The correct sequence of decreasing number of \(\pi\)-bonds in the structures of \(\mathrm{H_2SO_3}\), \(\mathrm{H_2SO_4}\) and \(\mathrm{H_2S_2O_7}\) is :
Show Hint
- \(\mathrm{H_2SO_3 > H_2SO_4 > H_2S_2O_7}\)
- \(\mathrm{H_2SO_4 > H_2S_2O_7 > H_2SO_3}\)
- \(\mathrm{H_2S_2O_7 > H_2SO_4 > H_2SO_3}\)
- \(\mathrm{H_2S_2O_7 > H_2SO_3 > H_2SO_4}\)
The Correct Option is C
Solution and Explanation
\(\mathrm{H_2SO_4} = \mathrm{(HO)_2S(=O)_2}\) has 2 S=O, so 2 \(\pi\)-bonds.
\(\mathrm{H_2S_2O_7} = \mathrm{(HO)(O=)_2S-O-S(=O)_2(OH)}\) has 4 S=O, so 4 \(\pi\)-bonds.
Decreasing order: \(\mathrm{H_2S_2O_7\ (4) > H_2SO_4\ (2) > H_2SO_3\ (1)}\).
Answer: (3) \(\mathrm{H_2S_2O_7 > H_2SO_4 > H_2SO_3}\)
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