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

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

Similar Questions

The number of unpaired electrons in $Ni(CO)_4$ is

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

Which of the following is not a square planar complex?

Identify the type of hybridization present in $[Ni(CN)_4]^{2-}$.

The magnetic moment of the high spin complex is $5.92 \ BM$. What is the electronic configuration?

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