Question

The oxidation number of Sulphur in ${S_2}{F_2}$ is:

Answer 1

Hint- We will let the oxidation state of $S$ in ${S_2}{F_2}$ be x. then we will add all the oxidation states and put them equal to the charge on the complex given to us by the question. We will get the oxidation state of $S$.

Complete answer:

Let x be the oxidation state of $S$ in ${S_2}{F_2}$​.
Since, the overall charge on the complex is 0, the sum of oxidation states of all elements in it should be equal to 0.
Therefore, \[2x + 2\left( { - 1} \right) = 0\]
\[x = + 1\]
Hence, the oxidation state of $S$ in ${S_2}{F_2}$​ is +1.

Note: Whenever there is no charge given on the formula of any complex, element etc, take the charge to be 0. Then solve the question further as done in the solution above. If there is a charge present like -1 etc, take the sum of all the oxidation states equal to that charge.