Question

What is the oxidation number of sulfur atoms in $L{i_2}S{O_4}$?
A) $+ 5$
B) $+ 6$
C) $- 6$
D) $- 5$

Answer 1

We need to remember that the oxidation number defines the count of electrons that can be lost, gain or even share while forming a chemical bond with other atoms or elements.

Complete answer:
Since $L{i_2}S{O_4}$ is a neutral compound that means it has no charge in it, therefore the sum of the oxidation number of all the atoms present in this compound will be $0$.
Since we need to find the oxidation number of $S$, therefore we must know or find the oxidation number for the rest of the atoms present in this compound.
As we know that the lithium belongs to group 1 i.e. Alkali metals, therefore it has an oxidation number of $+ 1$ whereas,
Oxygen belongs to group $16$ and it has an oxidation number of $- 2$.
For calculating the oxidation number of $S$ in $L{i_2}S{O_4}$ this procedure is followed:
Oxidation Number for $Li$=$+ 1$ since there are $2Li$ atoms
So it will be $+ 2$
Similarly for$O$, Oxidation number=$- 2$ since $4O$ atoms
So it will be $- 8$
Let the oxidation number of $S$ will be x in $L{i_2}S{O_4}$
$2 + x + \left( { - 8} \right) = 0$
On simplification we get,
$\Rightarrow x = 6$
Thus, Oxidation number of $S$ is + $$$$6.
Option A) this is an incorrect option as the solution for this question $+ 6$.
Option B) this is a correct option as explained above, therefore the oxidation number of $L{i_2}S{O_4}$ is $+ 6$.
Option C) this is an incorrect option you might get the wrong answer while solving this question. Sometimes while doing calculations in a hurry you can get $- 6$ but that’s not a correct option.
Option D) this is an incorrect option .

So Option B is correct.

Note:
Always check your calculations after completing your question, since the options given are tricky, sometimes in a hurry one can get value as $- 6$ which is not correct. Do solve the question carefully.