PYQ explainedUnit 13 · Coordination chemistry15 October 2026

Crystal field theory: why [CoF₆]³⁻ is paramagnetic but [Co(CN)₆]³⁻ is diamagnetic

Q: Why is [CoF6]3− paramagnetic while [Co(CN)6]3− is diamagnetic?

Both contain Co3+, a d6 ion, in an octahedral field. F− is a weak-field ligand: Δo (156 kJ/mol) is smaller than the pairing energy P (252 kJ/mol), so the electrons spread out as t2g4 eg2 with 4 unpaired electrons and a spin-only moment of 4.90 BM. CN− is a strong-field ligand: Δo (404 kJ/mol) is larger than P, so all six electrons pair in t2g6 and the complex is diamagnetic. This was asked in UPSC Civil Services 2024, Chemistry Paper I, Q7(c)(i), for 10 marks.

Q: What is pairing energy, and how does it decide high spin or low spin?

Pairing energy P is the extra energy needed to put a second electron into an orbital that already holds one, mainly the extra repulsion between the two. For a d4 to d7 ion in an octahedral field, each electron after the third either pairs in t2g (costing P) or goes up to eg (costing Δo). For d6, E(low spin) − E(high spin) = 2(P − Δo), so the complex is low spin when Δo is greater than P and high spin when it is smaller.

Q: Why is F− a weak-field ligand and CN− a strong-field ligand?

F− is a σ donor and also a π donor: its filled p orbitals push the t2g orbitals up, which makes Δo smaller. CN− is a strong σ donor and a π acceptor: its empty π* orbitals take electron density from t2g by back-bonding, which lowers t2g and makes Δo larger. That is why the spectrochemical series runs I− < Br− < Cl− < F− < OH− < H2O < NH3 < en < NO2− < CN− < CO.

Q: How do you calculate the spin-only magnetic moment of a complex?

Find the oxidation state and the number of d electrons, place them on the splitting diagram to get the number of unpaired electrons n, then use μ = √(n(n + 2)) BM. For n = 1, 2, 3, 4 and 5 this gives 1.73, 2.83, 3.87, 4.90 and 5.92 BM. In UPSC Civil Services 2023, Paper I, Q8(b), 10 marks, the four complexes give 3.87, 3.87, 2.83 and 3.87 BM.

Q: Which d-electron counts can be high spin or low spin?

In an octahedral field only d4, d5, d6 and d7 have two possible arrangements. For d1 to d3 and d8 to d10 there is only one way to fill t2g and eg, so the number of unpaired electrons does not depend on the ligand. Tetrahedral complexes are nearly always high spin, because Δt is only about 4/9 of Δo and is smaller than P.

Q: How often does UPSC ask crystal field theory and magnetic moments?

From 2021 to 2025 the topic appeared in 14 of 15 Chemistry Paper I question papers across UPSC Civil Services, Combined Geo-Scientist and IFoS, 23 questions in all. Seven were direct questions on magnetic behaviour or spin-only moments, for 5 to 15 marks. The only paper without it was IFoS 2021.

By Vimal Swarnkar (VKC) · UPSC Chemistry · Civil Services, Combined Geo-Scientist and IFoS

Same metal, same charge, same six ligands around it, and yet one complex is pulled into a magnet and the other is pushed out. UPSC asks some version of this almost every year. In this post I will show you how often it comes, which book to use, how I plan the answer, and how the answer should look in your booklet.

The question as set in UPSC Civil Services (Main) 2024, Chemistry Paper I, Q7(c)(i), 10 marks

On the basis of Crystal Field Theory, account for the following statement: while [CoF₆]³⁻ is paramagnetic, [Co(CN)₆]³⁻ is diamagnetic.

Short answer

Both contain Co³⁺, d⁶, in an octahedral field. F⁻ is a weak-field ligand: Δ_o = 156 kJ mol⁻¹ is smaller than the pairing energy P = 252 kJ mol⁻¹, so [CoF₆]³⁻ is high spin, t_2g⁴e_g², with 4 unpaired electrons and μ = √(4 × 6) = 4.90 BM: paramagnetic. CN⁻ is a strong-field ligand: Δ_o = 404 kJ mol⁻¹ is larger than P, so [Co(CN)₆]³⁻ is low spin, t_2g⁶, with no unpaired electron and μ = 0: diamagnetic. The rule behind it: E(low spin) − E(high spin) = 2(P − Δ_o), so a d⁶ ion goes low spin only when Δ_o > P.

