Animated Solution for Chemistry - Coordination Compounds: 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 $4.90\text{ BM}$. The geometry and the crystal field stabilisation energy of the complex is

Imagine you are a detective, and the magnetic moment is your first major clue. We are given a $d^6$ metal ion forming a complex with aqua ($H_{2}O$) ligands, and its spin-only magnetic moment is $4.90\text{ BM}$.

The spin-only magnetic moment formula is our trusty magnifying glass:

By substituting $\mu =4.90$, we get $n(n+2)\approx 24$. Solving this simple quadratic equation reveals that $n=4$. This means our metal ion has exactly 4 unpaired electrons.

Octahedral vs. Tetrahedral

Now, here is where many students fall into a trap. A $d^6$ ion with 4 unpaired electrons could theoretically form either a high-spin octahedral complex or a high-spin tetrahedral complex. Water is a weak field ligand, meaning the crystal field splitting energy is less than the pairing energy, so the electrons will avoid pairing up if possible.

Let's test the octahedral hypothesis first. In an octahedral field, the $d^6$ high-spin configuration is $t_{2g}^4{e}_g^2$. The Crystal Field Stabilization Energy (CFSE) would be:

If we look at our options, $-0.4{\Delta }_0$ is nowhere to be found! This brilliant question uses the options to guide our logic. Therefore, the complex must be tetrahedral, where the metal ion undergoes $s{p}^3$ hybridization with four water ligands.

Let's visualize the splitting in a tetrahedral field. The five degenerate $d$-orbitals split into a lower energy $e$ set (two orbitals) and a higher energy $t_2$ set (three orbitals).

Following Hund's rule for our high-spin $d^6$ ion, we first place one electron in each of the five orbitals. The sixth electron must then pair up in the lowest available energy level, which is the $e$ set.

This gives us the electronic configuration: $e^3{t}_2^3$.

Finally, let's calculate the CFSE for our tetrahedral complex. Each electron in the $e$ set stabilizes the complex by $-0.6{\Delta }_t$, and each electron in the $t_2$ set destabilizes it by $+0.4{\Delta }_t$.

Substituting our electron counts ($n_e=3$ and $n_{t_2}=3$):

Thus, the geometry is tetrahedral and the CFSE is $-0.6{\Delta }_t$, leading us confidently to the correct option.