Question

The coordination number of $N{i^{ + 2}}$ is $4$ .
$NiC{l_2} + KCN(excess) \to A$ (Cyano complex)
$NiC{l_2} + conc.HCl(excess) \to B$ (Chloro complex)
Predict the magnetic nature of A and B.
(a) Both are diamagnetic.
(b) A is diamagnetic and B is paramagnetic with one unpaired electron.
(c) A is diamagnetic and B is paramagnetic with two unpaired electrons.
(d) Both are paramagnetic.

Answer 1

First we have to write the complete reaction, then the electronic configuration of the resultant products and the presence of electrons will decide the nature of the complex whether the product is paramagnetic or diamagnetic.

Complete step-by-step answer: First, we complete the reactions,
$NiC{l_2} + KCN(excess) \to A$
$NiC{l_2} + KCN(excess) \to {\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$
Here, as given A is a cyano complex. A is ${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$ .
Now, completing the second reaction,
$NiC{l_2} + conc.HCl(excess) \to B$
$NiC{l_2} + conc.HCl(excess) \to {\left[ {NiC{l_4}} \right]^{2 - }}$
Here, as given B is a chloro complex. B is ${\left[ {NiC{l_4}} \right]^{2 - }}$ .
So, now writing the electronic configuration of the complexes,
As the oxidation number of $Ni$ is $+ 2$
Atomic number of $Ni$ is $28$
The electronic configuration of the complex ${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$ is:
$Ni = \left[ {Ar} \right]3{d^8}4{s^2}$
$N{i^{ + 2}} = \left[ {Ar} \right]3{d^8}$
As $CN$ is a strong ligand, it will fill two electrons in one orbital and then another orbitals are filled up. Hence, pair up all the electrons. As in the electrons are in even number and presence of strong ligand, all the electrons in ${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$ are paired up.
So, due to absence of unpaired electrons in the complex, this complex will be diamagnetic.
Hence, ${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$ is diamagnetic in nature.
As the oxidation number of $Ni$ is $+ 2$
Atomic number of $Ni$ is $28$
The electronic configuration of the complex ${\left[ {NiC{l_4}} \right]^{2 - }}$ is:
$Ni = \left[ {Ar} \right]3{d^8}4{s^2}$
$N{i^{ + 2}} = \left[ {Ar} \right]3{d^8}$
As $Cl$ is a weak ligand, it will first fill a single electron in all the orbitals and then the pairing occurs. Hence, only if the number of electrons are $10$ then all the electrons are paired up. As in the ${\left[ {NiC{l_4}} \right]^{2 - }}$ , there are eight electrons in the complex, due to presence of weak ligands, two electrons in the ${\left[ {NiC{l_4}} \right]^{2 - }}$ are unpaired. So, due to the presence of unpaired electrons in the complex, this complex ${\left[ {NiC{l_4}} \right]^{2 - }}$ will be paramagnetic.
Hence, ${\left[ {NiC{l_4}} \right]^{2 - }}$ is paramagnetic in nature.So, from the above discussions, A is diamagnetic and B is paramagnetic due to the presence of the two unpaired electrons.

Hence, the correct option in these question is (c) A is diamagnetic and B is paramagnetic with two unpaired electrons.

Note: If there is a presence of any unpaired electron in the complex, the complex is paramagnetic in nature and if there is no unpaired electron in the complex, the complex is diamagnetic in nature. Presence of any unpaired electron in the complex observed by its electronic configuration.