Question

What is the value of $x$ in complex ${{[Ni{{(CN)}_{4}}]}^{x}}$?

Answer 1

Hint: An atom's total number of lost or gained electrons is indicated by its oxidation number, which is a number. Any element that is pure has an oxidation number of zero. The oxidation number for mono atomic ions always equals the ion's corresponding net charge. The net charge of a polyatomic ion is equal to the total sum of all the oxidation numbers of the atoms that make up the ion.

Complete step by step solution:
The most stable oxidation state of nickel ($Ni$) is +2. The charge on one $CN$ligand is -1.
The value of $x$ in ${{[Ni{{(CN)}_{4}}]}^{x}}$can be calculated as:
$2+4(-1)=x$
$2-4=x$
$x=-2$
Hence, the value of $x$ in complex ${{[Ni{{(CN)}_{4}}]}^{x}}$is -2. That is, the given compound is written as ${{[Ni{{(CN)}_{4}}]}^{2-}}$.

Additional Information: When an ion atom loses electrons (one or more) during a chemical reaction, oxidation occurs. Oxygen is not necessarily required for this reaction. The term is essentially used when oxygen unexpectedly loses an electron during the process. The oxidation state of chemical species increases during the oxidation process. Whereas the oxidation state of chemical species increases during the reduction process.

Note: In contrast to the oxidation state, which indicates how oxidised an atom is in a molecule, the oxidation number describes the charge that the core metal atom will retain once all ligands have been removed. An oxidation state is represented by Arabic numbers, whereas an oxidation number is represented by Roman numbers.