Colour and spectra — d-d transitions, charge transfer; selection rules: Exercises

Two constants we will re-use every time — memorise the picture, not the digits:

Level 1 — Recognition

L1.1 — d-count and colour

Problem. Which of these ions can show a d–d band, and which are colourless from d–d alone? $Sc^{3 +}$ , $Ti^{3 +}$ , $Zn^{2 +}$ , $Cu^{2 +}$ .

Recall Solution

First find the d-electron count $d^{n}$ (see Oxidation States of Transition Metals).

$Sc^{3 +}$ : Sc is $[Ar] 3 d^{1} 4 s^{2}$ ; remove 3 electrons → $d^{0}$ . No electron to lift → colourless.
$Ti^{3 +}$ : Ti loses 3 → $d^{1}$ . One electron in $t_{2 g}$ that can jump to $e_{g}$ → d–d allowed by population.
$Zn^{2 +}$ : $d^{10}$ . The upper $e_{g}$ is completely full → no empty seat to jump into → colourless.
$Cu^{2 +}$ : $d^{9}$ (one hole in $e_{g}$ ) → d–d possible → coloured (pale blue). Answer: d–d bands for $Ti^{3 +}$ and $Cu^{2 +}$ ; colourless (d–d) for $Sc^{3 +}$ and $Zn^{2 +}$ .

L1.2 — Complementary colour

Problem. A complex absorbs at $500 nm$ . What colour does the eye see?

Recall Solution

$500 nm$ is green light. The transmitted light is white minus green. Your eye adds up the remainder → red/purple (the complementary colour). It does not look green.

Level 2 — Application

L2.1 — $\overset{ν}{ˉ}$ from $λ$

Problem. A complex absorbs at $λ = 625 nm$ . Compute the wavenumber $\overset{ν}{ˉ}$ in $cm^{- 1}$ .

Recall Solution

Step 1 (unit fix). Convert nm → cm: $625 nm = 625 \times 1 0^{- 7} cm = 6.25 \times 1 0^{- 5} cm$ . Why cm? Wavenumber is quoted per centimetre, the spectroscopist's standard. Step 2. $\overset{ν}{ˉ} = \frac{1}{6.25 \times 1 0 ^{- 5} cm} = 16000 cm^{- 1}$ . Answer: $\overset{ν}{ˉ} = 16000 cm^{- 1}$ .

L2.2 — $Δ_{o}$ in joules and per mole

Problem. For the band above ( $\overset{ν}{ˉ} = 16000 cm^{- 1}$ ), find $Δ_{o}$ in joules per photon and in kJ/mol. Take $N_{A} = 6.022 \times 1 0^{23} mol^{- 1}$ .

Recall Solution

Step 1. $Δ_{o} = h c \overset{ν}{ˉ} = (1.986 \times 1 0^{- 23} J\cdotcm) (16000 cm^{- 1})$ . $= 3.178 \times 1 0^{- 19} J$ per photon. Why this bridge? $\overset{ν}{ˉ}$ is proportional to energy, and $h c$ is the exact proportionality constant. Step 2 (per mole). $\times N_{A} = 3.178 \times 1 0^{- 19} \times 6.022 \times 1 0^{23} = 1.914 \times 1 0^{5} J/mol \approx 191 kJ/mol$ . Answer: $Δ_{o} \approx 3.18 \times 1 0^{- 19} J \approx 191 kJ/mol$ .

L2.3 — Which ligand absorbs bluer?

Problem. $[CoF_{6}]^{3 -}$ absorbs near $700 nm$ ; $[Co (NH_{3})_{6}]^{3 +}$ absorbs near $470 nm$ . Which has the larger $Δ_{o}$ , and is that consistent with the Spectrochemical Series?

Recall Solution

Shorter absorbed $λ$ ⇒ larger gap, because $Δ_{o} = h c / λ$ . $470 nm < 700 nm$ , so $[Co (NH_{3})_{6}]^{3 +}$ has the larger $Δ_{o}$ . In the spectrochemical series $NH_{3} > F^{-}$ in field strength, so a bigger gap for the ammine complex is exactly what we expect. Consistent. ✓

Figure — Colour and spectra — d-d transitions, charge transfer; selection rules

Level 3 — Analysis

L3.1 — Why is $MnO_{4}^{-}$ intense but $[Mn (H_{2} O)_{6}]^{2 +}$ faint?

