Question

In how many of the following compounds of sulphur, the oxidation state of sulphur atom is $+6$?
$\mathrm{H}{}_2\mathrm{S}{}_2\mathrm{O}{}_8,\mathrm{H}{}_2\mathrm{SO}{}_5,\mathrm{H}{}_2\mathrm{SO}{}_3,\mathrm{H}{}_2\mathrm{S}{}_2\mathrm{O}{}_7,\mathrm{SO}{}_2\mathrm{Cl}{}_2,\mathrm{SOCl}{}_2$

• A. 3
• B. 5
• C. 4
• D. 6

Answer 1

Answer: (C)

The oxidation state of sulphur in the given compounds can be found through their respective structures which are as follows:

(a) $H_2{S}_2{O}_8$



We can calculate oxidation state sulphur as:

$x+2(-2)+1(-1)+1(-1)=0$

$x=+6$

Thus, the sulphur is in $+6$ oxidation state.

(b) $H_{2}S{O}_5$



$\therefore x+2(-2)+1(-1)+1(-1)=0$

$x=+6$

Thus, the sulphur is in $+6$ oxidation state.

(c) $H_{2}S{O}_3$



$x+2(-1)+1(-2)=0$

$x=+4$

Thus, sulphur is present in $+4$ oxidation state.

(d) $H_{2}S{O}_4$



$x+2(-1)+1(-2)=0$

$x=+4$

Thus, sulphur is present in $+4$ oxidation state.

(e) $H_2{S}_2{O}_7$



$x+2(-2)+1(-1)+1(-1)=0$

$x=+6$

Thus, sulphur is present in $+6$ oxidation state.

(f) $S{O}_{2}C{l}_2$



or $x+2(-1)+2(-2)=0$

$x=+6$

Thus, sulphur has $+6$ oxidation state.

(g) $SOC{l}_2$



$x+1(-2)+2(-1)=0$

$x=+4$

Thus, sulphur has $+4$ oxidation state.

Compounds having $+6$ oxidation state of sulphur atom are

$H_2{S}_2{O}_8,H_{2}S{O}_5,H_2{S}_2{O}_7,S{O}_{2}C{l}_2$

Hence, the total number of molecules having $+6$

oxidation state of sulphur atom is $4$ .