JEE Mains Chemistry · Coordination Compounds

Spin-Only Magnetic Moment

The spin-only magnetic moment √(n(n+2)) BM depends only on the number of unpaired electrons n, so every question here is a count of unpaired electrons: oxidation state, d count, then high or low spin.

All Coordination Compounds notes JEE Mains Chemistry strategy

Why this matters

Thirty-five PYQs, twenty-three of them multiple choice, and five from 2026. Nineteen compute a spin-only moment or work back from a moment to the ion; eleven put complexes in order of moment or unpaired electrons; five count how many species in a list are paramagnetic. Twelve of the thirty-five ask for a number.

Concept 1 of 3: Spin-only magnetic moment from unpaired electrons

Each unpaired electron is a tiny magnet; paired electrons cancel. The spin-only formula turns the number of unpaired electrons n into a moment in Bohr magnetons. Learn the five values and the question becomes: how many unpaired electrons? Run it backwards too: a measured moment gives n, n gives the d count, and the d count gives the ion.

Definition

$\mu = \sqrt{n(n+2)}$ BM: n = 1 → 1.73, 2 → 2.83, 3 → 3.87, 4 → 4.90, 5 → 5.92.
A moment close to one of these (for example 6.06 BM or 3.95 BM) is read as the nearest n (5 or 3).
Steps: oxidation state → d count → strong or weak field → configuration → n → $\mu$ .
n = 0 means diamagnetic: d⁰ ( $\mathrm{V_2O_5}$ , $\mathrm{TiCl_4}$ ), d¹⁰ (Cu⁺, Zn²⁺, Ni(CO)₄), low-spin d⁶, square planar d⁸.
Cu²⁺ in any complex (Fehling's solution, $\mathrm{[Cu(NH_3)_4]^{2+}}$ ) has 1 unpaired electron: 1.73 BM.

Spin-only magnetic moment

\mu_{\text{spin-only}} = \sqrt{n(n+2)}\ \text{BM}

Worked example

Find the spin-only magnetic moment of

\mathrm{[V(H_2O)_6]^{3+}}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 29 Jan 2025 · Q101Moderate

Example 1 · Coordination Compounds · Spin-Only Magnetic Moment

The calculated spin-only magnetic moments of

K_{3}\left\lbrack Fe(OH)_{6} \right\rbrack

and

K_{4}\left\lbrack Fe(OH)_{6} \right\rbrack

respectively are :

Copper(I) is diamagnetic

CuI and

\mathrm{K_3[Cu(CN)_4]}

contain Cu⁺, which is d¹⁰, so their moment is 0, not 1.73 BM. Only copper(II) has one unpaired electron.

Watch the units the answer is asked in

2.83 BM written in units of

10^{-1}

BM is 28. Writing 3 (rounded BM) or 283 in that blank loses the mark.

The same oxidation state does not explain different moments

Fe is +3 in both

\mathrm{[Fe(CN)_6]^{3-}}

(1.73 BM) and

\mathrm{[Fe(H_2O)_6]^{3+}}

(5.92 BM). The difference comes from the ligand's field strength, so a reason that cites the common oxidation state is true but not the explanation.

Drill 18 more on spin-only magnetic moment from unpaired electrons

Concept 2 of 3: Unpaired electrons and moments of high-spin aqua ions

Most ordering questions use water or a halide, so the ions are high spin. Then the number of unpaired electrons rises from d¹ to d⁵ and falls again to d¹⁰. Know this ladder and any order of moments is a lookup; a cyanide complex drops to its low-spin count.

Definition

High spin (weak field): n rises 1, 2, 3, 4, 5 from d¹ to d⁵ and falls 4, 3, 2, 1, 0 from d⁶ to d¹⁰.
Low spin with CN⁻: d⁴ 2, d⁵ 1, d⁶ 0, d⁷ 1 unpaired.
Tetrahedral complexes are high spin: $\mathrm{[MnBr_4]^{2-}}$ 5, $\mathrm{[CoCl_4]^{2-}}$ 3, $\mathrm{[NiCl_4]^{2-}}$ 2.
The largest moment in a list usually belongs to a d⁵ ion with a weak ligand (Mn²⁺, Fe³⁺).

