Animated Solution for Chemistry - Coordination Compounds: The sum of the spin only magnetic moment values (in B.M.) of $[{\text{Mn(Br)}}_6{]}^{3-}$ and $[{\text{Mn(CN)}}_6{]}^{3-}$ is ______.

To solve this problem, we first need to understand the state of our central metal ion, Manganese (Mn), in both complexes: $[{\text{Mn(Br)}}_6{]}^{3-}$ and $[{\text{Mn(CN)}}_6{]}^{3-}$.

Both bromide (${\text{Br}}^{-}$) and cyanide (${\text{CN}}^{-}$) are monoanionic ligands, meaning they carry a charge of $-1$. Since there are six ligands in each complex and the overall charge of the coordination sphere is $-3$, we can set up a simple algebraic equation to find the oxidation state of Manganese ($x$):

Solving this gives $x=+3$. Thus, Manganese is in the $+3$ oxidation state in both complexes.

Manganese has an atomic number of 25. Its neutral ground-state electronic configuration is $[\text{Ar}]3d^{5}4s^2$. When it loses three electrons to form the ${\text{Mn}}^{3+}$ ion, it loses the two $4s$ electrons first, followed by one $3d$ electron. This leaves us with a $3d^4$ configuration.

Now, we have four $d$-electrons to place in the split $d$-orbitals. How they are placed depends entirely on the nature of the approaching ligands.

Weak vs. Strong

According to Crystal Field Theory (CFT), the five degenerate $d$-orbitals split into two distinct energy levels in an octahedral field: the lower energy $t_{2g}$ level and the higher energy $e_g$ level. The energy gap between them is denoted by ${\Delta }_o$.

For $[{\text{Mn(Br)}}_6{]}^{3-}$: Bromide is a weak field ligand. It causes a very small splitting, meaning ${\Delta }_o$ is less than the pairing energy ($P$). Because the gap is small, the fourth electron finds it easier to jump up to the $e_g$ level rather than pairing up in the $t_{2g}$ level. This results in a high-spin complex with the configuration $t_{2g}^3{e}_g^1$. We have exactly $n=4$ unpaired electrons.

Using the spin-only magnetic moment formula:

For $[{\text{Mn(CN)}}_6{]}^{3-}$: Cyanide, on the other hand, is a strong field ligand. It causes a massive splitting, making ${\Delta }_o$ much greater than the pairing energy ($P$). The energy gap is too large to cross, so the fourth electron is forced to pair up in the lower $t_{2g}$ level. This results in a low-spin complex with the configuration $t_{2g}^4{e}_g^0$. Here, we have one pair and $n=2$ unpaired electrons.

Applying the formula again:

The question asks for the sum of these two magnetic moment values.

This beautiful problem perfectly illustrates how different ligands can drastically alter the quantum mechanical properties of a molecule, changing its magnetic behavior entirely!