The hybridisation of Ni,shape,and number of u | vedclass

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(A) $1$. In $[NiCl_4]^{2-}$,the oxidation state of $Ni$ is $x + 4(-1) = -2$,so $x = +2$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.$2$.

Vedclass · Accepted Answer

$sp^3$,tetrahedral,$2$

Vedclass · Answer

(A) $1$. In $[NiCl_4]^{2-}$,the oxidation state of $Ni$ is $x + 4(-1) = -2$,so $x = +2$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.$2$. $Cl^-$ is a weak field ligand,so it does not cause pairing of electrons in the $3d$ orbitals.$3$. The $3d^8$ configuration has $2$ unpaired electrons.$4$. To accommodate $4$ ligands,$Ni^{2+}$ undergoes $sp^3$ hybridisation using one $4s$ and three $4p$ orbitals.$5$. $sp^3$ hybridisation results in a tetrahedral geometry.$6$. Therefore,the hybridisation is $sp^3$,the shape is tetrahedral,and the number of unpaired electrons is $2$.

The hybridisation of $Ni$,shape,and number of unpaired electrons present in $[NiCl_4]^{2-}$ are respectively:

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