Decreasing order of bond angle is

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

Question

(A) To determine the bond angle,we analyze the hybridization and geometry of each molecule:$1$. $BeCl_2$: It has $sp$ hybridization and a linear geome

Vedclass · Accepted Answer

$BeCl_2 > NO_2 > SO_2$

Vedclass · Answer

(A) To determine the bond angle,we analyze the hybridization and geometry of each molecule:$1$. $BeCl_2$: It has $sp$ hybridization and a linear geometry with a bond angle of $180^{\circ}$.$2$. $NO_2$: It has $sp^2$ hybridization with one unpaired electron on the central nitrogen atom. The bond angle is approximately $134^{\circ}$.$3$. $SO_2$: It has $sp^2$ hybridization with one lone pair on the central sulfur atom. The lone pair-bond pair repulsion is greater than the single electron-bond pair repulsion,leading to a smaller bond angle of approximately $119^{\circ}$.Therefore,the decreasing order of bond angle is $BeCl_2 (180^{\circ}) > NO_2 (134^{\circ}) > SO_2 (119^{\circ})$.

Similar Questions

Which of the following represents the correct increasing order of bond angles for the given species?

In the $OF_2$ molecule,what are the number of bonding electron pairs and lone pairs of electrons present,respectively?

Identify the correct statement regarding geometry and lone pair of electrons present in $CH_4$ and $SiCl_4$.

$A$ molecule of the type $AX_5$ has square pyramidal geometry. Hence,the number of lone pairs on '$A$' is:

The correct order of increasing bond angles is

Vedclass Test Series

Exam Paper Generator

Online Exam Module