The magnetic moment of [NiX_4]^{2-} ion is fo | vedclass

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(C) The magnetic moment of $[NiX_4]^{2-}$ is zero,which implies that all electrons in the $Ni^{2+}$ ion are paired. $Ni^{2+}$ has a $3d^8$ electronic

Vedclass · Accepted Answer

$dsp^2$ hybridised

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(C) The magnetic moment of $[NiX_4]^{2-}$ is zero,which implies that all electrons in the $Ni^{2+}$ ion are paired. $Ni^{2+}$ has a $3d^8$ electronic configuration. For the electrons to be paired in a $4$-coordinate complex,the $Ni^{2+}$ ion must undergo $dsp^2$ hybridization,resulting in a square planar geometry. This is characteristic of strong field ligands that force pairing of electrons.

List $-I$	List $-II$
$A$. $[PtCl_4]^{2-}$	$I$. $sp^3d$
$B$. $BrF_5$	$II$. $d^2sp^3$
$C$. $PCl_5$	$III$. $dsp^2$
$D$. $[Co(NH_3)_6]^{3+}$	$IV$. $sp^3d^2$

The magnetic moment of $[NiX_4]^{2-}$ ion is found to be zero. Then the ion is : ($X =$ monodentate anionic ligand)

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Match List $-I$ with List $-II$:

List $-I$ List $-II$

$A$. $[PtCl_4]^{2-}$ $I$. $sp^3d$

$B$. $BrF_5$ $II$. $d^2sp^3$

$C$. $PCl_5$ $III$. $dsp^2$

$D$. $[Co(NH_3)_6]^{3+}$ $IV$. $sp^3d^2$

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