Question

The oxidation number of sulphur in ${H_2}S{O_4}$​ is:
A. $ + 2$
B. $ + 3$
C. $ + 4$
D. $ + 6$
E. $ + 8$

Answer 1

Hint:In this question, students have to find the oxidation number i.e., oxidation state of sulphur. Oxidation number is defined as the absolute number of electrons that a particle either gains or loses to form a chemical bond with another molecule.

Complete answer:
There are some rules that students need to remember while figuring the oxidation states in a molecule.
Any Free component has an oxidation number equivalent to zero.
For monatomic particles the oxidation number consistently has a similar value as the net charge comparing to the particle
The hydrogen particle has an oxidation condition of $ + 1$. When the bonding takes place with less electronegativity it shows an oxidation number of $ - 1$
When oxygen is bonded with peroxides, its oxidation state is $ - 2$.
All group 1 elements and alkali metals have oxidation state $ + 1$.
Similarly, alkaline earth metals have oxidation state $ + 2$
Now, let us consider oxidation number of Sulphur be X
With the help of above rules, we can calculate oxidation number of Sulphur as follows:
$2( + 1) + X + 4( - 2) = 0$
Further solving the above equation, we have:
$
\Rightarrow 2 + X - 8 = 0 \\
\Rightarrow X - 6 = 0 \\
 \Rightarrow X = 6 \\
 $
Here $ + 1$ is oxidation number of Hydrogen
And $ - 2$ is oxidation number of Oxygen
Hence, from the above explanation we can find that oxidation number of Sulphur in ${H_2}S{O_4}$ is $ + 6$
Hence, the correct option is (D) .

Note:Students are advised to do all the calculations properly. Also, students should take care of the fact that the electronegativities differ for all the elements in different conditions, so they should abide by the rules thoroughly mentioned above while answering these types of questions. These questions are important for both boards as well as competitive exams.