Question

What is the oxidation number of Mn in $\text{MnC}{{\text{l}}_{2}}$?

Answer 1

Hint: Oxidation number (or oxidation state) refers to the total number of electrons that are gained or lost by an atom during chemical bond formation with another atom. In a neutral compound, oxidation numbers of each element must add up to zero.

Complete answer:
There are some rules we need to know before assigning oxidation number to elements in a compound:
(1) The oxidation number of an element in its free state is always zero.
(2) The oxidation number of a monatomic ion is the same as the charge on it.
(3) The sum of all oxidation numbers in a neutral compound is zero. And that in a polyatomic ion the sum is equal to the charge on the ion.
Now, in $\text{MnC}{{\text{l}}_{2}}$ we assume that the oxidation number of Mn is x. And we know that the oxidation state of chlorine is $-1$.
According to the third rule stated above, the sum of oxidation states of all elements in a neutral compound should be equal to 0. So, we can write it in the form of an equation as:
\[\text{x}+2\times \left( -1 \right)=0\]
On rearranging the equation, we get
\[\text{x}=+2\]
Hence, the oxidation number of Mn in $\text{MnC}{{\text{l}}_{2}}$ is +2.

Note:
The transition metals show variable oxidation states due to the participation of ns and (n-1)d orbitals in bonding. The energy difference between these two orbitals is very low and thus both can be used in bonding. Manganese is a transition metal and it shows variable oxidation states of +2, +3, +4, +6, and +7. Out of these, the most stable oxidation state is +2.