[PdCl_2(PMe_3)_2] is a diamagnetic complex of | vedclass

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(C) $Pd(II)$ is a $4d$ transition metal, and due to the high crystal field splitting energy $(\Delta_o)$, it forms square planar complexes with strong

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$2$

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(C) $Pd(II)$ is a $4d$ transition metal, and due to the high crystal field splitting energy $(\Delta_o)$, it forms square planar complexes with strong field ligands, resulting in $0$ unpaired electrons.$Ni(II)$ is a $3d$ transition metal with a $d^8$ configuration.In the complex $[NiCl_2(PMe_3)_2]$, $Cl^-$ is a weak field ligand.For $Ni(II)$ $(d^8)$ with weak field ligands, the complex typically adopts a tetrahedral geometry ($sp^3$ hybridization).In a tetrahedral field, the $d$-orbitals split as $e The $8$ electrons fill as $e^4 t_2^4$, resulting in $2$ unpaired electrons.

$LIST-I$	$LIST-II$
$P. dsp^2$	$1. [FeF_6]^{4-}$
$Q. sp^3$	$2. [Ti(H_2O)_3Cl_3]$
$R. sp^3d^2$	$3. [Cr(NH_3)_6]^{3+}$
$S. d^2sp^3$	$4. [FeCl_4]^{2-}$
	$5. Ni(CO)_4$
	$6. [Ni(CN)_4]^{2-}$

Hybridisation	Geometry	Complex structure
$(A) \ sp^3$	$(i) \ \text{Square planar}$	$(p) \ [Fe(CN)_6]^{3-}$
$(B) \ d^2sp^3$	$(ii) \ \text{Tetrahedral}$	$(q) \ [ZnCl_4]^{2-}$
$(C) \ dsp^2$	$(iii) \ \text{Octahedral}$	$(r) \ [Ni(NH_3)_4]^{2+}$
-	$(iv) \ \text{Linear}$	$(s) \ [Ag(CN)_2]^-$

$[PdCl_2(PMe_3)_2]$ is a diamagnetic complex of $Pd(II)$. How many unpaired electrons are present in the analogous complex of $Ni(II)$?

Similar Questions

The paramagnetic complex ion which has no unpaired electrons in $t_{2g}$ orbitals is

Match each set of hybrid orbitals from $LIST-I$ with complex(es) given in $LIST-II$. The correct option is

$LIST-I$ $LIST-II$

$P. dsp^2$ $1. [FeF_6]^{4-}$

$Q. sp^3$ $2. [Ti(H_2O)_3Cl_3]$

$R. sp^3d^2$ $3. [Cr(NH_3)_6]^{3+}$

$S. d^2sp^3$ $4. [FeCl_4]^{2-}$

$5. Ni(CO)_4$

$6. [Ni(CN)_4]^{2-}$

Aluminium chloride in acidified aqueous solution forms an ion having which geometry?

Pick out the correct statement with respect to $[Mn(CN)_6]^{3-}.$

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