[Pd(F)(Cl)(Br)(I)]^{2-} has n number of geome | vedclass

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(B) The complex $[Pd(F)(Cl)(Br)(I)]^{2-}$ is a square planar complex of the type $[M(abcd)]$. It has $3$ geometrical isomers. Thus,$n = 3$.The formula

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$1.73 \ BM$ and $-2.0 \ \Delta_{0}$

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(B) The complex $[Pd(F)(Cl)(Br)(I)]^{2-}$ is a square planar complex of the type $[M(abcd)]$. It has $3$ geometrical isomers. Thus,$n = 3$.The formula for the second complex becomes $[Fe(CN)_6]^{3- - 6} = [Fe(CN)_6]^{3-}$.In $[Fe(CN)_6]^{3-}$,the oxidation state of $Fe$ is $+3$. The electronic configuration of $Fe^{3+}$ is $3d^5$.Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals. The configuration becomes $t_{2g}^5 e_g^0$.Number of unpaired electrons $(n_{u}) = 1$.Spin-only magnetic moment $= \sqrt{n_{u}(n_{u}+2)} \ BM = \sqrt{1(1+2)} \ BM = \sqrt{3} \ BM \approx 1.73 \ BM$.$CFSE = [(-0.4 	imes n_{t2g}) + (0.6 	imes n_{eg})] \ \Delta_{0} = [(-0.4 	imes 5) + (0.6 	imes 0)] \ \Delta_{0} = -2.0 \ \Delta_{0}$.

$[Pd(F)(Cl)(Br)(I)]^{2-}$ has $n$ number of geometrical isomers. Then,the spin-only magnetic moment and crystal field stabilisation energy $[CFSE]$ of $[Fe(CN)_6]^{n-6}$ are
[Note : Ignore the pairing energy]

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$[Pd(F)(Cl)(Br)(I)]^{2-}$ has $n$ number of geometrical isomers. Then,the spin-only magnetic moment and crystal field stabilisation energy $[CFSE]$ of $[Fe(CN)_6]^{n-6}$ are[Note : Ignore the pairing energy]