Question

The oxidation state of sulphur in $\text{ NaHS}{{\text{O}}_{\text{4}}}\text{ }$?
A) $\text{ 0 }$
B) $\text{ + 2 }$
C) $\text{ }-2\text{ }$
D) $\text{ + 4 }$
E) $\text{ + 6 }$

Answer 1

The oxidation state is defined as the charge on the atom. The sum of the oxidation state of an atom in a molecule is equal to the charge on the molecule. The neutral molecules charge less species. It has zero charges.

Complete Solution :
- The oxidation number of an atom is defined as the charge on an atom that appears on the atom on forming an ionic bond. An atom which has higher electronegativity has a negative oxidation state and more electropositive elements have positive oxidation state.
- We are interested in determining the oxidation state of an atom in a molecule. We know that the net charge on a neutral atom or molecule is always equal to zero. Let’s determine the oxidation state of $\text{ S }$ in the $\text{ NaHS}{{\text{O}}_{\text{4}}}\text{ }$ .
- The oxidation number of alkali metal that is a sodium atom is equal to $\text{ +1 }$ . Similarly the hydrogen has an $\text{ +1 }$ oxidation state. Oxygen is an electronegative element. It has an $\text{ }-2\text{ }$ oxidation state. The oxidation states are as listed below:

Element	Oxidation state
$\text{ Na }$	$\text{ +1 }$
$\text{ H }$	$\text{ +1 }$
$\text{ O }$	$-2\text{ }$

We know that the sum of the oxidation state of an atom in a neutral molecule is equal to zero. Then we have,
$\text{ 0 = O}\text{.S}\text{. of Na + O}\text{.S}\text{. of H + O}\text{.S}\text{. of S + 4}\left( \text{O}\text{.S}\text{. of O} \right)$
Let’s substitute the values of oxidation state .we have,
$\begin{aligned} & \text{ 0 = 1 + 1 + O}\text{.S}\text{. of S + 4}\left( -2 \right) \\\ & \Rightarrow \text{O}\text{.S}\text{. of S}=\text{ 8}-2\text{ = +6 } \\\ \end{aligned}$
Therefore the oxidation state of sulphur in $\text{ NaHS}{{\text{O}}_{\text{4}}}\text{ }$ is equal to $\text{+6 }$. So, the correct answer is “Option E”.

Note: Note that, along with $\text{+6 }$ oxidation state sulphur exists in $\text{ }-2\text{ }$, 0, $\text{ +}2\text{ }$ and $\text{ +4 }$ oxidation state. The valence shell configuration of Sulphur is $\text{ 3}{{\text{s}}^{\text{2}}}\text{ 3}{{\text{p}}^{\text{4 }}}$ . Thus it exhibits a variable oxidation state. It loses six electrons in the outermost shell and exhibits $\text{ +6 }$ an oxidation state. Remember that oxygen atoms do not exhibit $\text{ +6 }$ an oxidation state.