Question

The oxidation State of ${\text{S}}$ in ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{4}}}{{\text{O}}_{\text{6}}}$is:

Answer 1

Oxidation state can also be referred to as oxidation number. It describes the degree of oxidation that is loss of electrons of an atom in a chemical compound.

Complete step by step answer:
Oxidation state that is the oxidation number is the loss of electrons of an atom basically.
To calculate the oxidation number of an atom the following points should be kept in mind.
1. The oxidation number of a free element is always ${\text{0}}$.
2. The oxidation number of a monatomic ion equals the charge of the ion.
3. The oxidation number of ${\text{H}}$ is ${\text{ + 1}}$, but it is ${\text{ - 1}}$ in when combined with less electronegative elements.
4.The oxidation number of ${\text{O}}$ in compounds is usually ${\text{ - 2}}$, but it is ${\text{ - 1}}$ in peroxides.
5.The oxidation number of a Group ${\text{1}}$ element in a compound is ${\text{ + 1}}$.
6.The oxidation number of a Group ${\text{2}}$ element in a compound is ${\text{ + 2}}$.
7.The oxidation number of a Group ${\text{17}}$ element in a binary compound is ${\text{ - 1}}$.
8.The sum of the oxidation numbers of all of the atoms in a neutral compound is ${\text{0}}$.
9.The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.
As asked in the question the oxidation number of ${\text{S}}$ in ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{4}}}{{\text{O}}_{\text{6}}}$is:
Oxidation number of of ${\text{S}}$ in ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{4}}}{{\text{O}}_{\text{6}}}$ is X.
Oxidation number of ${\text{Na}}$ in ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{4}}}{{\text{O}}_{\text{6}}}$ is ${\text{ + 1}}$.
Oxidation number of ${\text{O}}$ is ${\text{ - 2}}$.
Sum of all oxidation numbers is ${\text{2 X( + 1) + (4X) + 6 \times ( - 2)}}$.
Solving this equation, we get ${\text{4X - 10 = 0}}$
${\text{X = }}\dfrac{{{\text{10}}}}{{\text{4}}}$
${\text{X = 2}}{\text{.5}}$
This means that this compound has sulfur atoms with mixed oxidation states. Since, it is normal for sulfur to have oxidation states of ${\text{ - 2,0, + 2, + 4, + 6}}$. It is more likely that there are three sulfurs with a ${\text{ + 2}}$oxidation state and one sulfur having ${\text{ + 4}}$.
Our required answer is A.

Note:
Key Points:
-The oxidation number of a free element is always 0.
-The oxidation number of a monatomic ion equals the charge of the ion.
-The oxidation number of H is +1, but it is -1 in when combined with less electronegative elements.
-The oxidation number of O in compounds is usually -2, but it is -1 in peroxides.