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

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(D) The atomic number of Nickel $(Ni)$ is $28$. Its electronic configuration is $[Ar] 3d^8 4s^2$.In the complex $[NiX_4]^{-2}$,let the oxidation state

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$2$,tetrahedral

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(D) The atomic number of Nickel $(Ni)$ is $28$. Its electronic configuration is $[Ar] 3d^8 4s^2$.In the complex $[NiX_4]^{-2}$,let the oxidation state of $Ni$ be $y$.$y + 4(-1) = -2 \implies y = +2$.So,$Ni^{+2}$ has the configuration $[Ar] 3d^8$.In the $3d$ orbital,there are $8$ electrons. According to Hund's rule,these are distributed as $3$ paired and $2$ unpaired electrons.Since the complex is paramagnetic and the ligand $X^-$ is a weak field ligand (as it forms a tetrahedral complex),no pairing of electrons occurs in the $3d$ orbitals.Thus,there are $2$ unpaired electrons.The hybridization involved is $sp^3$,which corresponds to a tetrahedral geometry.

Nickel $(Z = 28)$ combines with a uninegative monodentate ligand $X^{-}$ to form a paramagnetic complex $[NiX_4]^{-2}.$ The number of unpaired electron$(s)$ in the nickel and geometry of this complex ion are,respectively $:$

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