Compare $F -^B r - O$ and $O -^B r - O$ in $F B r O_{3}$ molecule.

F -^B r - O > O -^B r - O

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F -^B r - O < O -^B r - O

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F -^B r - O = O -^B r - O

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None of these

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Solution

The correct option is D $F -^B r - O < O -^B r - O$
In molecule, $F -^B r - O$ bond angle is less than $O -^B r - O$ bond angle.
$F -^B r - O < O -^B r - O$
$F -^B r - O$ bond angle is $103^{0} C$ whereas $O -^B r - O$ bond angle is $115^{0} C$ . This is in accordance with Bent's rule. Since $F$ is more electronegative, the bond pair of electrons is more towards $F$ than towards $O$ . hence, the bond pair of electrons of $B r - F$ bond experiences less repulsion than the bond pair of electrons of $B r - O$ bond.

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The radii of $F, F^{-}, O$ and $O^{2 -}$ are in the order of

The radii of $F, F^{-}, O$ and $O^{- 2}$ are in the order of

O_{2} F_{2}

X - O - X

X - O - X

(X = H, F o r C l)

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