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(B) $1$. The atomic number of $Ni$ is $28$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.$2$. In the complex $[Ni(H_2O)_6]^{2+}$,the oxida

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$sp^3d^2$

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(B) $1$. The atomic number of $Ni$ is $28$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.$2$. In the complex $[Ni(H_2O)_6]^{2+}$,the oxidation state of $Ni$ is $+2$. Thus,the configuration of $Ni^{2+}$ is $[Ar] 3d^8$.$3$. $H_2O$ is a weak field ligand,so it does not cause pairing of electrons in the $3d$ orbitals.$4$. To accommodate six pairs of electrons from six $H_2O$ ligands,the metal ion uses one $4s$,three $4p$,and two $4d$ orbitals.$5$. This results in $sp^3d^2$ hybridization,which corresponds to an outer orbital octahedral complex.

The type of hybridization involved in the metal ion of the $[Ni(H_2O)_6]^{2+}$ complex is:

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