The bond angle of water is 104.5^{circ} due t | vedclass

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

Question

(A) In $H_2O$,the oxygen atom undergoes $sp^3$ hybridization,which would ideally result in a tetrahedral geometry with a bond angle of $109.5^{\circ}$

Vedclass · Accepted Answer

Repulsion between lone pair and bond pair

Vedclass · Answer

(A) In $H_2O$,the oxygen atom undergoes $sp^3$ hybridization,which would ideally result in a tetrahedral geometry with a bond angle of $109.5^{\circ}$.However,the presence of two lone pairs on the oxygen atom exerts strong repulsion on the two $O-H$ bond pairs.According to the $VSEPR$ theory,the order of repulsion is $lp-lp > lp-bp > bp-bp$.Due to the strong $lp-lp$ repulsion,the bond angle is compressed from the ideal $109.5^{\circ}$ to $104.5^{\circ}$.

The bond angle of water is $104.5^{\circ}$ due to

Similar Questions

Identify the molecule with linear geometry.

In which of the following pairs,the central atoms have the same number of lone pairs of electrons?

What is the geometry of the $IBr^{-}_{2}$ ion?

If lone pairs of electrons occupy axial positions,then which of the following statements about $ClF_3$ is incorrect?

The number of bond pairs of electrons and the total number of lone pairs of electrons in $XeOF_4$ are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module