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

The Palladium Complex

Let's start our journey by analyzing the first complex given in the problem: $[\text{Pd(F)(Cl)(Br)(I)}{]}^{2-}$. This is a fascinating molecule because it features a Palladium ion in the $+2$ oxidation state surrounded by four completely different halogen ligands.

Now, there is a golden rule in coordination chemistry that you must always remember: Palladium(II) and Platinum(II) ions always form square planar complexes, regardless of the nature of the ligands attached to them. This happens because they belong to the $4d$ and $5d$ transition series, where the crystal field splitting energy is intrinsically very high, forcing a low-spin $ds{p}^2$ hybridization.

Since we have a square planar geometry with four distinct ligands, we can classify this as an $[Mabcd]$ type complex. If you try to arrange four different groups around a central square plane, you will find exactly three unique geometrical isomers. You can visualize this by fixing one ligand's position and placing the other three trans to it, one by one. Therefore, the value of our mysterious variable $n$ is exactly $3$.

Finding the Iron Complex

With the value of $n$ securely in our grasp, we can now decode the second part of the problem. We are given the iron complex $[{\text{Fe(CN)}}_6{]}^{n-6}$.

By substituting $n=3$ into the formula, the complex reveals its true identity: $[{\text{Fe(CN)}}_6{]}^{3-}$. This is the famous hexacyanoferrate(III) ion.

To understand its magnetic and energetic properties, we first need to determine the oxidation state of the central iron atom. Let the oxidation state of iron be $x$. Since each cyanide ligand carries a $-1$ charge, we can set up a simple equation:

Solving this gives us $x=+3$. Iron is in the $+3$ oxidation state.

Iron has an atomic number of $26$, which gives the neutral atom an electronic configuration of $[\text{Ar}]3d^{6}4s^2$. When it loses three electrons to form the ${\text{Fe}}^{3+}$ ion, it loses the two $4s$ electrons and one $3d$ electron, leaving us with a $3d^5$ configuration.

Now, we must invoke Crystal Field Theory. The cyanide ion (${\text{CN}}^{-}$) is a notoriously strong field ligand because of its ability to engage in $\pi$-backbonding. This creates a massive energy gap (${\Delta }_o$) between the lower $t_{2g}$ and higher $e_g$ orbitals in the octahedral field.

Because ${\Delta }_o$ is much greater than the pairing energy ($P$), the five $d$-electrons will avoid the high-energy $e_g$ orbitals. Instead, they will crowd into the lower $t_{2g}$ orbitals, pairing up as much as possible. This results in the electronic configuration: $t_{2g}^5{e}_g^0$.

Magnetic Moment and CFSE

If we look closely at the $t_{2g}^5$ arrangement, we see two fully paired orbitals and exactly one unpaired electron.

We can now calculate the spin-only magnetic moment ($\mu$) using the formula:

Substituting $n=1$:

Finally, let's calculate the Crystal Field Stabilisation Energy (CFSE). The formula for an octahedral complex is:

Plugging in our electron counts:

Our calculated magnetic moment is $1.73\text{ BM}$ and the CFSE is $-2.0{\Delta }_o$, which perfectly matches option (a). This is a brilliant problem that elegantly bridges the concepts of geometrical isomerism and crystal field theory!