Correct order of bond angles for the following is:

N H_{3} > P C l_{3} > B C l_{3}

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B C l_{3} > C H_{4} > P C l_{3}

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B C l_{3} > P C l_{3} > N H_{3}

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P C l_{3} > B C l_{3} > N H_{3}

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Solution

The correct option is B $B C l_{3} > C H_{4} > P C l_{3}$
$B C l_{3}$ is $s p^{2}$ hybridised $C H_{4}$ is $s p^{3}$ hybridised $P C l_{3}$ is $s p^{3}$ hybridised.

As the character of p is more the bond angle of central atom is less.

However $C H_{4}$ and $P C l_{3}$ are having same p character but the lone pair on $P C l_{3}$ reduces the regular bond angle.

So the correct order is $B C l_{3} > C H_{4} > P C l_{3}$

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Question 49

Predict the shapes of the following molecules on the basis of hybridisation.

$B C l_{3}, C H_{4}, C O_{2}, N H_{3}$

N H_{3}

P C l_{5}

P C l_{3}

The correct order of hybridisation of the central atom in the following species $N H_{3}$ , ${[P t C l_{4}]}^{2 -}$ , $P C l_{5}$ and $B C l_{3}$

N H_{3}, [P t C l_{4}]^{2 -}, P C l_{5}

B C l_{3}

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