The correct increasing order for bond angles | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$ClF_3 < PF_3 < BF_3$

Vedclass · Answer

(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 $ClF_3 < PF_3 < BF_3$.

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

Similar Questions

In the $BrF_3$ molecule,the equatorial lone pairs minimize which of the following?

Statement $I$ : $SO_2$ and $H_2O$ both possess $V$-shaped structure.
Statement $II$ : The bond angle of $SO_2$ is less than that of $H_2O$.
In the light of the above statements,choose the most appropriate answer from the options given below:

Which of the following is the incorrect order of bond angle?

The correct order of bond angle is:

The shape of a molecule which has $EB_5L_2$ molecular formula
{$E =$ Central atom,$B =$ Terminal atom,$L =$ Lone pair}

Vedclass Test Series

Exam Paper Generator

Online Exam Module