Question

Which of these complexes has the lowest-energy `d-d` transition band in a UV-Vis spectrum?

`a)` `["CoBr"("NH"_3)_5]^(3+)`
`b)` `["Co"("H"_2"O")("NH"_3)_5]^(3+)`
`c)` `["Co"("CN"-kappa""C)("NH"_3)_5]^(2+)`
`d)` `["Co"("NH"_3)_6]^(3+)`

Answer 1

The lowest-energy `d-d` transition band would mean that the ligand field splitting energy for the octahedral complex, `Delta_o`, is small.

That is when the complex has a high-spin configuration, and the complex has few strong-field ligands or many weak-field ligands.

In the spectrochemical series, the ligand field strength of each ligand is:

`overbrace("Br"^(-))^("pi donor")` `"<<"` `overbrace("H"_2"O")^("mostly sigma donor") < overbrace("NH"_3)^"sigma donor"` `"<<"` `overbrace("CN"^(-))^"pi acceptor AND sigma donor"`

And so we have...

`barul(|stackrel(" ")(" "Delta_o(A) < Delta_o(B) < Delta_o(D) < Delta_o(C)" ")|)`

---

First off, let's define `pi` acceptors, `sigma` donors, and `pi` donors... [You may also want to read the this answer for a brief review of crystal field theory vs. ligand field theory.](https://socratic.org/questions/why-does-an-increase-in-oxidation-state-of-the-central-metal-ion-in-a-transition)

• `bbpi` acceptors interact in a backbonding interaction, where the metal `t_(2g)` orbitals donate electron density back into the ligand's antibonding orbitals. As a result, the `t_(2g)` orbitals are lowered in energy (because repulsions are lessened), increasing `Delta_o`.
• `bbpi` donors donate electron density into the `t_(2g)` orbitals (which are the triply-degenerate, `pi`-compatible orbitals), destabilizing those orbitals and slightly decreasing `Delta_o`.
• `bbsigma` donors donate electron density into the metal `sigma`-compatible orbitals, raising the energy of the metal antibonding orbitals, the `e_g^"*"` (in crystal field theory they may be labeled simply `e_g`), thus increasing `Delta_o`.

In short... having a lot of `pi` acceptors promotes low spin, having a lot of `pi` donors promotes high spin, and having a lot of `sigma` donors promotes low spin.

You can look at the spectrochemical series to check which ligands are strong-field and which are weak-field.
https://en.wikipedia.org/wiki/Spectrochemical_series

The one different ligand is highlighted in red and shall be considered.

`A)`

`["Co"color(red)("Br")("NH"_3)_5]^(2+)`, or pentamminebromocobalt(III), contains:

• `"Br"^(-)` is a weak-field ligand, because it is a `bbpi` donor.
• `"NH"_3` is a strong-field ligand, because it is a `bbsigma` donor.

Since `A` contains a weak-field ligand, it is expected to have a `Delta_o` that is the smallest among `A-D`.

`B)`

`["Co"color(red)("H"_2"O")("NH"_3)_5]^(3+)`, or pentammineaquacobalt(III), likewise has many strong-field ammine ligands, so this is going to be low spin, i.e. `Delta_o` is large.

In fact, water is a `sigma` donor (primarily), so `Delta_o` for this complex is larger than for `A`.

`C)`

`["Co"color(red)("CN")("NH"_3)_5]^(2+)`, or pentamminecyanocobalt(III), has five `"NH"_3` ligands, so you again know that this is going to have a large `Delta_o`. However, `"CN"^(-)`, cyanide, is both a `pi` acceptor and `sigma` donor, making it VERY strong-field.

Hence, this has the largest `Delta_o` among `A-C`.

`D)`

`["Co"color(red)("NH"_3)("NH"_3)_5]^(3+)`, or hexamminecobalt(III) (I wrote the formula like that on purpose), has all six `sigma` donors, and so it has a large `Delta_o`, being a low-spin complex.

Comparing, this has a larger `Delta_o` than `A` and `B`, but smaller than `C`.

Overall:

`color(blue)barul(|stackrel(" ")(" "Delta_o(A) < Delta_o(B) < Delta_o(D) < Delta_o(C)" ")|)`