The correct order of increasing bond angles i | vedclass

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Question

(B) The bond angles for the given molecules are as follows:$OF_2$: $103^\circ$$H_2O$: $104.5^\circ$$Cl_2O$: $111^\circ$$ClO_2$: $117.5^\circ$In $OF_2$

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$OF_2 < H_2O < Cl_2O < ClO_2$

Vedclass · Answer

(B) The bond angles for the given molecules are as follows:$OF_2$: $103^\circ$$H_2O$: $104.5^\circ$$Cl_2O$: $111^\circ$$ClO_2$: $117.5^\circ$In $OF_2$,$H_2O$,and $Cl_2O$,the central oxygen atom is $sp^3$ hybridized with $2$ lone pairs. The bond angle decreases as the electronegativity of the surrounding atom increases because the bonding electron pairs are pulled closer to the central atom,increasing lone pair-lone pair repulsion.$Cl_2O$ has a larger bond angle than $H_2O$ and $OF_2$ due to the steric repulsion between the bulky chlorine atoms.$ClO_2$ has a different electronic structure (it is a radical with an odd electron on the central atom),which results in a larger bond angle compared to the others.Thus,the increasing order of bond angles is: $OF_2 < H_2O < Cl_2O < ClO_2$.

The correct order of increasing bond angles is

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