# What is the term symbol for "Cr" in ["Cr"("CN")_6]^(4-)?

Mar 5, 2017

By recognizing that ${\text{CN}}^{-}$ is a $\pi$ acceptor, we would find that it is a strong-field ligand.

What that means is that ${\text{CN}}^{-}$ accepts electron density from the ${d}_{x y}$, ${d}_{x z}$, and/or ${d}_{y z}$ orbitals of $\text{Cr}$ into its ${\pi}^{\text{*}}$ molecular orbitals. Here is an example with $\text{CO}$:

Therefore, since it's a strong-field ligand and there are six ${\text{CN}}^{-}$ ligands surrounding the $\text{Cr}$, the free-ion field of $\text{Cr}$ is perturbed so that it becomes an octahedral field in which the ${e}_{g}$ and ${t}_{2 g}$ orbitals are far apart in energy, relative to $\sigma$ donors and $\pi$ donors:

This means the electron configuration of $\text{Cr}$ will be such that the electrons pair in the ${t}_{2 g}$ orbitals first before filling the ${e}_{g}$ orbitals. This is called low-spin.

$\text{Cr}$ in ["Cr"("CN")_6]^(4-) has an oxidation state of $+ 2$, which makes it a ${d}^{4}$ metal. Refer to a periodic table to see that. Since it's a ${d}^{4}$ metal, it means it has four $d$ electrons to distribute in its ${e}_{g}$ and ${t}_{2 g}$ orbitals.

For low-spin, we thus have this ${t}_{2 g}$ configuration:

$\underline{\uparrow \downarrow} \text{ "ul(uarr color(white)(darr))" } \underline{\uparrow \textcolor{w h i t e}{\downarrow}}$

As for the term symbol, I assume since this is a molecule, a molecular term symbol would be appropriate. But apparently, since you want the atomic term symbol, I'll give that instead...

""^(2S+1) L_J

• $S$ is the total spin angular momentum, and is calculated as $S = | {M}_{S} | = | {\sum}_{i} {m}_{s , i} |$ for electron $i$.
• $L$ is the total orbital angular momentum, and is calculated as $L = | {M}_{L} | = | {\sum}_{i} {m}_{l , i} |$ for electron $i$. $L = 0 , 1 , 2 , 3 , 4 , 5 , . . .$ corresponds to $S , P , D , F , G , H , . . .$.
• $J = \left\{| L + S | , | L + S - 1 | , . . . , | L - S + 1 | , | L - S |\right\}$ is the total angular momentum, and ranges from the magnitude of $L - S$ to the magnitude of $L + S$, including integer values in between.

From the above configuration, we take the doubly-occupied orbital to have ${m}_{l} = - 2$, and the two singly-occupied orbitals to have ${m}_{l} = - 1$ and $0$, respectively. So:

$S = \frac{1}{2} - \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 1$

$2 S + 1 = 2 \left(1\right) + 1 = 3$ $\implies$ triplet

$L = | {\sum}_{i} {m}_{l , i} | = | - 2 - 2 - 1 + 0 | = 5$ $\implies H$

Thus, we do have color(blue)(""^3 H).

We don't need $J$ in a free-ion field, but in the presence of a magnetic field, we would, as spin-orbit coupling would be introduced. So:

$J = \left\{5 - 1 , 5 + 0 , 5 + 1\right\} = \left\{4 , 5 , 6\right\}$

That would give us, in a magnetic field, three levels, each individually triply-degenerate:

${\text{^(3) H_4, ""^(3) H_5, }}^{3} {H}_{6}$

From Hund's rules, we maximize $S$ and then $L$, and if those are already maximized with the determined term symbols, then we look to see if the subshell is more or less than half-filled.

As an approximation, these orbitals have significant metal contributions (significant ionic character), so we assume they are essentially equivalent to metal $d$ orbitals when it comes to their degeneracies.

Therefore, for this determination, we approximate them as:

${\underbrace{\underline{\uparrow \downarrow} \text{ "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(color(white)(uarr darr))" } \underline{\textcolor{w h i t e}{\uparrow \downarrow}}}}_{3 d}$

As a result, it's less than half-filled, so we take the smallest $J$ for the ground-state term symbol. Therefore, the ground-state term symbol in a magnetic field is color(blue)(""^3 H_4).

Had this been high-spin ${d}^{4}$, repeat the process and you should get a ""^5 D ground state term. It would also be less than half-filled.