Question

Among the following, which complex compound is diamagnetic?
(i)- $[Co{{(N{{H}_{3}})}_{6}}]C{{l}_{3}}$
(ii)- $[Ni{{(N{{H}_{3}})}_{6}}]C{{l}_{2}}$
(iii)- $[Cr{{({{H}_{2}}O)}_{6}}]C{{l}_{3}}$
(iv)- $[Fe{{({{H}_{2}}O)}_{6}}]C{{l}_{2}}$

(a)- (i)
(b)- (ii)
(c)- (iii)
(d)- (iv)

Answer 1

If the compound has unpaired electrons then the compound is paramagnetic and if it has all the electrons in pairs then it is diamagnetic. In a complex compound, the diamagnetic nature is calculated by finding the oxidation number of the central metal atom and then its configuration.

Complete answer:
Let us study all the complexes one by one:
(i)- $[Co{{(N{{H}_{3}})}_{6}}]C{{l}_{3}}$
First, we have to calculate the oxidation number of cobalt ions.
The oxidation number of $N{{H}_{3}}$ is zero because it is a neutral ligand and the oxidation number of $Cl$ is -1. So, the oxidation number of cobalt is:
$x\text{ + 6(0) + 3(-1) = 0}$
$x\text{ = +3}$
So, the oxidation number of cobalt is +3.
The ground state electronic configuration of cobalt (27) is $4{{s}^{2}}3{{d}^{7}}$ and in $C{{o}^{3+}}$ state $4{{s}^{0}}3{{d}^{6}}$
It has 6 electrons which are paired.
Hence, $[Co{{(N{{H}_{3}})}_{6}}]C{{l}_{3}}$is diamagnetic.
(ii)- $[Ni{{(N{{H}_{3}})}_{6}}]C{{l}_{2}}$
First, we have to calculate the oxidation number of nickel ions.
The oxidation number of $N{{H}_{3}}$ is zero because it is a neutral ligand and the oxidation number of $Cl$ is -1. So, the oxidation number of nickel is:
$x\text{ + 6(0) + 2(-1) = 0}$
$x\text{ = +2}$
So, the oxidation number of nickel is +2.
The ground state electronic configuration of nickel (28) is $4{{s}^{2}}3{{d}^{8}}$ and in $N{{i}^{2+}}$ state $4{{s}^{0}}3{{d}^{8}}$
It has 2 unpaired electrons.
Hence, $[Ni{{(N{{H}_{3}})}_{6}}]C{{l}_{2}}$is paramagnetic.
(iii)- $[Cr{{({{H}_{2}}O)}_{6}}]C{{l}_{3}}$
First, we have to calculate the oxidation number of chromium ion.
The oxidation number of ${{H}_{2}}O$ is zero because it is a neutral ligand and the oxidation number of $Cl$ is -1. So, the oxidation number of chromium is:
$x\text{ + 6(0) + 3(-1) = 0}$
$x\text{ = +3}$
So, the oxidation number of chromium is +3.
The ground state electronic configuration of chromium (24) is $4{{s}^{1}}3{{d}^{5}}$ and in $C{{r}^{3+}}$ state $4{{s}^{0}}3{{d}^{3}}$
It has 3 unpaired electrons.
Hence, $[Cr{{({{H}_{2}}O)}_{6}}]C{{l}_{3}}$ is paramagnetic.
(iv)- $[Fe{{({{H}_{2}}O)}_{6}}]C{{l}_{2}}$
First, we have to calculate the oxidation number of iron ions.
The oxidation number of ${{H}_{2}}O$ is zero because it is a neutral ligand and the oxidation number of $Cl$ is -1. So, the oxidation number of iron is:
$x\text{ + 6(0) + 2(-1) = 0}$
$x\text{ = +2}$
So, the oxidation number of iron is +2.
The ground state electronic configuration of iron (26) is $4{{s}^{2}}3{{d}^{6}}$ and in $F{{e}^{2+}}$ state $4{{s}^{0}}3{{d}^{6}}$
It has 4 unpaired electrons.
Hence, $[Fe{{({{H}_{2}}O)}_{6}}]C{{l}_{2}}$ is paramagnetic.

So, the correct option is (a)- (i)

Note:
If the complex compound has a strong field ligand like $N{{O}_{2}}^{-}, C{{N}^{-}}, CO$, etc they will pair up the unpaired electrons and if the compound has a weak field ligand ${{H}_{2}}O, N{{H}_{3}},{{F}^{-}}$, etc, they will not pair up the unpaired electrons.