Question

What is the oxidation number of carbon in $$C{N}^{-}$$? (Give the absolute value)

Answer 1

Hint: Oxidation number can be calculated by calculating the number of electrons carbon needs to gain, lose or share, in order to form the compound $$C{N}^{-}$$. The compound given to us known as Cyanide. It consists of a carbon atom triple bonded to nitrogen.

Complete step-by-step solution -

Oxidation number is also known as Oxidation state.
Oxidation number is defined as the number of electrons an atom should lose or gain to complete its octet (or duet – in case of Hydrogen, Lithium, Beryllium, Boron).
When an atom needs to
gain electrons, the oxidation number is negative
lose electrons, the oxidation number is positive
neither gain nor lose an electron, the oxidation number is zero
Since Nitrogen has 5 electrons in its valence shell, it requires three electrons to fill its octet. Therefore, the oxidation number of Nitrogen is -3.
Also, as we can see, the total charge on the cyanide ion is equal to -1.
Let us assume that the oxidation number of Carbon is ‘x’.
Therefore,
Oxidation number of Carbon + Oxidation number of Nitrogen = Charge on the Cyanide ion
$$\begin{array}{rl}& \Rightarrow x+(-3)=-1\\ & \Rightarrow x=+2\end{array}$$ $$$$
Hence, we can say that the oxidation number of Carbon in Cyanide ion is +2.

Additional Information:
Cyanide is widely used for the mining of gold and silver because it has the ability to dissolve these metals and their ores. Not only this, Cyanide is also used in electroplating.

Note: While calculating oxidation numbers, we can keep the following points in mind-
(1). The oxidation number of an atom in a neutral molecule (that contains atoms of only one element) is zero.
(2). The oxidation number of simple ions (containing only one atom) is equal to the charge on the ion. For example, the oxidation number of $$C{l}^{-}$$is -1.