Question

In how many of the following compounds of sulphur, the oxidation state of sulphur atom is $+6$?
$\ce{H_2 S_2 O_8 , H_2SO_5 , H_2SO_3 , H_2S_2O_7 , SO_2Cl_2 , SOCl_2}$

• 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) $H_2SO_3$



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

$x = +4$

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

(d) $H_2SO_4$



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

$x = + 4$

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

(e) $H_2S_2O_7$



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

$x = +6$

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

(f) $SO_2Cl_2$



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

$x = +6$

Thus, sulphur has $+6$ oxidation state.

(g) $SOCl_2$



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

$x = +4$

Thus, sulphur has $+4$ oxidation state.

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

$H _{2} S _{2} O _{8}, H _{2} SO _{5}, H _{2} S _{2} O _{7}, SO _{2} Cl _{2}$

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

oxidation state of sulphur atom is $4$ .