The compound that has the largest H-M-H bond angle (M = N, O, S, C) is :
Show Hint
- \(\mathrm{CH_4}\)
- \(\mathrm{H_2S}\)
- \(\mathrm{NH_3}\)
- \(\mathrm{H_2O}\)
The Correct Option is A
Solution and Explanation
\(\mathrm{CH_4}\): sp3, no lone pair, H-C-H = 109.5 degrees
\(\mathrm{NH_3}\): one lone pair, H-N-H about 107 degrees
\(\mathrm{H_2O}\): two lone pairs, H-O-H about 104.5 degrees
\(\mathrm{H_2S}\): two lone pairs; S is larger and less electronegative, so the bonding pairs stay closer to H and the bond angle drops to about 92 degrees.
The largest angle is therefore in \(\mathrm{CH_4}\).
Answer: (1) \(\mathrm{CH_4}\)
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