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 $4{{s}^{2}}3{{d}^{6}}$ and in $F{{e}^{3+}}$ state $4{{s}^{0}}3{{d}^{5}}$
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 $4{{s}^{2}}3{{d}^{7}}$ and in $C{{o}^{3+}}$ state $4{{s}^{0}}3{{d}^{6}}$
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 $4{{s}^{2}}3{{d}^{7}}$ and in $C{{o}^{3+}}$ state $4{{s}^{0}}3{{d}^{6}}$
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 $4{{s}^{2}}3{{d}^{8}}$ and in $N{{i}^{2+}}$ state $4{{s}^{0}}3{{d}^{8}}$
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 $6{{s}^{1}}5{{d}^{9}}$ and in $P{{t}^{2+}}$ state $6{{s}^{0}}5{{d}^{8}}$
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 $4{{s}^{2}}3{{d}^{10}}$ and in $Z{{n}^{2+}}$ state $4{{s}^{0}}3{{d}^{10}}$
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.