Question

# The correct order of bond angles (smallest first) for H2S,NH3,BF3 and SiH4 is:

A
H2S<SiH4<NH3<BF3
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B
NH3<H2S<SiH4<BF3
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C
H2S<NH3<SiH4<BF3
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D
H2S<NH3<BF3<SiH4
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Solution

## The correct option is C H2S<NH3<SiH4<BF3BF3 is sp2 hybridised with 3 bond pairs and hence, the bond angle is 120∘. SiH4 and NH3 are sp3 hybridised. NH3 has one lone pair whereas SiH4 has none. So, the bond angle of NH3 is smaller than that of SiH4 because, the lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion. If the central atom belongs to the groups 13, 14, 15 or 16 (and is from third or higher period) and is surrounded by atoms having electronegativity less than 2.5, it does not involve hybridization. The lone pair of electrons are present in the pure s-orbital of the atom. Hence, the bond angle is around 90o for such compounds. They are called Drago's molecules. H2S is drago's molecule and does not involve hybridization. Hence, the bond angle is 92.6∘ Therefore, the correct order of bond angle is: H2S<NH3<SiH4<BF3 92.6∘<107∘<109∘28′<120∘

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