Select the pairs(s) in which bond angle of first member is higher than second one.

B r O_{3}

^{-}, C l O_{3}

^{-}

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A s I_{3}, S b I_{3}

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S b B r_{3}, S b I_{3}

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N F_{3}, N H_{3}

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Solution

The correct option is C $A s I_{3}, S b I_{3}$
(A) The bond angle decreases with decrease in electronegativities and increase in size of central atom. Hence, the correct order of $O - X - O$ bond angle is:
$C l O_{3}$ $^{-} > B r O_{3}^{-}$
$C l O_{3}$ $^{-}$ $107^{o}$ ; $B r O_{3}$ $^{-}$ $104^{o}$
(B) $A s I_{3}$ ${100.02}^{o}$ ; $S b I_{3}$ $90^{o}$
(C) $S b B r_{3}$ ${98.2}^{o}$ ; $S b l_{3}$ $99^{o}$
(D) $N H_{3}$ ${106.6}^{o}$ ; $N F_{3}$ ${102.2}^{o}$
Hence, the correct option is B.

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Similar questions

(X - A - X)

(A) O S F_{2}

O S C l_{2}

O S B r_{2}

(B) S b C l_{3}

S b B r_{3}

S b I_{3}

(C) P I_{3}

A s I_{3}

S b I_{3}

Iodide	$P I_{3}$	$A s I_{3}$	$S b I_{3}$
Bond angle	102	$100^{o} 2^{'}$	$99^{o}$

P I_{3}

C l O_{3}^{-}, B r O_{3}^{-}, I O_{3}^{-}

N H_{3}

N F_{3}

P H_{3}

P F_{3}

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