When $S O_{2}$ gas is bubbled in acidified (by $H_{2} S O_{4}$ ) aqueous solution of $K_{2} C r_{2} O_{7}$ , a green coloured solution is formed above $50^{o}$ C, which on evaporation crystalizes into a violet crystal which is used in the tanning of leather and in gelatin emulsions in photographic film as hardener. Find the oxidation state of sulphur in the complex formed, if oxidation state of chromium in the same complex is $+ 3$ .

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Solution

The corresponding reaction will be,
$K_{2} C r_{2} O_{7} + H_{2} S O_{4} + 3 S O_{2} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + H_{2} O$
On evaporation crystal formed is Chrome alum, $K_{2} S O_{4} . C r_{2} (S O_{4})_{3} .12 H_{2} O$
Thus, the oxidation state of $S$ is $+ 6$

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The corresponding reaction will be, K2Cr2O7+H2SO4+3SO2→K2SO4+Cr2(SO4)3+H2O On evaporation crystal formed is Chrome alum, K2SO4.Cr2(SO4)3.12H2O Thus, the oxidation state of S is +6

The corresponding reaction will be,
$K_{2} C r_{2} O_{7} + H_{2} S O_{4} + 3 S O_{2} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + H_{2} O$
On evaporation crystal formed is Chrome alum, $K_{2} S O_{4} . C r_{2} (S O_{4})_{3} .12 H_{2} O$
Thus, the oxidation state of $S$ is $+ 6$