Question

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

Answer 1

By recognizing that `"CN"^(-)` is a `pi` acceptor, we would find that it is a strong-field ligand.

What that means is that `"CN"^(-)` accepts electron density from the `d_(xy)`, `d_(xz)`, and/or `d_(yz)` orbitals of `"Cr"` into its `pi^"*"` molecular orbitals. Here is an example with `"CO"`:


Therefore, since it's a strong-field ligand and there are six `"CN"^(-)` ligands surrounding the `"Cr"`, the free-ion field of `"Cr"` is perturbed so that it becomes an octahedral field in which the `e_g` and `t_(2g)` orbitals are far apart in energy, relative to `sigma` donors and `pi` donors:


This means the electron configuration of `"Cr"` will be such that the electrons pair in the `t_(2g)` orbitals first before filling the `e_g` orbitals. This is called low-spin.

`"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_(2g)` orbitals.

For low-spin, we thus have this `t_(2g)` configuration:

`ul(uarr darr)" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))`

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 = {|L+S|, |L+S-1|, . . . , |L-S+1|, |L-S|}` 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 = 1/2-1/2+1/2+1/2 = 1`

`2S+1 = 2(1) + 1 = 3` `=>` triplet

`L = |sum_i m_(l,i)| = |-2-2-1+0| = 5` `=> 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 = {5-1, 5+0, 5+1} = {4,5,6}`

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

`""^(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(ul(uarr darr)" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr)))_(3d)`

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)`.

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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.