GATE Transition-Metal Chemistry — Oxidation States, Magnetic Moments and Periodic Trends
Most GATE questions on the d block start the same way: work out the oxidation state, convert it to a d-electron count, then answer whatever is actually being asked — magnetic moment, geometry, colour, stability or distortion. That first step is mechanical and worth practising until it is automatic, because everything downstream depends on it and a wrong dn makes a perfectly reasoned answer wrong.
This article covers oxidation-state assignment, the spin-only magnetic moment with the arithmetic worked in full, high- versus low-spin decisions, Jahn–Teller distortion, ground-state term symbols and the trends that separate the 3d series from 4d and 5d. Crystal-field stabilisation energy has its own article in this knowledge base and is not repeated here.
Step 1 — oxidation state and d-electron count
For the 3d series the group number equals the total of 3d + 4s electrons in the neutral atom, so n = group number − oxidation state works directly. Iron is in group 8, so Fe(III) is 8 − 3 = d5.
| Common ligand | Charge | Common ligand | Charge |
|---|---|---|---|
| NH3, H2O, CO, en, py, PPh3 | 0 | F−, Cl−, Br−, I−, OH−, CN−, NO2−, SCN− | −1 |
| acac−, glycinate | −1 | ox2− (oxalate), CO32−, S2− | −2 |
| EDTA4− | −4 | NO+ (linear nitrosyl, usual convention) | +1 |
Assign the oxidation state and dn count for each.
[Cr(NH3)4Cl2]+: four neutral NH3 plus two Cl− gives −2 from the ligands. x + (−2) = +1 → x = +3. Chromium is group 6, so n = 6 − 3 = d3.
[Fe(CN)6]3−: six CN− gives −6. x − 6 = −3 → x = +3. Iron is group 8 → d5.
[Ni(CO)4]: all ligands neutral, complex neutral → x = 0. Nickel is group 10 → d10. Ten d electrons, no unpaired spins, so this complex is colourless and diamagnetic — both facts follow from one line of arithmetic.
K2[MnO4]: the anion is MnO42−; four O2− gives −8. x − 8 = −2 → x = +6. Manganese is group 7 → d1. Contrast permanganate MnO4−, where x − 8 = −1 → x = +7 → d0: no d electrons at all, which is why its intense purple colour cannot be a d–d transition.
Step 2 — high spin or low spin
Only d4 to d7 octahedral complexes have a choice. Compare the splitting Δo with the pairing energy P: if Δo > P the electrons pair up (low spin); if Δo < P they spread out (high spin).
| dn | High-spin config | Unpaired (HS) | Low-spin config | Unpaired (LS) |
|---|---|---|---|---|
| d4 | t2g3eg1 | 4 | t2g4 | 2 |
| d5 | t2g3eg2 | 5 | t2g5 | 1 |
| d6 | t2g4eg2 | 4 | t2g6 | 0 |
| d7 | t2g5eg2 | 3 | t2g6eg1 | 1 |
Three practical rules decide which column applies. Strong-field ligands (CN−, CO, NO2−, phen, bipy, NH3 at the borderline) push towards low spin; weak-field ligands (I−, Br−, Cl−, F−, H2O, OH−) push towards high spin. A higher metal oxidation state increases Δo. And 4d and 5d metals are essentially always low spin, because their more diffuse orbitals give a much larger Δ.
Tetrahedral complexes are the important exception: Δt is only about four-ninths of Δo for the same metal and ligands, which is never enough to beat the pairing energy. Treat every tetrahedral complex as high spin.
Step 3 — the spin-only magnetic moment
Full form including orbital contribution: µS+L = √( 4S(S + 1) + L(L + 1) ) BM
| n | n(n + 2) | µso (BM) |
|---|---|---|
| 1 | 3 | 1.73 |
| 2 | 8 | 2.83 |
| 3 | 15 | 3.87 |
| 4 | 24 | 4.90 |
| 5 | 35 | 5.92 |
(a) Compare [Fe(H2O)6]3+ and [Fe(CN)6]3−.
Both are Fe(III), d5. Water is weak field → high spin,
t2g3eg2, n = 5:
µ = √(5 × 7) = √35 = 5.92 BM
Cyanide is strong field → low spin, t2g5, n = 1:
µ = √(1 × 3) = √3 = 1.73 BM
Same metal, same oxidation state, same geometry — and a magnetic moment differing by more than a factor of three, decided entirely by the ligand. Magnetic measurement is therefore a direct experimental test of field strength.
(b) [CoF6]3− versus [Co(NH3)6]3+.
Both Co(III), group 9 → d6. Fluoride is weak field → high spin, n = 4:
µ = √(4 × 6) = √24 = 4.90 BM, paramagnetic.
Ammonia here is strong enough for Co(III) → low spin t2g6, n = 0:
µ = 0 BM, diamagnetic. The complex is also kinetically inert, which is the usual
follow-up point about low-spin d6.
(c) Working backwards: a complex has µ = 3.87 BM. How many unpaired electrons?
n(n + 2) = 3.872 = 14.98
n2 + 2n − 14.98 = 0
n = (−2 + √(4 + 59.92)) / 2 = (−2 + √63.92) / 2 = (−2 + 7.995) / 2 = 3.00 → n = 3
Verify: √(3 × 5) = √15 = 3.873 ✓. Three unpaired electrons is consistent with octahedral d3 (Cr3+), high-spin d7, or tetrahedral d7 — the moment alone does not identify the ion, so a question expecting a unique answer must give you the metal or the geometry as well.
