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Last updated date: 05th Dec 2023
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# The coordination compound ${K_4}\left[ {Ni{{\left( {CN} \right)}_4}} \right]$, the oxidation state of Ni is:A) $- 1$B) $0$C) $+ 1$D) $+ 2$

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Hint: In a coordination complex, there are metal, coordination complexes and ligands attached to it. The metal, ligands and counterions together form a coordination complex. The nature of the ligand can be unidendate, bidentate, trident and so on depending upon the atoms that are directly bonded to the central atom.

We have to know that the coordination number is derived for the central atom present in any complex, that central atom is bonded to many ligands around it. So the number of atoms, molecules or ions which are bonded to central atoms defines the coordination number of any complex.
For example: A complex $[CuC{l_2}]$- has a coordination number two since it has two chlorine atoms.
Now, what is the oxidation state? Oxidation state can be defined on the basis of gaining and losing an electron. When an atom loses or gains an electron so as to form a chemical bond with another atom in a molecule, it defines oxidation state.
For example: A complex $Mn{O_2}$ has $+ 2$ oxidation state.
Solving oxidation state of Nickel in complex ${K_4}\left[ {Ni{{\left( {CN} \right)}_4}} \right]$
Cyanide ion carries $- 1$ charge on it thus there are four atoms of cyanide thus total charge will be $- 4$. Potassium has $+ 1$charge since it is an alkali metal belonging to group one and there are a total four potassium atoms thus total oxidation state for potassium will be $4$.
Consider the oxidation state of Nickel to be $x$:
$4 + x + \left( { - 4} \right) = 0$

Thus the correct option for this question will be option (B).

Note:
We must have to know that $CN$ in this complex is a cyanide ion. When an atom loses or gains an electron so as to form a chemical bond with another atom in a molecule defines oxidation state. Ni is the atomic symbol of the nickel element.