{\circ}$.$2$. $PF_3$: Phosphorus is $sp^3$ hybridized with one lone pair,resulting in a trigonal pyramidal geometry. Due to the lone pair-bond pair repulsion,the bond angle is approximately $97^{\circ}$.$3$. $ClF_3$: Chlorine is $sp^3d$ hybridized with two lone pairs,resulting in a $T$-shaped geometry. Due to the repulsion from two lone pairs,the bond angle is distorted to approximately $87.5^{\circ}$.Comparing these values: $87.5^{\circ} (ClF_3) Therefore,the correct increasing order is $ClF_3 < PF_3 < BF_3$.

$S_8$	$A$
$P_4$	$B$
$S_6$	$C$
$O_3$	$D$

The correct increasing order for bond angles among $BF_3$,$PF_3$ and $ClF_3$ is:

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