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Wedge and Dash

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Question

Choose the correct increasing order of bond angles in $H_{2} S, N H_{3}, B F_{3}$ and $S i H_{4}$ :

A

H_{2} S < S i H_{4} < N H_{3} < B F_{3}

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B

N H_{3} < H_{2} S < S i H_{4} < B F_{3}

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C

H_{2} S < N H_{3} < B F_{3} < S i H_{4}

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D

H_{2} S < N H_{3} < S i H_{4} < B F_{3}

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Solution

The correct option is D $H_{2} S < N H_{3} < S i H_{4} < B F_{3}$
The repulsion order is: $l o n e p a i r - l o n e p a i r > b o n d p a i r - l o n e p a i r > b o n d p a i r - b o n d p a i r$

$H_{2} S$ - has 2 lone pair of electrons and has a bent shape.

$N H_{3}$ - $s p^{3}$ hybridised, pyramidal shape

$S i H_{4}$ - $s p^{3}$ hybridised, tetrahedral shape

$B F_{3}$ - $s p^{2}$ hybridised, trigonal planar shape

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