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

Hint: 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 $4s^{2}3d^7$ and in $C{o}^{3+}$ state $4s^{0}3d^6$
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 $4s^{2}3d^8$ and in $N{i}^{2+}$ state $4s^{0}3d^8$
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 $4s^{1}3d^5$ and in $C{r}^{3+}$ state $4s^{0}3d^3$
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 $4s^{2}3d^6$ and in $F{e}^{2+}$ state $4s^{0}3d^6$
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.