Full topic notes: Crystal field theory · two-page short notes

A word from me

When I check answers on this question, almost everyone writes “F⁻ is a weak-field ligand, so high spin; CN⁻ is a strong-field ligand, so low spin” and stops. That line is correct, but it is a slogan, not an explanation. The question says “on the basis of Crystal Field Theory”, and the theory says something sharper: every electron after the third has a choice. It can pair up in t_2g and pay the pairing energy P, or it can go up to e_g and pay Δ_o. It takes the cheaper road. Once you write that comparison with numbers, the examiner sees that you understand the theory, and the marks follow.

My advice is to practise with one book and one table. Take the table of P and Δ_o in Book 1 below, and make your own small table of ten complexes: metal ion, d count, ligand, Δ_o against P, configuration, unpaired electrons and μ. Do it once by hand and this whole family of questions, from “calculate the spin-only moment” to “why is this one diamagnetic”, becomes the same question. If something does not work out, write to me on Telegram or at [email protected]. I would rather you ask than lose these marks.

How often UPSC asks crystal field theory and magnetic moments

From 2021 to 2025, crystal field splitting and the magnetic behaviour it explains came in 14 of 15 papers across Civil Services (Paper I), Combined Geo-Scientist (Paper I) and IFoS (Paper I): every year in Civil Services, every year in Geo-Scientist, and in IFoS from 2022 to 2025. That is 23 questions in five years. It came 7 times directly, as “explain the magnetic behaviour by CFT” or “calculate the spin-only magnetic moment” of given complexes, for 5 to 15 marks: Civil Services 2023 and 2024, Geo-Scientist 2021, 2022 and 2024, and IFoS 2022 and 2024. The other 16 times it came indirectly: the factors that decide Δ_o, splitting diagrams for octahedral, tetrahedral and square-planar fields (20 marks in Geo-Scientist 2023), the order of ligand field strength, high spin and colour, CFSE in spinels, and the d-block moments set against the lanthanides.

Syllabus and books

In Civil Services it is Paper I, Unit 13, Coordination chemistry, part (i). IFoS and the Geo-Scientist exam ask it in their Chemistry Paper I. The Civil Services syllabus reads:

“Bonding in transition metal complexes. Valence bond theory, crystal field theory and its modifications; applications of theories in the explanation of magnetism and electronic spectra of metal complexes.”

Two phrases in it decide this question: crystal field theory and explanation of magnetism.

You do not need all of these books. Any one of them is enough for this question. I have listed three so that you can use whichever you already have.

Book 1: Geoffrey A. Lawrance, Introduction to Coordination Chemistry, Chapter 7, Section 7.3.3 “A Magnetic Personality? – Paramagnetism and Diamagnetism”, with Table 7.7, which compares P and Δ_o for Co³⁺ with F⁻, NH₃ and CN⁻. These are the numbers used in this post. If you have this book, this is the one I would use.
Book 2: Shriver and Atkins’ Inorganic Chemistry (5th edition), Chapter 20, Section 20.1(b) “Ligand-field stabilization energies”, with Example 20.1 on high-spin and low-spin d⁶, and Section 20.1(c) “Magnetic measurements”, with Table 20.3 of spin-only moments.
Book 3: Miessler, Fischer and Tarr, Inorganic Chemistry (5th edition), Section 10.3 “Ligand Field Theory”, with Table 10.7 of ligand field stabilization and pairing energies.

Keyword map: how to build the answer

Read the tree from left to right. Every leaf is a point to write, in that order. The highlighted words are the ones the examiner looks for.

[CoF₆]³⁻ paramagnetic, [Co(CN)₆]³⁻ diamagnetic: account by CFT

Introduction

Given information

x − 6 = −3, so Co³⁺, d⁶ in both
Six ligands, octahedral

Assumptions

Ligands as point charges
t_2g at −0.4Δ_o, e_g at +0.6Δ_o (barycentre)
Electrons take the lower total energy
Orbital moment quenched

Key law

Low spin if Δ_o > P
μ = √n(n+2) BM

Body

Calculation

High spin t_2g⁴e_g²: −0.4Δ_o
Low spin t_2g⁶: −2.4Δ_o + 2P
Difference = 2(P − Δ_o)

Values

P = 252; Δ_o(F⁻) = 156; Δ_o(CN⁻) = 404
Spectrochemical series: F⁻ weak (π donor), CN⁻ strong (π acceptor)
μ = 4.90 BM and 0