When the spin-only formula is not enough. It ignores orbital angular momentum. For ions whose ground term is an A or E term the orbital contribution is largely quenched by the ligand field and the spin-only value works well. For ions with a T ground term — high-spin octahedral Co(II) and Fe(II) are the standard cases — the orbital contribution is not quenched and observed moments run noticeably above the spin-only figure. Lanthanides are a different matter entirely: the 4f electrons are shielded, spin–orbit coupling dominates, and the moment must be computed as µ = gJ√(J(J + 1)). Applying the spin-only formula to a lanthanide is simply the wrong physics.
Jahn–Teller distortion
The theorem states that a non-linear molecule in an orbitally degenerate electronic state distorts to remove the degeneracy. In practice the question is only ever: is the eg set unevenly occupied?
| Configuration | eg occupancy | Distortion | Example |
|---|---|---|---|
| High-spin d4 | eg1 | Strong | Cr2+, Mn3+ |
| d9 | eg3 | Strong | Cu2+ — the textbook case |
| Low-spin d7 | eg1 | Strong | Ni3+, low-spin Co2+ |
| d1, d2, low-spin d4, d5 | t2g uneven | Weak | Rarely detected structurally |
| d3, d5 HS, d6 LS, d8, d10 | even | None | Regular octahedron |
The distortion is strong when eg is uneven because those orbitals point directly at the ligands, so moving a ligand changes the energy a great deal. An uneven t2g set gives only a weak effect because those orbitals point between the ligands. That physical distinction is the answer to "why is d9 strongly distorted but d1 not?" — a very common short question.
Ground-state term symbols for free ions
Apply Hund's rules in order: maximise S, then maximise L for that S, then take J = |L − S| for a less than half-filled shell and J = L + S for more than half-filled.
| dn | Ground term | dn | Ground term |
|---|---|---|---|
| d1, d9 | 2D | d4, d6 | 5D |
| d2, d8 | 3F | d5 | 6S |
| d3, d7 | 4F | d10 | 1S |
Derive the ground term for d2.
Maximum S: two electrons in different orbitals with parallel spins, S = ½ + ½ = 1, so the multiplicity 2S + 1 = 3 (a triplet).
Maximum L consistent with that: the two electrons must occupy different ml values, so the largest total is ml = +2 and +1, giving L = 3 = F.
Ground term = 3F ✓, matching the table. Note the pairing between dn and d10−n: the hole formalism means d8 has the same term.
Trends across and down the block
| Property | Across the 3d series | Down a group (3d → 4d → 5d) |
|---|---|---|
| Atomic radius | Decreases slowly, then rises slightly at the end | 4d > 3d, but 5d ≈ 4d because of the lanthanide contraction |
| Maximum oxidation state | Rises to Mn (+7), then falls | Higher states far more accessible for 4d and 5d |
| Δ (ligand-field splitting) | Rises with oxidation state | Increases substantially — 4d/5d are almost always low spin |
| Metal–metal bonding | Limited | Common and strong for 4d/5d (cluster chemistry) |
| Ionisation energy | Rises gradually | 5d higher than 4d, again from the contraction |
The lanthanide contraction is the single most quoted consequence: the poor shielding by 4f electrons means that after the lanthanides the 5d metals are no larger than their 4d partners. Zirconium and hafnium therefore have almost identical radii and almost identical chemistry, which is precisely why separating them is famously difficult. Niobium and tantalum behave the same way.
Redox potentials across the 3d series are irregular rather than smooth, and the irregularities track electronic stability. Compare two M(III)/M(II) couples that differ only by one place in the series. Reducing Mn(III) gives Mn(II), which is high-spin d5 — a half-filled shell — so the reduction is electronically favourable and Mn(III) is a comparatively strong oxidant. Reducing Fe(III) does the opposite: Fe(III) is itself the d5 ion, and reduction to d6 destroys that half-filled arrangement, so the couple is much milder. The general method for these questions is always the same — identify which side of the couple carries the half-filled or fully filled configuration, and argue from there.
Common mistakes
- Forgetting ligand charges. Six CN− contribute −6; six NH3 contribute nothing. Getting this wrong changes dn and every answer after it.
- Using the total d count in the moment formula. µ = √(n(n + 2)) uses unpaired electrons only.
- Treating a tetrahedral complex as possibly low spin. Δt is far too small; tetrahedral means high spin.
- Assuming 4d and 5d complexes behave like 3d ones. They are essentially always low spin, and they reach much higher oxidation states.
- Applying the spin-only formula to lanthanides. Use gJ√(J(J + 1)) there.
- Expecting Jahn–Teller distortion for every dn. Check the eg occupancy; d3, high-spin d5 and d8 are regular.
- Claiming a d0 or d10 ion is coloured by d–d transitions. It has no d–d transitions at all; any colour is charge transfer.
- Reading a magnetic moment as identifying a unique ion. It gives n, and several configurations share the same n.
Where this appears in GATE Chemistry
| Question type | What is tested | Usual form |
|---|---|---|
| Assign oxidation state and dn | Ligand charge bookkeeping | Numerical or multiple choice |
| Predict µso | High/low spin decision, then √(n(n+2)) | Numerical answer in BM |
| Deduce n from a measured µ | Reversing the formula | Numerical answer (an integer) |
| Identify Jahn–Teller cases | Uneven eg occupancy | Select all that distort |
| Ground-state term symbol | Hund's rules | Multiple choice |
| Explain a trend or an anomaly | Half-filled stability, lanthanide contraction | Conceptual |
The relative weight of inorganic topics is revised between syllabus editions, so confirm the current scope and question pattern from the official GATE notification for your exam year rather than from any secondary summary, this article included.
Get the d-electron count right and the rest follows. The electron configuration tool writes out the ground-state configuration of an element or ion, so you can confirm that Fe(III) really is d5 and Co(III) really is d6 before you commit to a high- or low-spin diagram. That one check prevents the most expensive error on this topic — a well-argued answer built on the wrong dn.
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