Aqua ion (high spin)	d count	Unpaired electrons	Spin-only moment (BM)
$\mathrm{Ti^{3+}}$	d¹	1	1.73
$\mathrm{V^{3+}}$	d²	2	2.83
$\mathrm{V^{2+}}$ , $\mathrm{Cr^{3+}}$	d³	3	3.87
$\mathrm{Cr^{2+}}$ , $\mathrm{Mn^{3+}}$	d⁴	4	4.90
$\mathrm{Mn^{2+}}$ , $\mathrm{Fe^{3+}}$	d⁵	5	5.92 The maximum: a d⁵ ion with a weak-field ligand.
$\mathrm{Fe^{2+}}$ , $\mathrm{Co^{3+}}$ (with F⁻)	d⁶	4	4.90
$\mathrm{Co^{2+}}$	d⁷	3	3.87
$\mathrm{Ni^{2+}}$	d⁸	2	2.83
$\mathrm{Cu^{2+}}$	d⁹	1	1.73
$\mathrm{Zn^{2+}}$	d¹⁰	0	0

Cr³⁺ is 3.87 BM with every ligand, because d³ has only one octahedral configuration.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 27 Jan 2024 · Q38Moderate

Example 2 · Coordination Compounds · Spin-Only Magnetic Moment

Consider the following complex ions

P =\left\lbrack FeF_{6} \right\rbrack^{3 -}

Q =\left\lbrack V\left( H_{2}O \right)_{6} \right\rbrack^{2 +}

R =\left\lbrack Fe\left( H_{2}O \right)_{6} \right\rbrack^{2 +}

The correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is:

Fe²⁺ and Fe³⁺ aqua ions differ by one unpaired electron

High-spin Fe³⁺ (d⁵) has 5 unpaired electrons and Fe²⁺ (d⁶) has 4. The extra electron in d⁶ pairs up, so adding an electron here LOWERS the moment.

Change of ligand can reverse an order

\mathrm{[FeF_6]^{3-}}

(5 unpaired) is above

\mathrm{[CoF_6]^{3-}}

(4), but

\mathrm{[Fe(CN)_6]^{3-}}

(1) is below

\mathrm{[Mn(CN)_6]^{3-}}

(2). Decide the spin state for each complex before ordering.

Drill 10 more on unpaired electrons and moments of high-spin aqua ions

Concept 3 of 3: Counting paramagnetic species in a list

A species is paramagnetic if it has at least one unpaired electron. For a list, go through each species in turn: oxidation state, d count, spin state, n. Mark it paramagnetic if n > 0. The traps are the diamagnetic ones that look paramagnetic: low-spin d⁶, square planar d⁸, d⁰ and d¹⁰.

Definition

Paramagnetic: n ≥ 1. Diamagnetic: n = 0.
Always diamagnetic: d⁰ ( $\mathrm{V_2O_5}$ , $\mathrm{[MnO_4]^-}$ ), d¹⁰, low-spin d⁶ ( $\mathrm{[Fe(CN)_6]^{4-}}$ , $\mathrm{[Co(NH_3)_6]^{3+}}$ , $\mathrm{[Co(C_2O_4)_3]^{3-}}$ ), square planar d⁸ ( $\mathrm{[Ni(CN)_4]^{2-}}$ , Pt(II)).
Always paramagnetic: any odd number of d electrons, since odd electrons cannot all pair.
The answer is a number of SPECIES, not of electrons: count each paramagnetic entry once, however many unpaired electrons it has.

