How many unpaired electrons are present in Ni | vedclass

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(B) The atomic number of $Ni$ is $28$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.When $Ni$ forms $Ni^{2+}$ ion,it loses two electrons f

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$2$

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(B) The atomic number of $Ni$ is $28$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.When $Ni$ forms $Ni^{2+}$ ion,it loses two electrons from the $4s$ orbital.Thus,the electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.In the $3d$ subshell,there are $5$ orbitals. According to Hund's rule,the $8$ electrons are filled as follows: $\uparrow\downarrow, \uparrow\downarrow, \uparrow\downarrow, \uparrow, \uparrow$.There are $2$ unpaired electrons in the $3d$ orbitals.

List-$I$ (Complex)	List-$II$ (Hybridization)
$A. Ni(CO)_4$	$I. sp^3d^2$
$B. [Ni(CN)_4]^{2-}$	$II. d^2sp^3$
$C. [Co(NH_3)_6]^{3+}$	$III. dsp^2$
$D. [CoF_6]^{3-}$	$IV. sp^3$

How many unpaired electrons are present in $Ni^{2+}$?

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Match the complexes in List-$I$ with their hybridization in List-$II$.
List-$I$ (Complex) List-$II$ (Hybridization)
$A. Ni(CO)_4$ $I. sp^3d^2$
$B. [Ni(CN)_4]^{2-}$ $II. d^2sp^3$
$C. [Co(NH_3)_6]^{3+}$ $III. dsp^2$
$D. [CoF_6]^{3-}$ $IV. sp^3$

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