1
MCQ (Single Correct Answer)

AIPMT 2003

The oxidation states of sulphur in the anions

SO3$${_3^{2 - }}$$, S2O$${_4^{2 - }}$$ and S2O$${_6^{2 - }}$$ follow the order
A
S2O$${_4^{2 - }}$$ < SO$${_3^{2 - }}$$ < S2O$${_6^{2 - }}$$
B
SO$${_3^{2 - }}$$ < S2O$${_4^{2 - }}$$ < S2O$${_6^{2 - }}$$
C
S2O$${_4^{2 - }}$$ < S2O$${_6^{2 - }}$$ < SO$${_3^{2 - }}$$
D
S2O$${_6^{2 - }}$$ < S2O$${_4^{2 - }}$$ < SO$${_3^{2 - }}$$

Explanation

SO32– : oxidation state of ‘S’ is +4

S2O42– : oxidation state of ‘S’ is +3.

S2O62– : oxidation state of ‘S’ is +5.

So, the order is S2O$${_4^{2 - }}$$ < SO$${_3^{2 - }}$$ < S2O$${_6^{2 - }}$$.

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