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The oxidation state of cobalt in the complex compound $[Co{{(N{{H}_{3}})}_{6}}]C{{l}_{3}}$
(A) +3
(B) +6
(C) +5
(D) +2

Answer
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163.2k+ views
Hint: To solve this problem, let’s suppose the oxidation state of Cobalt to be x. We must first add the oxidation states of the remaining elements until the total is equal to the net charge in the molecule.

Complete Step by Step Solution:
In the give question we have to find the oxidation state of Cobalt. We first must know that the given compound is a coordination compound in which the $N{{H}_{3}}$ molecule is acting as a ligand.
The hypothetical charge that an atom would hold if all of its links to other atoms were entirely ionic in nature is what is known as an atom's oxidation state.
To proceed further let’s suppose the oxidation number of Co to be x.
Now if we assume the complete dissociation of the compound into its ionic form then the compounds that will form would be
$[Co{{(N{{H}_{3}})}_{6}}]C{{l}_{3}}\to {{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}}+3C{{l}^{-}}$
So, we can see that the complex has a net charge of +3 and we know that $N{{H}_{3}}$ is a neutral molecule that has 0 charge.
So, oxidation state of Co can be given as
$x+6(0)=+3$ Where, x is the oxidation state of Co
After solving it we get $x=+3$ and hence, the oxidation state of Cobalt in the given compound is$+3$.
Hence, the correct option is A. +3 i.e. the oxidation state of Co is +3.

Note: In the given coordination compound the ligand is $N{{H}_{3}}$ which is a neutral ligand. Hence we need to remember that the charge on the same is 0. Another thing to take care of is to properly distribute the charge when dissociating the compound into the ions. If there is no attacking group (like Cl in this case) then the compound is neutral and neutral compounds have a net charge of 0.