How will you account for the 104.5^{circ} bon | vedclass

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

Question

(N/A) In water $(H_2O)$,the oxygen atom undergoes $sp^3$ hybridization,which theoretically predicts a tetrahedral geometry with a bond angle of $109^{

Vedclass · Accepted Answer

Vedclass · Answer

(N/A) In water $(H_2O)$,the oxygen atom undergoes $sp^3$ hybridization,which theoretically predicts a tetrahedral geometry with a bond angle of $109^{\circ} 28^{\prime}$.
However,the oxygen atom in water is bonded to two hydrogen atoms and also possesses two lone pairs of electrons.
According to the $VSEPR$ theory,the repulsion between lone pair-lone pair $(lp-lp)$ is stronger than the repulsion between lone pair-bond pair $(lp-bp)$,which in turn is stronger than bond pair-bond pair $(bp-bp)$ repulsion.
Due to the strong repulsion exerted by the two lone pairs on the two $O-H$ bond pairs,the bond angle is compressed from the ideal tetrahedral angle of $109^{\circ} 28^{\prime}$ to $104.5^{\circ}$.

How will you account for the $104.5^{\circ}$ bond angle in water?

Similar Questions

Which among the following species does not have an angular shape?
$I.$ $ICl_2^+$
$II.$ $NH_2^-$
$III.$ $CS_2$
$IV.$ $H_2Se$

The correct order of bond angles of the following compounds is

Consider the following reactions:
$2 Cu_2S + 3 O_2 \rightarrow 2 Cu_2O + 2 SO_2$
$2 Cu_2O + Cu_2S \rightarrow 6 Cu + SO_2 \uparrow$
Here,$X$ is $Cu_2S$ and $Y$ is $SO_2$. The shape of the molecule $Y$ is:

Which of the following species has tetrahedral geometry?

The molecular shapes of $SF_4$,$CF_4$,and $XeF_4$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module