Question

What is the oxidation number of the sulphur atom in $L{{i}_{2}}S{{O}_{4}}$ ?
A. +4
B. +6
C. +2
D. -2

Answer 1

Hint: The oxidation state or oxidation number of an element in the molecule is going to give information about the number of electrons gained or donated by a particular element in the given molecule can be determined.

Complete answer:
- In the question it is asked to calculate the oxidation number of the sulphur in $L{{i}_{2}}S{{O}_{4}}$ .
- We know that lithium belongs to alkali earth metal and by losing one electron it will get stable electronic configuration.
- Therefore the oxidation number of one lithium atom is +1.
- We know that oxygen has the capability to donate two electrons from others to get octet fully filled electronic configuration.
- Therefore each oxygen atom is going to carry a ‘-2’ charge.
- Therefore the oxidation of sulphur in lithium sulphate can be calculated as follows.
Oxidation number of sulphur (S) = 2 (Lithium ions) + x + 4 (oxygen atoms)
$\Rightarrow 2 (+1) + x + 4 (-2) = 0$
$ +2 + x - 8 = 0$
$ x = 8 – 2 = 6.$
Therefore the oxidation number of sulphur in lithium sulphate ($L{{i}_{2}}S{{O}_{4}}$ ) is ‘+6’.

Note:
Sulphur is going to form six bonds with the four oxygen atoms and two lithium atoms in lithium sulphate. The maximum oxidation state is going to be shown by the sulphur atom is ‘+6’ and sulphur will get an octahedral structure by forming 6 bonds.