Problem. Both contain manganese. One is a deep, staining purple; the other is barely pink. Explain using d-count and selection rules, and predict which has $ε \sim 1 0^{4}$ vs $ε \sim 0.05$ .

Recall Solution

$MnO_{4}^{-}$ : Mn is $+ 7$ → $d^{0}$ . There are no d-electrons, so a d–d transition is impossible. Its colour is LMCT: an oxygen lone-pair electron leaps onto Mn. CT is Laporte-allowed (it changes parity properly), so it is fully allowed → $ε \sim 1 0^{4}$ (intense, staining). $[Mn (H_{2} O)_{6}]^{2 +}$ : Mn is $+ 2$ → high-spin $d^{5}$ ; every d-orbital holds one electron, all spins parallel. Any $t_{2 g} \to e_{g}$ jump must land in an orbital where it would have to pair with an opposite spin → the number of unpaired electrons changes → spin-forbidden, and it is Laporte-forbidden too. Doubly forbidden → $ε \sim 0.05$ (pale pink). Answer: $MnO_{4}^{-}$ has $ε \sim 1 0^{4}$ ; $[Mn (H_{2} O)_{6}]^{2 +}$ has $ε \sim 0.05$ .

L3.2 — Concentration from Beer–Lambert

Problem. A d–d band has $ε = 10 L mol^{- 1} cm^{- 1}$ . A CT band has $ε = 1 0^{4} L mol^{- 1} cm^{- 1}$ . Using Beer–Lambert Law ( $A = ε c ℓ$ , $ℓ = 1 cm$ ), what concentration of each gives an absorbance $A = 1$ ? Comment on the ratio.

Recall Solution

Rearrange: $c = \frac{A}{ε ℓ}$ .

d–d: $c = \frac{1}{10 \times 1} = 0.1 mol/L$ .
CT: $c = \frac{1}{1 0 ^{4} \times 1} = 1 \times 1 0^{- 4} mol/L$ . Ratio: the CT band reaches the same absorbance at $1000 \times$ lower concentration. That is precisely why a trace of permanganate stains water deep purple while a d–d complex must be concentrated to show strong colour.

Level 4 — Synthesis

L4.1 — From spectrum to ligand identity

Problem. An octahedral Ti(III) complex ( $d^{1}$ ) shows its single d–d band at $\overset{ν}{ˉ} = 20300 cm^{- 1}$ . (a) Give $Δ_{o}$ in kJ/mol. (b) A second Ti(III) complex absorbs at $13000 cm^{- 1}$ . Which complex has the weaker-field ligand, and roughly where should each appear on the Spectrochemical Series relative to $H_{2} O$ (whose Ti(III) band sits near $20000 cm^{- 1}$ )?

Recall Solution

(a) $Δ_{o} = h c \overset{ν}{ˉ} = 1.986 \times 1 0^{- 23} \times 20300 = 4.032 \times 1 0^{- 19} J$ . Per mole: $\times N_{A} = 2.428 \times 1 0^{5} J/mol \approx 243 kJ/mol$ . (b) Smaller $\overset{ν}{ˉ}$ ⇒ smaller gap ⇒ weaker field. The $13000 cm^{- 1}$ complex has the weaker-field ligand (well below H₂O's ~ $20000 cm^{- 1}$ — e.g. a halide-type ligand). The $20300 cm^{- 1}$ complex is essentially H₂O-strength or slightly stronger. Answer: (a) $\approx 243 kJ/mol$ ; (b) the $13000 cm^{- 1}$ complex is weaker-field.

L4.2 — Colour prediction from a shift

Problem. Replacing $H_{2} O$ by $CN^{-}$ in a complex moves the absorption from $500 nm$ to $310 nm$ . (a) Compute both $Δ_{o}$ values in $cm^{- 1}$ . (b) What colour does each complex appear? (c) Explain the direction of the shift.

Recall Solution

(a) Water: $\overset{ν}{ˉ} = 1/ (500 \times 1 0^{- 7} cm) = 20000 cm^{- 1}$ . Cyanide: $\overset{ν}{ˉ} = 1/ (310 \times 1 0^{- 7} cm) = 32258 cm^{- 1}$ (≈ $3.23 \times 1 0^{4}$ ). (b) Water absorbs green (500 nm) → looks red/purple. Cyanide absorbs at 310 nm — that is ultraviolet, outside the visible window → the cyanide complex looks colourless (or very pale) because it no longer removes any visible wavelength. (c) $CN^{-}$ is a strong-field ligand (high in the spectrochemical series) → larger $Δ_{o}$ → higher-energy (shorter- $λ$ ) absorption. Push the ligand field high enough and the band leaves the visible range entirely.

