Exercises — Colour of complexes — d-d transitions

The rule: a d-d transition needs both at least one $d$-electron and at least one empty $d$-slot. That means the ion must be $d^1$ through $d^9$.

How to get the count: $Sc$ is group 3, atomic config $[Ar]3d^{1}4s^2$; removing 3 electrons ($4s^2$ then $3d^1$) leaves $d^0$. $Ti$: $[Ar]3d^{2}4s^2\to T{i}^{3+}=d^1$. $Zn$: $[Ar]3d^{10}4s^2\to Z{n}^{2+}=d^{10}$. $Cu$: $[Ar]3d^{10}4s^1\to C{u}^{2+}=d^9$. $Mn$: $[Ar]3d^{5}4s^2\to M{n}^{2+}=d^5$.

• Upper set = ==$e_g$== $=d_{x^2-y^2},d_{z^2}$ — they point straight at the ligands, so they are repelled hardest and pushed up.
• Lower set = ==$t_{2g}$== $=d_{xy},d_{yz},d_{zx}$ — they point between ligands, feel less repulsion, sit lower.

See figure s01: the electron jumps from the lower rail up to the higher rail across the gap ${\Delta }_o$.

(a) Use ${\Delta }_o=hc/\lambda$. Why this formula: we are handed a colour and asked for an energy — exactly what it converts. ${\Delta }_o=\frac{(6.626\times 10^{-34})(3.0\times 10^8)}{500\times 10^{-9}}=\frac{1.9878\times 10^{-25}}{5.0\times 10^{-7}}\approx 3.98\times 10^{-19}\text{J}.$ This is the energy carried by one photon = the gap for one ion.

(b) Multiply by $N_A$ to get per mole, then divide by 1000 for kJ: ${\Delta }_o=(3.98\times 10^{-19})(6.022\times 10^{23})\approx 2.40\times 10^5{\text{J mol}}^{-1}\approx \boxed{240{\text{kJ mol}}^{-1}}.$

Rearrange the toolbox formula so wavelength is the subject (we want a colour out): $\lambda =\frac{hc}{{\Delta }_o}=\frac{(6.626\times 10^{-34})(3.0\times 10^8)}{3.0\times 10^{-19}}=\frac{1.9878\times 10^{-25}}{3.0\times 10^{-19}}\approx 6.63\times 10^{-7}\text{m}=\boxed{663\text{nm}}.$ $663\text{nm}$ is in the red region. Absorbing red, we see its complement — green/blue-green (see Complementary Colours & the Colour Wheel).

${\Delta }_1=\frac{hc}{{\lambda }_1}=\frac{1.9878\times 10^{-25}}{800\times 10^{-9}}\approx 2.48\times 10^{-19}\text{J}.$ ${\Delta }_2=\frac{hc}{{\lambda }_2}=\frac{1.9878\times 10^{-25}}{600\times 10^{-9}}\approx 3.31\times 10^{-19}\text{J}.$ ${\Delta }_2>{\Delta }_1$, so $N{H}_3$ produces the bigger gap ⇒ $N{H}_3$ is the stronger field ligand. This matches the Spectrochemical Series: $H_{2}O<N{H}_3$. Ratio: ${\Delta }_2/{\Delta }_1={\lambda }_1/{\lambda }_2=800/600\approx 1.33$ — the gap grows by about $33\%$.

${\Delta }_t=\frac{4}{9}(2.7\times 10^{-19})=1.2\times 10^{-19}\text{J}.$ Why $\frac{4}{9}$: a tetrahedron has only 4 ligands (vs 6) and none point straight at a d-orbital, so the splitting is much weaker — see Octahedral vs Tetrahedral Splitting. ${\lambda }_t=\frac{hc}{{\Delta }_t}=\frac{1.9878\times 10^{-25}}{1.2\times 10^{-19}}\approx 1.66\times 10^{-6}\text{m}=1656\text{nm}.$ That is deep infrared — beyond visible. The absorbed wavelength moved to the red/IR side (longer). Because the gap shrank, the photon needed is less energetic, i.e. longer $\lambda$. (In practice such a complex may look nearly colourless in the visible if its main absorption falls outside $400$–$700\text{nm}$.)

Strongest field = biggest $\Delta$ = shortest absorbed $\lambda$ (since $\Delta \propto 1/\lambda$). So order by increasing wavelength: $\underset{\text{strongest}}{\underbrace{450\text{nm}}}<\underset{\text{middle}}{\underbrace{560\text{nm}}}<\underset{\text{weakest}}{\underbrace{620\text{nm}}}.$ ==Strongest → weakest: the $450$ nm ligand, then $560$ nm, then $620$ nm.== No arithmetic needed — but you must remember the inverse link, or you'll rank them exactly backwards.

