If Hund's rule is violated,then select the CO | vedclass

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(B) The atomic number of $Ni$ is $28$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.If $Hund's$ rule is violated,the $8$ electrons in the

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$d^2sp^3$,diamagnetic

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(B) The atomic number of $Ni$ is $28$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.If $Hund's$ rule is violated,the $8$ electrons in the $3d$ orbitals would pair up in the lower energy orbitals first.Specifically,the $3d$ orbitals would be filled as: $(d_{xy})^2 (d_{yz})^2 (d_{zx})^2 (d_{x^2-y^2})^2 (d_{z^2})^0$.This leaves two inner $d$-orbitals ($d_{x^2-y^2}$ and $d_{z^2}$) vacant.These two vacant $d$-orbitals,along with one $4s$ and three $4p$ orbitals,undergo $d^2sp^3$ hybridization.Since all electrons are paired in this configuration,the complex becomes diamagnetic.Therefore,the correct statement is $d^2sp^3$ and diamagnetic.

List-$I$ (Compound)	List-$II$ (Hybridization of Central atom)
$A. [Ni(NH_3)_6]^{2+}$	$1. sp^3$
$B. [PtCl_4]^{2-}$	$2. sp^3d^2$
$C. [Ni(CO)_4]$	$3. dsp^2$
$D. [Co(ox)_3]^{3-}$	$4. d^2sp^3$

If $Hund's$ rule is violated,then select the $CORRECT$ statement regarding $[Ni(NH_3)_6]^{2+}$:

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