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JEE Main2017Chemistryp Block Elements (Group 15, 16, 17 & 18)Actual

The correct sequence of decreasing number of π -bonds in the structure of H 2 S O 3 , H 2 S O 4 and H 2 S 2 O 7 is:

Options

  1. AH 2 S 2 O 7 > H 2 S O 3 > H 2 S O 4
  2. BH 2 S 2 O 7 > H 2 S O 4 > H 2 S O 3
  3. CH 2 S O 4 > H 2 S 2 O 7 > H 2 S O 3
  4. DH 2 S O 3 > H 2 S O 4 > H 2 S 2 O 7

Correct answer

B. H 2 S 2 O 7 > H 2 S O 4 > H 2 S O 3

Step-by-step solution

Number of π -bonds H 2 S 2 O 7 = HO − S | | | | O O − O − S | | | | O O − OH 4 H 2 So 4 = OH − S | | | | O O − OH 2 H 2 SO 3 = HO − S . . | | O − OH 1 Therefore, the correct sequence of decreasing number of pi-bonds is H 2 S 2 O 7 ​ ( 4 )   > H 2 SO 4   ( 2 )   > H 2 SO 3 ​ ( 1 ) . The numbers in bracket indicate the number of pi bonds.

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