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

Hint: 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.