Level 5 — Mastery

L5.1 — Full diagnosis of an unknown

Problem. An unknown octahedral complex is deep, staining orange with $ε \approx 8000$ and its band peaks at $460 nm$ . Diagnose completely: (a) $\overset{ν}{ˉ}$ ; (b) is the band d–d or charge transfer, and why; (c) what does this imply about the metal's oxidation state; (d) name the likely transition type (LMCT vs MLCT) if the metal is in a high oxidation state; (e) contrast with a hypothetical genuine d–d complex of the same colour — how would its $ε$ and concentration-to-stain differ?

Recall Solution

(a) $\overset{ν}{ˉ} = 1/ (460 \times 1 0^{- 7} cm) = 21739 cm^{- 1} \approx 2.17 \times 1 0^{4} cm^{- 1}$ . (b) $ε \approx 8000$ is enormous — d–d bands are Laporte-forbidden and sit at $ε \sim 1$ – $100$ . Only a fully allowed charge-transfer transition reaches $ε \sim 1 0^{3}$ – $1 0^{4}$ . So this is a CT band, not d–d. (c) Intense CT (especially LMCT) is typical of a metal in a high oxidation state (easily reduced), or a $d^{0}$ / $d^{10}$ centre where d–d is impossible anyway. (d) In a high oxidation state the transition is most likely LMCT — a ligand lone pair jumps onto the metal. (e) A real d–d complex of the same orange (also absorbing ~460 nm) would have $ε \sim 10$ instead of $8000$ — roughly $800 \times$ weaker. From Beer–Lambert, to reach $A = 1$ it would need $c = 1/ (10 \times 1) = 0.1 mol/L$ , whereas the CT complex needs $c = 1/ (8000 \times 1) = 1.25 \times 1 0^{- 4} mol/L$ — about $800 \times$ less. The CT species stains at trace level; the d–d species must be concentrated.

L5.2 — Two forbidden-ness levels compared

Problem. Complex A ( $[Ti (H_{2} O)_{6}]^{3 +}$ , $d^{1}$ ) has $ε \approx 5$ . Complex B ( $[Mn (H_{2} O)_{6}]^{2 +}$ , high-spin $d^{5}$ ) has $ε \approx 0.03$ . (a) Which selection rules does each violate? (b) Explain quantitatively (in words) why B is roughly $100 \times$ weaker. (c) What single physical effect lets A absorb at all despite being Laporte-forbidden?

Recall Solution

(a) A ( $d^{1}$ ): the single electron flips $t_{2 g} \to e_{g}$ keeping its spin — $Δ S = 0$ , so it is only Laporte-forbidden (spin rule satisfied). B ( $d^{5}$ high-spin): all five spins are parallel; promoting one to $e_{g}$ demands it pair against an opposite spin, changing the unpaired count — so B is both Laporte- and spin-forbidden. (b) Each forbidding factor suppresses $ε$ by roughly one to two orders of magnitude. A carries one penalty (Laporte) → $ε \sim$ a few. B carries two penalties (Laporte + spin) → an extra ~ $100 \times$ suppression → $ε \sim 0.03$ . Ratio $5/0.03 \approx 167$ , i.e. about $100$ – $200 \times$ , consistent with one extra forbidding rule. (c) Vibronic coupling — an asymmetric vibration momentarily destroys the centre of symmetry, mixing p-character into the d-orbitals and making the Laporte-forbidden jump partly allowed. Without it, A would be truly dark.

Recall ladder

Recall One-line answers (hide and recite)

$\overset{ν}{ˉ}$ from $625 nm$ ::: $16000 cm^{- 1}$ $Δ_{o}$ at $16000 cm^{- 1}$ per mole ::: $\approx 191 kJ/mol$ Larger $Δ_{o}$ absorbs which way ::: shorter $λ$ (bluer, higher energy) Why $MnO_{4}^{-}$ is intense ::: Laporte-allowed LMCT, $ε \sim 1 0^{4}$ Why $[Mn (H_{2} O)_{6}]^{2 +}$ is pale ::: doubly forbidden (Laporte + spin), $ε \sim 0.05$ What relaxes the Laporte rule ::: asymmetric vibration (vibronic coupling) $c$ for $A = 1$ with $ε = 1 0^{4}, ℓ = 1$ cm ::: $1 \times 1 0^{- 4} mol/L$