JEE Main 2017 · 9 Apr · Shift 1 · Chemistry Q49

ChemistrySingle correct+4 / -1Question 49 of 90

The correct sequence of decreasing number of π-bonds in the structures of $\ce{H2SO3}$ , $\ce{H2SO4}$ and $\ce{H2S2O7}$ is :

\ce{H2SO3}>\ce{H2SO4}>\ce{H2S2O7}

\ce{H2SO4}>\ce{H2S2O7}>\ce{H2SO3}

\ce{H2S2O7}>\ce{H2SO4}>\ce{H2SO3}

\ce{H2S2O7}>\ce{H2SO3}>\ce{H2SO4}

Answer and worked solution

Official answer: C

Counting S=O bonds: $\ce{H2S2O7}$ has 4, $\ce{H2SO4}$ has 2 and $\ce{H2SO3}$ has 1. So $\ce{H2S2O7}>\ce{H2SO4}>\ce{H2SO3}$ .

JEE Main 2017 · 9 Apr · Shift 1, Chemistry question 49