When $S O_{2}$ is passed in acidified $K_{2} C r_{2} O_{7}$ solution oxidation state of $S$ changed from?

+ 4

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

+ 4

+ 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

+ 4

+ 6

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

+ 6

+ 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $+ 4$ to $+ 6$
Reaction is:

$S O_{2} + K_{2} {C r}_{2} O_{7} + 3 H_{2} S O_{4} \to C r_{2} {(S O_{4})}_{3} + K_{2} S O_{4} + 3 H_{2} O$

Oxidation state of $S$ in $S O_{2}$ : $x + 2 \times (- 2) = 0$
$x = + 4$

Oxidation state of $S$ in $S O_{4}^{2 -}$ : $x + 4 \times (- 2) = - 2$
$x = + 6$

Therefore oxidation state of $S$ changes from $+ 4 t o + 6$ .

Hence, the correct option is $C$

Suggest Corrections

Similar questions

When SO₂ is passed through an acidified solution of K2Cr2O7, then chromium sulphate is formed. Change in the oxidation state of Cr is from.

S O_{2}

S

S O_{2}

When $H_{2} S O_{3}$ is converted into $H_{2} S O_{4}$ the change in the oxidation state of sulphur is from:

H_{2} O_{2}

K_{4} [F e {(C N)}_{6}]

Join BYJU'S Learning Program