Check

Observed μ of [CoF₆]³⁻ close to 5 BM

Sketch

Two splitting diagrams with arrows

Conclusion

Reading the result

Same ion: the ligand decides the spin
4 unpaired: paramagnetic; none: diamagnetic

Principle

Δ_o against P for d⁴ to d⁷

Application

[Fe(H₂O)₆]²⁺ high spin, [Fe(CN)₆]⁴⁻ low spin
Haemoglobin: spin change on oxygenation

Final answer

Configuration, n, μ, double underlined

Diagrams that prove it

Figure 1. The same Co³⁺ ion in two fields, drawn to scale. The yellow bar beside each gap is the pairing energy P = 252 kJ mol⁻¹. With F⁻ the gap is shorter than the bar, so it is cheaper to go up to e_g: four unpaired electrons. With CN⁻ the gap is longer than the bar, so it is cheaper to pair in t_2g: no unpaired electron.

Figure 2. Energy of the six d electrons in each arrangement as Δ_o grows, with P fixed at 252 kJ mol⁻¹. The low-spin line falls five times faster and crosses the high-spin line exactly where Δ_o = P. F⁻ sits on the high-spin side, NH₃ just past the crossing and CN⁻ far beyond it. This is why [Co(NH₃)₆]³⁺ is also diamagnetic, and why [CoF₆]³⁻ is one of the very few high-spin Co(III) complexes.

What gains marks

The oxidation state worked out in one line, x + 6(−1) = −3, and the d count, d⁶, for both complexes.
The octahedral splitting with the barycentre values −0.4Δ_o for t_2g and +0.6Δ_o for e_g.
The words pairing energy and the comparison Δ_o against P, with numbers, not only “weak” and “strong”.
Both configurations, t_2g⁴e_g² and t_2g⁶, drawn with arrows.
The spin-only formula and the two values, 4.90 BM and 0.
Why F⁻ is weak and CN⁻ is strong: π donor against σ donor and π acceptor, with the spectrochemical series.
One application from the syllabus, such as Fe(II) with water and with cyanide, or haemoglobin.

Match the length to the marks

I always tell my students to let the marks decide the length. For 5 marks, as in Geo-Scientist 2021 Q3(a), I would write the d count, a small table of configuration, unpaired electrons and μ, and one line on Δ_o against P, about half a page. For 10 marks, as in Civil Services 2024, write the numbered assumptions, the two energy expressions, the values, the table and the conclusion, the two pages below. For 15 marks, as in IFoS 2024 Q6(a), add the tetrahedral splitting (Δ_t = 4/9 Δ_o), why the observed moment can differ from the spin-only value, and a line on the colour.

Model answer, as written in the booklet

This is the full 10-mark answer, in the order of the tree above. Page 1 sets up the theory and the two energies, and page 2 puts in the numbers and explains the result.

Page 1 of 2 · Introduction and calculationGiven, assumptions, splitting, two energies

Q.)On the basis of Crystal Field Theory, account for: while [CoF₆]³⁻ is paramagnetic, [Co(CN)₆]³⁻ is diamagnetic. [CSE 2024], 10M

Ans.Given: [CoF₆]³⁻ and [Co(CN)₆]³⁻. Oxidation state of Co: x + 6(−1) = −3, so x = +3 in both.

Co (Z = 27): [Ar] 3d⁷4s² → Co³⁺: [Ar] 3d⁶. Six ligands, so the field is octahedral.

For an octahedral complex, assume that

i) the ligands are point negative charges on the ±x, ±y, ±z axes; bonding is purely electrostatic.
ii) d_x²−y² and d_z² point at the ligands and are raised (e_g, +0.6Δ_o); d_xy, d_yz, d_xz point between them and are lowered (t_2g, −0.4Δ_o), keeping the barycentre.
iii) the electrons take the arrangement of lower total energy, counting the pairing energy P for every extra pair.
iv) the orbital moment is quenched, so μ comes from spin only.

For d⁶, three electrons enter t_2g. The 4th, 5th and 6th can either pair in t_2g (cost P each) or go up to e_g (cost Δ_o each). So we get two possible arrangements:

High spin, t_2g⁴e_g² (one pair, as in the free ion, so no extra P):

E_{HS} = 4 (- 0.4 Δ_{o}) + 2 (+ 0.6 Δ_{o}) = - 0.4 Δ_{o}

— — (i)

Low spin, t_2g⁶ (three pairs, two more than in the free ion):

E_{LS} = 6 (- 0.4 Δ_{o}) + 2 P = - 2.4 Δ_{o} + 2 P

— — (ii)

So, subtracting (i) from (ii), we get

E_{LS} - E_{HS} = 2 (P - Δ_{o})

— — (iii)

i.e. low spin if Δ_o > P, high spin if Δ_o < P. The number of unpaired electrons n then gives

μ = \sqrt{n (n + 2)} BM

— — (iv)

All other symbols have their usual meaning.

