Question

Out of \[Ni{(CO)_4}\], ${[Ni(CN)_4^{}]^{2 - }}$ and ${[NiCl_4^{}]^{2 - }}$ select paramagnetic and diamagnetic.
A.\[Ni{(CO)_4}\] and ${[Ni(CN)_4^{}]^{2 - }}$ are diamagnetic
B.${[NiCl_4^{}]^{2 - }}$ is paramagnetic
C.Both (A) and (B) correct
D.None of these

Answer 1

Hint:
We know that the Valence Bond Theory abbreviated as VBT, this theory is given by Linus Pauling, this theory is useful for explaining the chemical bonding. VBT is based on covalent interactions between the central metal and the ligands in the complex. VBT also explains the geometry, magnetic behavior and the formation of a complex compound.
We need to know that if the electrons in outer shell are filled which is said to be diamagnetic whereas the electrons in outer shell have unpaired electron then which is known as paramagnetic.

Complete step by step answer:
Let us see how the Valence Bond Theory explains the magnetic behavior of \[Ni{(CO)_4}\]${[Ni(CN)_4^{}]^{2 - }}$ and ${[NiCl_4^{}]^{2 - }}$ ions.
\[Ni{(CO)_4}{\text{,}}\] tetracarbonylnickel. Nickel has an atomic number of \[28\] and outer most electronic configuration is $3{d^8}4{s^2}$.
 

At the ground state of nickel two 3d-orbitals are singly filled. In the complex, \[Ni{(CO)_4}\] the ligand $CO$ is a neutral ligand, so electron pair takes place at $3d$ and $4s$ orbitals and nickel has $zero(0)$ oxidation state,

The four CO ligands approach 4s and 4p orbital leads to form sp3 hybridisation which is given as,

In \[Ni{(CO)_4}{\text{,}}\] there are no unpaired electrons, so this complex is diamagnetic.
Now we discuss about ${[Ni(CN)_4^{}]^{2 - }}$, tetracyanonickelate $\left( {II} \right)$ ion as,
As we know that the Nickel has atomic number of \[28\] and outer most electronic configuration is $3{d^8}4{s^2}$.

Two orbitals are singly filled. In the complex, ${[Ni(CN)_4^{}]^{2 - }}$ the ligand $CN$ is a strong ligand, so electron pair takes place and nickel has $2 + $oxidation state,

The CN ligand approach one 3d, one 4s and two 4p orbitals to generate dsp2 hybridisation which is given below as,

In ${[Ni(CN)_4^{}]^{2 - }}$ complex, there are no unpaired electrons, so this is diamagnetic.
${[NiCl_4^{}]^{2 - }}$, tetrachloronickelate $\left( {II} \right)$ ion, atomic number of nickel is \[28\] and outer most electronic configuration is $3{d^8}4{s^2}$.

Two orbitals are singly filled. In the complex, ${[NiCl_4^{}]^{2 - }}$ the ligand $Cl$ is a weak ligand, so no electron pair takes place and nickel has $2 + $ oxidation state,

In this case, the weak ligand nature of Cl, the unpaired electrons in d orbital do not get paired. Thus they remains unpaired so this ligand approach one 4s and three 4p orbitals to create sp3 hybridisation as,

From above information the complex, ${[NiCl_4^{}]^{2 - }}$ is paramagnetic, because there are two unpaired electrons.
From above data \[Ni{(CO)_4}\],${[Ni(CN)_4^{}]^{2 - }}$ are diamagnetic but ${[Ni(Cl)_4^{}]^{2 - }}$ is paramagnetic.

The correct option is C. Both (A) and (B) correct.

Note: \[Ni{(CO)_4}\], ${[Ni(CN)_4^{}]^{2 - }}$ and ${[Ni(Cl)_4^{}]^{2 - }}$ structure is tetrahedral, but hybridization is $s{p^3}$, $ds{p^2}$and$s{p^3}$ respectively. VBT's important aspect is the condition of maximum overlap, which leads to the formation of the strongest possible bonds. We must remember that in the coordination compound, the ligand is an ion or molecule which forms a bond with the central metal ion or atom through donating a pair of electrons. Ligands are otherwise known as Lewis base.