Question

Among the following complexes, the diamagnetic complexes are:
(K)- $K_3[Fe{(CN)}_6]$
(L)- $[Co{(N{H}_3)}_6]C{l}_3$
(M)- $N{a}_3[Co{(oxalate)}_3]$
(N)- $[Ni{(H_{2}O)}_6]C{l}_2$
(O)- $K_2[Pt{(CN)}_4]$
(P)- $[Zn{(H_{2}O)}_6]{(N{O}_3)}_2$

(A) K, L, M, N
(B) K, M, O, P
(C) L, M, O, P
(D) L, M, N, O

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 step by step solution:
Let us study all the complexes one by one:
(K)- $K_3[Fe{(CN)}_6]$
First, we have to calculate the oxidation number of the iron atom.
The oxidation number of $CN$ is -1 because it is a negative ligand and the oxidation number of $K$ is +1. So, the oxidation number of iron is:
$(+1)3+x+6(-1)=0$
$\Rightarrow x=+3$
The ground state electronic configuration of iron (26) is $4s^{2}3d^6$ and in $F{e}^{3+}$ state $4s^{0}3d^5$
It has 5 unpaired electrons.
This compound has a strong field ligand, but 1 electron is unpaired so it is paramagnetic.
(L)- $[Co{(N{H}_3)}_6]C{l}_3$
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:
$x\text{ + 6(0) + 3(-1) = 0}$
$\Rightarrow x\text{ = +3}$
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 that are paired because of the presence of a strong field ligand.
Hence, $[Co{(N{H}_3)}_6]C{l}_3$ is diamagnetic.
(M)- $N{a}_3[Co{(oxalate)}_3]$
First, we have to calculate the oxidation number of cobalt ions.
The oxidation number of Oxalate is -2 because it is a negative ligand and the oxidation number of $Na$ is +1. So, the oxidation number of cobalt is:
$3(+1)+x+3(-2)=0$
$\Rightarrow x=+3$
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 that are paired because of the presence of a strong field ligand.
Hence, $N{a}_3[Co{(oxalate)}_3]$ is diamagnetic.
(N)- $[Ni{(H_{2}O)}_6]C{l}_2$
First, we have to calculate the oxidation number of nickel 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 nickel is:
$x\text{ + 6(0) + 2(-1) = 0}$
$\Rightarrow x\text{ = +2}$
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 and it has a weak field ligand.
Hence, $[Ni{(H_{2}O)}_6]C{l}_2$is paramagnetic.
(O)- $K_2[Pt{(CN)}_4]$
First, we have to calculate the oxidation number of the platinum atom.
The oxidation number of $CN$ is -1 because it is a negative ligand and the oxidation number of $K$ is +1. So, the oxidation number of platinum is:
$2(+1)+x+4(-1)=0$
$\Rightarrow x=+2$
The ground state electronic configuration of platinum (78) is $6s^{1}5d^9$ and in $P{t}^{2+}$ state $6s^{0}5d^8$
It has 8 paired electrons and has strong field ligands.
It is diamagnetic.
(P)- $[Zn{(H_{2}O)}_6]{(N{O}_3)}_2$
First, we have to calculate the oxidation number of the zinc ion.
The oxidation number of $H_{2}O$ is zero because it is a neutral ligand and the oxidation number of $N{O}_3$ is -1. So, the oxidation number of zinc is:
$x+(0)6+(-1)2=0$
$\Rightarrow x\text{ = +2}$
So, the oxidation number of zinc is +2.
The ground state electronic configuration of zinc (30) is $4s^{2}3d^{10}$ and in $Z{n}^{2+}$ state $4s^{0}3d^{10}$
It has 10 paired electrons and it has a weak field ligand.
Hence, $[Zn{(H_{2}O)}_6]{(N{O}_3)}_2$ is diamagnetic.

Therefore the correct answer is an option (C)- L, M, O, P.

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, $H_{2}O,F^{-}$, etc, they will not pair up the unpaired electrons.