Page 2 of 2 · Values and conclusionNumbers, table, check, sketch, reading, principle, application

Values for Co³⁺ (kJ mol⁻¹): P = 252; Δ_o(F⁻) = 156; Δ_o(CN⁻) = 404.

Position in the spectrochemical series: I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO. F⁻ is a π donor, which pushes t_2g up and makes Δ_o small; CN⁻ is a strong σ donor and a π acceptor (back-bonding), which pulls t_2g down and makes Δ_o large.

So, putting the values in (i), (ii) and (iii), we get

[CoF₆]³⁻: E_HS = −0.4 × 156 = −62.4; E_LS = −2.4 × 156 + 504 = +129.6; 2(P − Δ_o) = +192 → high spin.

[Co(CN)₆]³⁻: E_HS = −0.4 × 404 = −161.6; E_LS = −2.4 × 404 + 504 = −465.6; 2(P − Δ_o) = −304 → low spin.

Complex	Δ_o vs P	Config.	n	μ (BM)	Nature
[CoF₆]³⁻	156 < 252	t_2g⁴e_g²	4	√24 = 4.90	para
[Co(CN)₆]³⁻	404 > 252	t_2g⁶	0	0	dia

Check: observed μ of [CoF₆]³⁻ is close to 5 BM, as expected for 4 unpaired electrons ✓.

Reading the result:

i) The metal ion is the same; only the ligand changes Δ_o, and with it the spin state.
ii) [CoF₆]³⁻: 4 unpaired electrons, so it is attracted into a magnetic field: paramagnetic.
iii) [Co(CN)₆]³⁻: all electrons paired (S = 0), so it is weakly repelled: diamagnetic.

Principle: for d⁴ to d⁷ octahedral ions the spin state is decided by Δ_o against P. The same splitting also gives the colour: [CoF₆]³⁻ absorbs in the red and looks blue, [Co(CN)₆]³⁻ absorbs near the ultraviolet and is pale yellow.

Application: Fe²⁺ (d⁶, P = 211): [Fe(H₂O)₆]²⁺ (Δ_o = 125) is high spin, while [Fe(CN)₆]⁴⁻ (Δ_o = 395) is low spin and diamagnetic. In haemoglobin, high-spin Fe(II) of deoxy-Hb becomes low spin on binding O₂.

So, [CoF₆]³⁻: t_2g⁴e_g², n = 4, μ = 4.90 BM, paramagnetic; [Co(CN)₆]³⁻: t_2g⁶, n = 0, μ = 0, diamagnetic

The same question in other exams

Combined Geo-Scientist 2021, Paper I, Q3(a), 5 marks: magnetic properties by CFT of [Co(NH₃)₆]³⁺, [Co(en)₃]³⁺, [Co(NO₂)₆]³⁻, [CoF₆]³⁻ and [Co(H₂O)₃F₃]. All are Co³⁺, d⁶. NH₃, en and NO₂⁻ give Δ_o > P: t_2g⁶, diamagnetic. The two fluoride complexes give Δ_o < P: t_2g⁴e_g², 4 unpaired, 4.90 BM.

Civil Services 2023, Paper I, Q8(b), 10 marks: spin-only moments of [Fe(H₂O)₅NO]²⁺, [Cr(NCS)₆]³⁻, [V(H₂O)₆]³⁺ and [Co(bpy)₃]²⁺. Count n and use (iv): 3.87, 3.87, 2.83 and 3.87 BM.

Combined Geo-Scientist 2022, Paper I, Q3(c), 10 marks: [Fe(NH₃)₆]³⁺, [CoBr₆]³⁻ and [CuCl₄]²⁻. Fe³⁺ d⁵ with NH₃, taken as low spin: n = 1, 1.73 BM. Co³⁺ d⁶ with Br⁻, a weaker ligand than F⁻: high spin, n = 4, 4.90 BM. Cu²⁺ d⁹: n = 1 whatever the field, 1.73 BM.

