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Oxidation number of carbon in $(CN)_{ 2 }$ is +3. If true enter 1, else enter 0.

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Hint: You have to consider the oxidation state of carbon as x and you know that nitrogen will be having -3 oxidation state as it is making a triple bond with carbon. Now, try to answer this question accordingly.

Complete step by step answer:

We should know that Cyanogen is the chemical compound with the formula $(CN)_{ 2 }$. It is a colorless, toxic gas with a pungent odor. The molecule is a pseudohalogen.

Cyanogen molecules consist of two CN groups – analogous to diatomic halogen molecules, such as $Cl_{ 2 }$, but far less oxidizing.

In cyanogen, the two cyano groups are bonded together at their carbon atoms: $$N\equiv C−C\equiv N$$

In the compound $(CN)_{ 2 }$
Let the oxidation state of Carbon = x
So we can write,
2(x - 3) = 0
x - 3 = 0
x = +3

Hence, we can say the oxidation number of carbon in $(CN)_{ 2 }$ is +3. So, this statement is true.

Therefore, the correct answer to this question is 1.

Note: We should also know about the isomers of cyanogen. Cyanogen is NCCN. There are less stable isomers in which the order of the atoms differs. Isocyanogen (or cyanoisocyanogen) is NCNC, diisocyanogen is CNNC, and diazodicarbon is CCNN.
Cyanogen produces the second hottest known natural flame (after carbon subnitride) with a temperature of over 4,525 $^{ o }C$ (8,177 $^{ o }F$) when it burns in oxygen.