Question

Give a reason for the statement. $'{\left[ {NiC{N_4}} \right]^{2 - }}'$ is diamagnetic while $'{\left[ {NiC{l_4}} \right]^{2 - }}'$ is paramagnetic in nature.
(a) In ${\left[ {NiC{l_4}} \right]^{2 - }}$, no unpaired electrons are present while in ${\left[ {NiC{N_4}} \right]^{2 - }}$, two unpaired electrons are present.
(b) In ${\left[ {NiC{N_4}} \right]^{2 - }}$, no unpaired electrons are present while in ${\left[ {NiC{l_4}} \right]^{2 - }}$, two unpaired electrons are present.
(c) ${\left[ {NiC{l_4}} \right]^{2 - }}$ shows $ds{p^2}$ hybridization, so it is paramagnetic.
(d) ${\left[ {NiC{N_4}} \right]^{2 - }}$ shows $s{p^3}$ hybridization, hence it is diamagnetic.

Answer 1

Diamagnetic is the property in which the total spin of each orbital is zero, and in paramagnetic substances, the total spin is not zero. Each orbital can have maximum two electrons with opposite spins, that is, $+ \dfrac{1}{2}$ and $- \dfrac{1}{2}$. Hence, a diamagnetic substance will have full filled orbitals, and a paramagnetic substance will have half-filled orbitals.

Complete step by step answer:
For compound ${\left[ {NiC{l_4}} \right]^{2 - }}$:
The overall charge of the complex is $- 2$, so, net charge on the central atom Nickel, ${\text{Ni}}$ is $+ 2$, by neutrality concept. Atomic number of Nickel is ${\text{28}}$, so, its ground state electronic configuration is ${\left[ {Ar} \right]_{18}}4{s^2}3{d^8}$, and, after the ligand $C{l^ - }$ approaches it, it goes to the excited state as $N{i^{2 + }}$ with electronic configuration of ${\left[ {Ar} \right]_{18}}4{s^0}3{d^8}$. Since, ${\text{4s}}$ has one orbital, and ${\text{3d}}$ has five orbitals with each orbital having the capacity of keeping maximum two electron, so, ${\text{Ni}}$ has one empty ${\text{4s}}$ orbital, two half-filled ${\text{3d}}$ orbitals, and the rest orbitals are full filled. As $C{l^ - }$ is a weak field ligand, when it approaches $N{i^{2 + }}$, it does not help in pairing of electrons. Thus, Nickel has unpaired electrons. Hence, ${\left[ {NiC{l_4}} \right]^{2 - }}$ is paramagnetic.
For compound ${\left[ {NiC{N_4}} \right]^{2 - }}$:
The overall charge of the complex is $- 2$, so, net charge on the central atom Nickel, ${\text{Ni}}$ is $+ 2$, by neutrality concept. Atomic number of Nickel is ${\text{28}}$, so, its ground state electronic configuration is ${\left[ {Ar} \right]_{18}}4{s^2}3{d^8}$, and, after the ligand $C{N^ - }$ approaches it, it goes to the excited state as $N{i^{2 + }}$ with electronic configuration of ${\left[ {Ar} \right]_{18}}4{s^0}3{d^8}$. So, $Ni$ has one empty ${\text{4s}}$ orbital, two half-filled ${\text{3d}}$ orbitals, and the rest orbitals are fully filled. As $C{N^ - }$ is a strong field ligand, when it approaches $N{i^{2 + }}$, it helps in pairing of electrons. Now, after the ligand metal interaction, Nickel is paired with electrons. So, ${\left[ {NiC{N_4}} \right]^{2 - }}$ is diamagnetic in nature.

So, the correct answer is Option B .

Note:
While writing the electronic configuration of the excited state of Nickel, $N{i^{2 + }}$, take out the electrons from $4{s^2}$ orbital first, and then from $3{d^8}$ orbital because ${\text{4s}}$ is the outermost orbital after ${\text{3d}}$. Ligand strength is given by electrochemical series.