Question

The oxidation number of $C$ in $HNC$ is:
(A) $ + 2$
(B) $ - 3$
(C) $ + 3$
(D) $0$

Answer 1

Hint:The oxidation number of a free element is always 0.
-The oxidation number of a monatomic ion equals the charge of the ion.
-The oxidation number of H is +1, but it is -1 in when combined with less electronegative elements.
-The oxidation number of O in compounds is usually -2, but it is -1 in peroxides.

Complete step by step answer:
Oxidation state, also called oxidation number describes the degree of oxidation (loss of electrons) of an atom in a chemical compound.

Oxidation number of hydrogen is $ + 1$. As nitrogen is more electronegative than carbon so its oxidation no. is $ - 3$.
We know that net charge on the compound is zero.
Let the oxidation number of carbon is $x$.
$1 + \left( { - 3} \right) + x = 0$
$x = 2$
Hence, option A is the correct answer.

Additional information:
Oxidation numbers are typically represented by integers which can be positive, negative or zero. In some cases, the average oxidation number of an element is fraction such as $\dfrac{8}{3}$.
For iron in magnetite $\left( {F{e_3}{O_4}} \right)$.
The highest known oxidation state is reported to be $ + 9$ in the tetrox iridium $\left( {IX} \right)$cation $\left( {IrO_4^ + } \right)$.
It is predicted that even a $ + 10$ oxidation state may be achievable by platinum in tetrox platinum cation $\left( {PtO_4^{2 + }} \right)$.
The lowest oxidation state is $ - 5$, as for boron in \[A{l_3}BC\].

Note:
The increase in oxidation number of an atom, through a chemical reaction is known as an oxidation. A decrease in oxidation state is known as a reduction.