Spin-Forbidden Bands and the Spectrum of Manganese(II)

Introduction

Many Mn²⁺ aqua solutions are very pale even though manganese is a transition metal with five d electrons. The explanation is not that no excited states exist. Rather, the high-spin d⁵ ground state has no low-lying excited sextet d–d state, so the metal-centred transitions that are energetically accessible change spin multiplicity. In an ideal octahedral complex those transitions also have the usual g→g parity restriction. Their combined weak probability gives an instructive example of how selection rules shape observed colour.

Core explanation

Mn²⁺ is d⁵. With a weak-to-moderate octahedral field, the five electrons occupy t₂g³e g² with parallel spins, giving S=5/2 and multiplicity six. Its free-ion ^6S ground term has L=0, and the octahedral component is ^6A₁g. The ^6S term contains only one orbital component, so the ligand field does not split it into a partner sextet target. Other d⁵ excited terms in the relevant range have quartet or doublet multiplicities. A transition from ^6A₁g to one of them changes S and violates the simple spin selection rule.

The parity rule adds a second restriction. Both the ground state and metal-centred excited d⁵ states are gerade in ideal O h. An electric-dipole g→g transition is Laporte-forbidden. Spin–orbit coupling mixes a little different-spin character, and odd vibrations momentarily remove inversion symmetry; together these mechanisms make faint absorption possible. The observed pale pink colour of many Mn²⁺ aqua solutions comes from very weak bands and can disappear visually at low concentration or short path length.

“Pale” should be interpreted with measurement conditions. Beer–Lambert absorbance is proportional to concentration and path length for one species, so a concentrated Mn²⁺ solution can show a detectable spectrum even if ε is small. Conversely, impurities or a small amount of a different oxidation state can dominate perceived colour. Manganese in a high oxidation state, such as permanganate, has intense charge-transfer absorption and cannot be explained by the weak high-spin Mn²⁺ d–d picture. Oxidation state and speciation must be established before invoking selection rules.

A high-spin d⁵ Tanabe–Sugano diagram has ^6A₁g ground on its weak-field side, with no same-spin low-energy d–d curves. The dashed or differently styled quartet curves indicate spin-forbidden targets. A strong-field low-spin d⁵ system has a different ground spin and is analysed on the other side of the crossover. The “all d⁵ complexes are pale” slogan is therefore false; the statement specifically concerns high-spin d⁵ metal-centred transitions in a geometry where other intense processes are absent.

Band shapes can help. Spin-forbidden features are often much weaker and sometimes sharper than ordinary spin-allowed d–d bands, although vibrations and multiple overlapping terms still broaden them. Exact term labels depend on the diagram and field strength; assigning them by arbitrary wavelength order is unsafe. Magnetic moment provides a strong complementary test: high-spin d⁵ has five unpaired electrons and a spin-only moment √[5(5+2)]≈5.92 μ B. A substantially different spin count calls for re-examining the high-spin assumption.

Step-by-step reasoning

Determine Mn oxidation state and d count. For Mn²⁺ d⁵ in a weak octahedral field, fill t₂g³e g² with five parallel spins and label ground ^6A₁g. Search the d⁵ diagram for accessible sextet excited d–d terms; none appears in the simple high-spin scheme. Classify quartet/doublet arrows as spin-forbidden and g→g arrows as parity-forbidden. Then use concentration, path length, charge-transfer possibilities and magnetic data to interpret the observed pale spectrum.

Visual explanation

Draw five up-spin arrows, one in each of the three t₂g and two e g boxes. Beside them place a ground line ^6A₁g and several higher quartet lines. Draw faint dotted arrows from the sextet line to quartet lines, labelled ΔS≠0; mark g→g below. In a separate small comparison sketch, draw a very tall charge-transfer absorption to show why an intensely purple manganese species requires a different mechanism.

Real-world analogy

A theatre has several upper balconies, but the normal lift requires passengers to keep the same access category. If all available balconies have a different category, only exceptional side routes allow entry and very few people take them. High-spin Mn²⁺ has real excited states, yet ordinary electric-dipole absorption has no straightforward same-spin route to them.

Real-world example

An aqueous solution of MnCl₂ can look nearly colourless or faintly pink depending on concentration, path length and lighting. Measuring its absorbance can still reveal weak ligand-field features. The dramatically stronger purple colour of MnO₄⁻ reflects a different oxidation state and charge-transfer transitions, demonstrating why “contains manganese” is insufficient to predict intensity.

Why?

Why does the ^6A₁g ground state have no simple spin-allowed d–d partner? The high-spin d⁵ free-ion sextet arises from a single ^6S term with L=0. In the basic ligand-field manifold, no other sextet term is available for a low-energy excitation while conserving S=5/2.

Common misconception

“Mn²⁺ is pale because its d orbitals cannot split.” Octahedral t₂g and e g levels do split, and many-electron excited states exist. Weak intensity comes from spin and parity selection rules, not from an absence of energy separation.

Worked example

An octahedral Mn²⁺ sample gives a weak band and a measured effective moment near 5.9 μ B. Five unpaired electrons are consistent with high-spin d⁵ and ^6A₁g ground. The weak band cannot be assigned to a sextet-to-sextet ligand-field transition in the simple diagram; a quartet target borrowed weak spin intensity is plausible. Because both states are d-derived and g, vibronic parity borrowing is also required. A very strong additional band should be checked for charge transfer or another species.

Quick check

1. What are S and multiplicity for high-spin octahedral d⁵? Answer: Five parallel spins give S=5/2 and multiplicity 2S+1=6.

Exam focus

State Mn²⁺→d⁵→high-spin ^6A₁g before discussing colour. Apply spin and parity rules separately. Do not generalise to low-spin d⁵ or to other manganese oxidation states; strong charge-transfer colour is a different phenomenon.

Advanced insight

Spin–orbit coupling mixes electronic states with different nominal S, so a spin-forbidden transition can borrow intensity from an allowed state. Its strength grows with the mixing matrix element and depends on energy separation from the borrowing state. Vibronic coupling supplies an independent parity-breaking channel, making the observed intensity the product of several small effects rather than an absolute zero.

Summary

High-spin Mn²⁺ is d⁵ with ^6A₁g ground. Its accessible d–d excited states have different spin and, in ideal octahedral symmetry, the same g parity. Weak borrowed transition intensity explains the pale colour; oxidation state, concentration and charge transfer must be checked in real samples.

Practice questions

1. Predict the spin-only moment for five unpaired electrons. Answer: μ so=√[n(n+2)] μ B=√35 μ B≈5.92 μ B, appropriate as a first estimate for high-spin d⁵. 2. Why is the intense colour of permanganate not evidence against the Mn²⁺ explanation? Answer: MnO₄⁻ contains manganese in a different oxidation state and has strong charge-transfer transitions; it is not a high-spin Mn²⁺ d–d spectrum. 3. Name two mechanisms that allow weak octahedral Mn²⁺ bands to be observed. Answer: Spin–orbit coupling lends intensity to nominally spin-forbidden arrows, and odd vibrations or distortion relax the g→g Laporte restriction. 4. Why can the pale colour become more visible in a longer cuvette? Answer: Beer–Lambert absorbance A=εcl rises with path length even if the intrinsic molar absorptivity ε of a forbidden band remains small.