Paramagnetic test

n \geq 1 \Rightarrow \text{paramagnetic} \qquad n = 0 \Rightarrow \text{diamagnetic}

Worked example

How many of these are paramagnetic:

\mathrm{[Fe(CN)_6]^{4-}}

\mathrm{[Cr(CN)_6]^{3-}}

\mathrm{[Zn(NH_3)_4]^{2+}}

\mathrm{[CuCl_4]^{2-}}

\mathrm{[Co(NH_3)_6]^{3+}}

\mathrm{[MnCl_4]^{2-}}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 1 · Q34Moderate

Example 3 · Coordination Compounds · Spin-Only Magnetic Moment

Given below are two statements Statements-I : The number of paramagnetic species among

\left\lbrack {CoF}_{6} \right\rbrack^{3 -},\ \left\lbrack {TiF}_{6} \right\rbrack^{3 -},\ V_{2}O_{5}

and

\left\lbrack Fe(CN)_{6} \right\rbrack^{3 -}

is 3 . Statement-II :

K_{4}\left\lbrack Fe(CN)_{6} \right\rbrack < K_{3}\left\lbrack Fe(CN)_{6} \right\rbrack < \left\lbrack Fe\left( H_{2}O \right)_{6} \right\rbrack{SO}_{4} \cdot H_{2}O < \left\lbrack Fe\left( H_{2}O \right)_{6} \right\rbrack{Cl}_{3}

is the correct order in terms of number of unpaired electron(s) in the complexes. In the light of the above statements, choose the correct answer from the options given below.

Low-spin d⁶ is diamagnetic

\mathrm{[Fe(CN)_6]^{4-}}

\mathrm{[Co(NH_3)_6]^{3+}}

and

\mathrm{[Co(C_2O_4)_3]^{3-}}

have all six electrons paired in t₂g. Counting them as paramagnetic because 'iron and cobalt are magnetic' is the commonest slip in these lists.

Ferricyanide has one unpaired electron

\mathrm{[Fe(CN)_6]^{3-}}

is low-spin d⁵,

t_{2g}^5

: one unpaired electron, so it is paramagnetic, while ferrocyanide is not.

Drill 4 more on counting paramagnetic species in a list

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Spin-only magnetic moment from unpaired electrons
Spin-only magnetic moment
$\mu_{\text{spin-only}} = \sqrt{n(n+2)}\ \text{BM}$
Counting paramagnetic species in a list
Paramagnetic test
$n \geq 1 \Rightarrow \text{paramagnetic} \qquad n = 0 \Rightarrow \text{diamagnetic}$

Reference tables (1)

Unpaired electrons and moments of high-spin aqua ions10 rows

Aqua ion (high spin)	d count	Unpaired electrons	Spin-only moment (BM)
$\mathrm{Ti^{3+}}$	d¹	1	1.73
$\mathrm{V^{3+}}$	d²	2	2.83
$\mathrm{V^{2+}}$ , $\mathrm{Cr^{3+}}$	d³	3	3.87
$\mathrm{Cr^{2+}}$ , $\mathrm{Mn^{3+}}$	d⁴	4	4.90
$\mathrm{Mn^{2+}}$ , $\mathrm{Fe^{3+}}$	d⁵	5	5.92 The maximum: a d⁵ ion with a weak-field ligand.
$\mathrm{Fe^{2+}}$ , $\mathrm{Co^{3+}}$ (with F⁻)	d⁶	4	4.90
$\mathrm{Co^{2+}}$	d⁷	3	3.87
$\mathrm{Ni^{2+}}$	d⁸	2	2.83
$\mathrm{Cu^{2+}}$	d⁹	1	1.73
$\mathrm{Zn^{2+}}$	d¹⁰	0	0

Cr³⁺ is 3.87 BM with every ligand, because d³ has only one octahedral configuration.

Watch out for (7)

Copper(I) is diamagnetic
→ Spin-only magnetic moment from unpaired electrons
Watch the units the answer is asked in
→ Spin-only magnetic moment from unpaired electrons
The same oxidation state does not explain different moments
→ Spin-only magnetic moment from unpaired electrons
Fe²⁺ and Fe³⁺ aqua ions differ by one unpaired electron
→ Unpaired electrons and moments of high-spin aqua ions
Change of ligand can reverse an order
→ Unpaired electrons and moments of high-spin aqua ions
Low-spin d⁶ is diamagnetic
→ Counting paramagnetic species in a list
Ferricyanide has one unpaired electron
→ Counting paramagnetic species in a list

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