Question

Calculate the oxidation number of sulphur in $N{{a}_{2}}{{S}_{4}}{{O}_{6}}$.

Answer 1

Hint: Oxidation number of any atom represents its capability to oxidize itself by losing electrons. Oxidation number depends on other bonded atoms in a molecule. Oxidation number for any atom in a molecule can be calculated by the formula that the net charge on a molecule is equal to 0. So, all the charges (oxidation states) equal to 0.

Complete answer:
The oxidation number or oxidation state of any atom is the ability of that atom to gain or lose or share electrons. The oxidation state is in the charge form on the respective atoms bonded in a molecule. The charge is negative when an atom has the tendency to take electrons, while the charge is positive when an atom has an ability to donate electrons.
As the oxidation state is calculated in the form of individual atoms, so the oxidation state of any electrically neutral molecule is zero. So any molecule without a charge has a 0 oxidation number. Thus, by adding the respective oxidation numbers of atoms we can take out the oxidation number of any particular element to be positive or negative.
The given molecule is $N{{a}_{2}}{{S}_{4}}{{O}_{6}}$ it has sodium, sulphur and oxygen. As we know that sodium being an alkali metal of group 1 has an oxidation number of +1, while oxygen consists of a -2 oxidation number. So, oxidation number of sulphur in $N{{a}_{2}}{{S}_{4}}{{O}_{6}}$ is calculated as:
$2\left( N{{a}^{1+}} \right)$+ ${{S}_{4}}$+$6\left( {{O}^{2-}} \right)$ = 0
${{S}_{4}}$ = 12 – 2
${{S}_{4}}$= 10
$S=\dfrac{10}{4}$
S = 2.5
Hence, the oxidation number of sulphur in $N{{a}_{2}}{{S}_{4}}{{O}_{6}}$ is 2.5.

Note:
The oxidation states of some particular elements are, alkali metals have a +1 oxidation state, alkaline earth metals have +2 oxidation state, all halogens have a -1 oxidation state, while oxygen in oxide has -2, in peroxide has -1 and in superoxide has -2. Electronegative elements have a negative oxidation number.