Question

The oxidation state of sulphur in $H_{2}S{O}_5$ and of chromium in $K_{2}C{r}_2{O}_7$ respectively is

• A. $8,6$
• B. $4,6$
• C. $8,8$
• D. $6,6$

Answer 1

Answer: (D)

(i) The structure of $H_{2}S{O}_5$ is $H-O-\underset{\begin{array}{c}||\\ O\end{array}}{\stackrel{\begin{array}{c}O\\ ||\end{array}}{S}}-O-O-H$ As it is clear by the structure that two oxygen atom has peroxide linkage and their oxidation state is -1. Thus, we can calculate oxidation number (state) of sulphur in $H_{2}S{O}_5$ as $\underset{(forH)}{2\times (+1)}+\underset{(forS)}{x}+\underset{(forO-O)}{2\times (-1)}$ $\underset{\begin{array}{c}(forother\\ threeOatoms)\end{array}}{+3\times (-2)=0}$ or $x=+6$ (ii) Oxidation state (number) of $Cr$ in $K_{2}C{r}_2{O}_7$ $\underset{(forK)}{2\times (+1)}\underset{(forCr)}{+2\times x}\underset{(forOatoms)}{+7\times (-2)=0}$ or $x=+6$