Match the List-I with List-II:

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Question

(A) $[Ni(CO)_4]$: $Ni$ is in $0$ oxidation state. Configuration: $[Ar] 3d^8 4s^2$. Due to strong field ligand $CO$,electrons pair up. Hybridization: $

Vedclass · Accepted Answer

$A-III, B-II, C-I, D-IV$

Vedclass · Answer

(A) $[Ni(CO)_4]$: $Ni$ is in $0$ oxidation state. Configuration: $[Ar] 3d^8 4s^2$. Due to strong field ligand $CO$,electrons pair up. Hybridization: $sp^3$. Shape: Tetrahedral. Magnetic moment: $0 \ BM$ (Diamagnetic). Matches $III$.
$(B)$ $[Ni(CN)_4]^{2-}$: $Ni$ is in $+2$ oxidation state. Configuration: $[Ar] 3d^8$. Due to strong field ligand $CN^-$,electrons pair up. Hybridization: $dsp^2$. Shape: Square planar. Magnetic moment: $0 \ BM$ (Diamagnetic). Matches $II$.
$(C)$ $[NiCl_4]^{2-}$: $Ni$ is in $+2$ oxidation state. Configuration: $[Ar] 3d^8$. Weak field ligand $Cl^-$. Hybridization: $sp^3$. Shape: Tetrahedral. Two unpaired electrons,$\mu = \sqrt{n(n+2)} = \sqrt{2(4)} = \sqrt{8} \approx 2.8 \ BM$. Matches $I$.
$(D)$ $[MnBr_4]^{2-}$: $Mn$ is in $+2$ oxidation state. Configuration: $[Ar] 3d^5$. Weak field ligand $Br^-$. Hybridization: $sp^3$. Shape: Tetrahedral. Five unpaired electrons,$\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.9 \ BM$. Matches $IV$.
Therefore,the correct matching is $A-III, B-II, C-I, D-IV$.

List-$I$ (Complex/Species)	List-$II$ (Shape and magnetic moment)
$A$. $[Ni(CO)_4]$	$I$. Tetrahedral,$2.8 \ BM$
$B$. $[Ni(CN)_4]^{2-}$	$II$. Square planar,$0 \ BM$
$C$. $[NiCl_4]^{2-}$	$III$. Tetrahedral,$0 \ BM$
$D$. $[MnBr_4]^{2-}$	$IV$. Tetrahedral,$5.9 \ BM$

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

Similar Questions

Statement $1$: $[Ti(H_2O)_6]^{4+}$ is coloured while $[Sc(H_2O)_6]^{3+}$ is colourless.
Statement $2$: $d-d$ transition is not possible in $[Sc(H_2O)_6]^{3+}$.

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

List $-I$ List $-II$

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

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

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

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

Consider that a $d^{6}$ metal ion $(M^{2+})$ forms a complex with aqua ligands,and the spin-only magnetic moment of the complex is $4.90 \ BM.$ The geometry and the crystal field stabilization energy $(CFSE)$ of the complex are:

What is true for $[Fe(CN)_6]^{3-}$ and $[FeF_6]^{3-}$?

Assertion : $[FeF_6]^{3-}$ is a low spin complex.
Reason : Low spin complexes have a lesser number of unpaired electrons.

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