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:

JEE Main 20179 AprilOnline

Exam: JEE Main
Year: 2017
Date: 9 April
Mode: Online
Subject: Chemistry
Chapter: p Block Elements (Group 15, 16, 17 & 18)

JEE Main 2017 · 9 April · Online · Chemistry · p Block Elements (Group 15, 16, 17 & 18)

The correct sequence of decreasing number of π -bonds in the...

The correct sequence of decreasing number of

\pi

-bonds in the structure of

{H}_{2}S{O}_{3}, {H}_{2}S{O}_{4}

and

{H}_{2}{S}_{2}{O}_{7}

is:

Options

A ${\text{H}}_{2}{\text{S}}_{2}{\text{O}}_{7} > {\text{H}}_{2}\text{S}{\text{O}}_{3} > {\text{H}}_{2}\text{S}{\text{O}}_{4}$
B ${\text{H}}_{2}{\text{S}}_{2}{\text{O}}_{7} > {\text{H}}_{2}\text{S}{\text{O}}_{4} > {\text{H}}_{2}\text{S}{\text{O}}_{3}$
C ${\text{H}}_{2}\text{S}{\text{O}}_{4} > {\text{H}}_{2}{\text{S}}_{2}{\text{O}}_{7} > {\text{H}}_{2}\text{S}{\text{O}}_{3}$
D ${\text{H}}_{2}\text{S}{\text{O}}_{3} > {\text{H}}_{2}\text{S}{\text{O}}_{4} > {\text{H}}_{2}{\text{S}}_{2}{\text{O}}_{7}$

Correct answer

B. ${\text{H}}_{2}{\text{S}}_{2}{\text{O}}_{7} > {\text{H}}_{2}\text{S}{\text{O}}_{4} > {\text{H}}_{2}\text{S}{\text{O}}_{3}$

Step-by-step solution

Number of $\pi$ -bonds

${H}_{2}{S}_{2}{O}_{7} = HO - \mathop{\mathop{\overset{| |}{S}}_{| |}O}_{O} - O - \mathop{\mathop{\overset{| |}{S}}_{| |}O}_{O} - OH$ $4$

${H}_{2}{So}_{4} = OH - \mathop{\mathop{\overset{| |}{S}}_{| |}O}_{O} - OH$ $2$

${H}_{2}{SO}_{3} = HO - \mathop{\mathop{\overset{. .}{S}}_{| |}}_{O} - OH$ $1$

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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