Combined Geo-Scientist 2024, Paper I, Q1(b), 5 marks: K₃[TiF₆] shows 1.70 BM. Ti³⁺ is d¹, n = 1, so μ_s.o. = 1.73 BM. The small difference comes from spin–orbit coupling: the t_2g¹ electron keeps a little orbital moment, and for a shell less than half full it lowers μ below the spin-only value.

IFoS 2022, Paper I, Q6(c), 15 marks: octahedral and tetrahedral splitting, and Ni²⁺ (d⁸) in both. Δ_t = 4/9 Δ_o (fewer ligands, none pointing straight at an orbital). Both have n = 2, 2.83 BM; the tetrahedral one is observed much higher, because its ground state keeps an orbital contribution.

IFoS 2024, Paper I, Q6(a), 15 marks: [CoCl₄]²⁻ (e⁴t₂³) and [Co(H₂O)₆]²⁺ (t_2g⁵e_g²): both n = 3, 3.87 BM. The octahedral ion deviates more (orbital contribution), and it is the less intensely coloured one, because its d–d bands are Laporte forbidden.

Every version is the same four steps: find the oxidation state and the d count, place the electrons on the right splitting diagram, compare Δ with P if the ion is d⁴ to d⁷ in an octahedral field, then count n and use μ = √n(n+2). Tetrahedral complexes are almost always high spin, because Δ_t is only about half of Δ_o.

Questions students ask

Why is [CoF6]3− paramagnetic while [Co(CN)6]3− is diamagnetic?

Both contain Co3+, a d6 ion, in an octahedral field. F− is a weak-field ligand: Δo (156 kJ/mol) is smaller than the pairing energy P (252 kJ/mol), so the electrons spread out as t2g4 eg2 with 4 unpaired electrons and a spin-only moment of 4.90 BM. CN− is a strong-field ligand: Δo (404 kJ/mol) is larger than P, so all six electrons pair in t2g6 and the complex is diamagnetic. This was asked in UPSC Civil Services 2024, Chemistry Paper I, Q7(c)(i), for 10 marks.

What is pairing energy, and how does it decide high spin or low spin?

Pairing energy P is the extra energy needed to put a second electron into an orbital that already holds one, mainly the extra repulsion between the two. For a d4 to d7 ion in an octahedral field, each electron after the third either pairs in t2g (costing P) or goes up to eg (costing Δo). For d6, E(low spin) − E(high spin) = 2(P − Δo), so the complex is low spin when Δo is greater than P and high spin when it is smaller.

Why is F− a weak-field ligand and CN− a strong-field ligand?

F− is a σ donor and also a π donor: its filled p orbitals push the t2g orbitals up, which makes Δo smaller. CN− is a strong σ donor and a π acceptor: its empty π* orbitals take electron density from t2g by back-bonding, which lowers t2g and makes Δo larger. That is why the spectrochemical series runs I− < Br− < Cl− < F− < OH− < H2O < NH3 < en < NO2− < CN− < CO.

How do you calculate the spin-only magnetic moment of a complex?

Find the oxidation state and the number of d electrons, place them on the splitting diagram to get the number of unpaired electrons n, then use μ = √(n(n + 2)) BM. For n = 1, 2, 3, 4 and 5 this gives 1.73, 2.83, 3.87, 4.90 and 5.92 BM. In UPSC Civil Services 2023, Paper I, Q8(b), 10 marks, the four complexes give 3.87, 3.87, 2.83 and 3.87 BM.

Which d-electron counts can be high spin or low spin?

In an octahedral field only d4, d5, d6 and d7 have two possible arrangements. For d1 to d3 and d8 to d10 there is only one way to fill t2g and eg, so the number of unpaired electrons does not depend on the ligand. Tetrahedral complexes are nearly always high spin, because Δt is only about 4/9 of Δo and is smaller than P.

How often does UPSC ask crystal field theory and magnetic moments?

From 2021 to 2025 the topic appeared in 14 of 15 Chemistry Paper I question papers across UPSC Civil Services, Combined Geo-Scientist and IFoS, 23 questions in all. Seven were direct questions on magnetic behaviour or spin-only moments, for 5 to 15 marks. The only paper without it was IFoS 2021.

Read the complete answers

CSE 2024 Q7(c)(i) solution CSE 2023 Q8(b) solution CSE 2025 Q5(f) solution All coordination chemistry questions Paper I book

By VKC, your mentor. Integrated M.Sc. Chemistry, IIT Roorkee. CSIR NET Chemical Sciences, All India Rank 4. New posts are announced first on Telegram.