$[N i C l_{4}]$ is paramagnetic while $[N i C O_{4}]$ is diamagnetic though both are tetrahedral. Explain.

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Solution

Though both $[N i C l_{4}]^{2 -}$ and $[N i (C O)_{4}]$ are tetrahedral, their magnetic characters are different. This is due to a difference in the nature of ligangds. $C l^{-}$ is a weak field ligand and it does not cause the pairing of unpaired 3d electrons. Hence, $[N i C l_{4}]^{2 -}$ is paramgnetic.

In $N i (C O)_{4}$ , Ni is in the zero oxidation state i.e., it has a configuration of $3 d^{8} 4 s^{2}$

But CO is a strong field ligand. Therefore, it causes the pairing of unpaired 3d electrons. Also, it causes the 4s electrons to shift to the 3d orbital, thereby giving rise to $s p^{3}$ hybridization. Since no unpaired electrons are present in this case, $[N i (C O)_{4}]$ is diamagnetic.

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Similar questions

[NiCl₄]²⁻ is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why?

[C o {(N H_{3})}_{5} C l] S O_{4}

[N i {(C O)}_{4}]

{[N i {C l}_{4}]}^{2 -}

[C o (e n)_{2} C l_{2}]^{+} .

d^{4}

Δ_{0} > P .

[N i C l_{4}]^{2 -}

[N i (C O)_{4}]

(i) Draw the geometrical isomers of complex $[C o (e n)_{2} C l_{2}]^{+}$ .

(ii) On the basis of crystal field theory, write the electronic configuration for d⁴ ion if $Δ_{0} > P$ .

(iii) $[N i C l_{4}]^{2 -}$ is paramagnetic, while $[N i (C O)_{4}]$ is diamagnetic, though both are tetrahedral. Why? (Atomic number of Ni = 28)

Explain on the basis of valence bond theory that [Ni(CN)₄]²⁻ ion with square

planar structure is diamagnetic and the [NiCl₄]²⁻ ion with tetrahedral geometry is paramagnetic.

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