Nickel (Z = 28) combines with a uninegative m | vedclass

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(B) The atomic number of Nickel $(Ni)$ is $28$. Its electronic configuration is $[Ar] 3d^8 4s^2$.In the complex $[NiL_4]^{2-}$,let the oxidation state

Vedclass · Accepted Answer

$dsp^2$,zero

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(B) The atomic number of Nickel $(Ni)$ is $28$. Its electronic configuration is $[Ar] 3d^8 4s^2$.In the complex $[NiL_4]^{2-}$,let the oxidation state of $Ni$ be $x$. Since $L$ is a uninegative ligand,its charge is $-1$.$x + 4(-1) = -2 \implies x = +2$.So,$Ni^{2+}$ has the configuration $[Ar] 3d^8$.For a complex to be diamagnetic,all electrons must be paired.In a $d^8$ system,if the complex is diamagnetic,the ligand must be a strong field ligand that causes pairing of electrons in the $3d$ orbitals,resulting in $dsp^2$ hybridisation.With $dsp^2$ hybridisation,the $3d$ orbitals are arranged such that there are no unpaired electrons.Therefore,the hybridisation is $dsp^2$ and the number of unpaired electrons is $0$.

Nickel $(Z = 28)$ combines with a uninegative monodentate ligand to form a diamagnetic complex $[NiL_4]^{2-}$. The hybridisation involved and the number of unpaired electrons present in the complex are respectively

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