• $[Mn(H_{2}O{)}_6{]}^{2+}$ is $M{n}^{2+}$, $d^5$ high-spin. Its d-d transitions are both Laporte-forbidden and spin-forbidden, so very little light is absorbed ⇒ very pale colour. It is a d-d transition, just a weak one.
• $KMn{O}_4$ has $Mn$ in the $+7$ state, which is $d^0$ — no d-d transition is even possible! Its intense purple comes from a completely different mechanism: a charge-transfer transition, where an electron jumps from an oxygen ligand into an empty metal orbital. These are allowed and hence hundreds of times stronger. See Charge-Transfer Spectra (why KMnO4 is intensely coloured).

Lesson: intensity tells you the type of transition; d-d = faint, charge-transfer = blazing.

(a) Purple is what remains after removing its complement from white light. The complement of purple is green-yellow, so the complex absorbs green-yellow ($\sim 520\text{nm}$). (b) Given: $\lambda \approx 520\text{nm}$. (c) ${\Delta }_o=\frac{hc}{\lambda }=\frac{1.9878\times 10^{-25}}{520\times 10^{-9}}\approx 3.82\times 10^{-19}\text{J/photon}.$ Per mole: $(3.82\times 10^{-19})(6.022\times 10^{23})\approx 2.30\times 10^5{\text{J mol}}^{-1}\approx \boxed{230{\text{kJ mol}}^{-1}}.$ (d) Weaker field ⇒ smaller $\Delta$ ⇒ longer absorbed $\lambda$ (shifts from green toward red/orange). Absorbing more red means the seen colour shifts toward the green/blue end. So the solution would look more green (its transmitted colour moves opposite to the absorbed shift).

What is a wavenumber? $\bar{\nu }=1/\lambda$, with $\lambda$ in cm. It is popular because it is directly proportional to energy ($E=hc\bar{\nu }$). $\lambda =\frac{1}{\bar{\nu }}=\frac{1}{20000{\text{cm}}^{-1}}=5.0\times 10^{-5}\text{cm}=5.0\times 10^{-7}\text{m}=\boxed{500\text{nm}}.$ Then ${\Delta }_o=hc/\lambda =3.98\times 10^{-19}\text{J}$ (identical to L2.1 — good, two roads, same city).

(a) Violet-looking complex absorbs the shorter wavelength ($570<650$ nm), so it needs the larger ${\Delta }_o$, hence the stronger field ligand. (b) ${\Delta }_{\text{green}}=\frac{hc}{650\times 10^{-9}}=\frac{1.9878\times 10^{-25}}{6.5\times 10^{-7}}\approx 3.06\times 10^{-19}\text{J}\Rightarrow \times N_A\approx \boxed{184{\text{kJ mol}}^{-1}}.$ ${\Delta }_{\text{violet}}=\frac{hc}{570\times 10^{-9}}=\frac{1.9878\times 10^{-25}}{5.7\times 10^{-7}}\approx 3.49\times 10^{-19}\text{J}\Rightarrow \times N_A\approx \boxed{210{\text{kJ mol}}^{-1}}.$ (c) For the weaker/green target use water: $[Cr(H_{2}O{)}_6{]}^{3+}$ (indeed violet-grey/green). For the stronger/violet target use a stronger ligand such as ammonia $[Cr(N{H}_3{)}_6{]}^{3+}$, since $H_{2}O<N{H}_3$ in the series. The stronger ligand raises ${\Delta }_o$, shifting the absorption to shorter $\lambda$ exactly as computed.

1. Ligands approach → their lone pairs repel the metal d-electrons (Crystal Field Theory).
2. Orbitals pointing at ligands ($e_g$) rise; those between ($t_{2g}$) fall → the set splits by ${\Delta }_o$.
3. Ligand strength / oxidation state / geometry set the size of ${\Delta }_o$ (Spectrochemical Series, Octahedral vs Tetrahedral Splitting).
4. A photon with $E={\Delta }_o$ is absorbed, promoting an electron $t_{2g}\to e_g$ — the d-d transition. Tool: Planck's $E=h\nu$ ⇒ ${\Delta }_o=hc/\lambda$ (Planck's Equation E=hν).
5. That one colour is subtracted from white light.
6. My eye receives "white minus absorbed" = the complementary colour (Complementary Colours & the Colour Wheel). Absorb green-yellow → see purple.

Mnemonic that packs it: "Ligands Split, Light Splits, You See the Opposite."

Absorbs $500$ nm — is ${\Delta }_o$ bigger or smaller than a complex absorbing $600$ nm? ::: Bigger — shorter $\lambda$ means larger gap ($\Delta \propto 1/\lambda$). ${\Delta }_t=\frac{4}{9}{\Delta }_o$ — so tetrahedral complexes tend to absorb at longer or shorter $\lambda$? ::: Longer (smaller gap, less energetic photon). Two Mn species, one faint one blazing — what distinguishes intensity? ::: Whether the transition is allowed; d-d = faint, charge-transfer = intense. Seen colour of a complex absorbing at $650$ nm (red)? ::: Green (the complement).