Find $n$, the number of geometrical isomers of $[Pd(F)(Cl)(Br)(I)]^{2-}$

• Oxidation state of Pd: $x-1-1-1-1=-2 \Rightarrow x=+2$
• So the complex is $[PdX_4]^{2-}$ with $Pd^{2+}$.
• $Pd^{2+}$ is a $d^8$ ion.
• For $4$-coordinate $d^8$ metal ion like $Pd^{2+}$, the preferred geometry is square planar.

Now, in a square planar complex of type $[MABCD]$ (all four ligands different), geometrical isomerism depends on which ligands are trans to each other.

Distinct trans pairings are:

• $F$ trans $Cl$, and $Br$ trans $I$
• $F$ trans $Br$, and $Cl$ trans $I$
• $F$ trans $I$, and $Cl$ trans $Br$

Hence, the number of geometrical isomers is $n=3$

Interpret the iron complex

Given complex: $[Fe(CN)_6]^{n-6}$ Since $n=3$, $[Fe(CN)_6]^{3-6}=[Fe(CN)_6]^{-3}=[Fe(CN)_6]^{3-}$

Find oxidation state and $d$-electron count of Fe

Let oxidation state of Fe be $x$: $x+6(-1)=-3$ $x-6=-3$ $x=+3$

So iron is $Fe^{3+}$.

Atomic number of Fe = 26 $Fe: [Ar]3d^64s^2$ $Fe^{3+}: 3d^5$

Nature of ligand and spin state

• $CN^-$ is a strong field ligand.
• Therefore, octahedral $Fe^{3+}(d^5)$ becomes low spin.

Electronic configuration in octahedral field: $t_{2g}^5e_g^0$

Calculate spin-only magnetic moment

For $t_{2g}^5$, number of unpaired electrons $n_u=1$.

Spin-only magnetic moment: $\mu=\sqrt{n_u(n_u+2)}\,BM$ $\mu=\sqrt{1(1+2)}=\sqrt{3}\approx 1.73\,BM$

Calculate CFSE

In octahedral field:

• each $t_{2g}$ electron contributes $-0.4\Delta_o$
• each $e_g$ electron contributes $+0.6\Delta_o$

For $t_{2g}^5e_g^0$: $CFSE=5(-0.4\Delta_o)+0(+0.6\Delta_o)$ $CFSE=-2.0\Delta_o$

(Pairing energy is ignored as instructed.)

Match with options

We obtained:

• Magnetic moment $=1.73\,BM$
• $CFSE=-2.0\Delta_o$

This matches Option A.

Comparison with stored correct answer

Stored correct answer: A

My derived answer: A

Hence, they agree.