JEE Mains Chemistry · The d- and f-Block Elements

Magnetic Moment and Colour

An ion's unpaired d electrons give its spin-only magnetic moment, √(n(n+2)) BM, and a partly filled d shell lets it absorb visible light, so d⁰ and d¹⁰ ions are colourless and diamagnetic.

All The d- and f-Block Elements notes JEE Mains Chemistry strategy

Why this matters

Twenty-five PYQs, nineteen of them multiple choice, and three from 2026. Sixteen turn an ion into a count of unpaired electrons and a spin-only moment, or run that backwards from a value in BM to the ion; nine ask which ions are coloured, which share a colour, and why permanganate is purple with no d electrons at all. Moments also finish many questions on other pages, so this formula pays several times.

Concept 1 of 2: The spin-only magnetic moment

Each unpaired electron is a tiny magnet; paired electrons cancel. So the moment depends only on n, the number of unpaired electrons. For a free 3d ion, the first five d electrons go in singly and the next five pair them up, so n rises to 5 at d⁵ and falls back to 0 at d¹⁰.

Definition

$\mu = \sqrt{n(n+2)}$ BM, where n is the number of unpaired electrons.
n = 0, 1, 2, 3, 4, 5 gives μ = 0, 1.73, 2.83, 3.87, 4.90, 5.92 BM. For f ions, n = 6 gives 6.93 and n = 7 gives 7.94.
Free ion, dˣ: n = x for d¹ to d⁵, and n = 10 − x for d⁶ to d¹⁰.
Remove 4s electrons first: $\mathrm{Mn^{2+}}$ is 3d⁵ (n = 5), $\mathrm{Ni^{2+}}$ is 3d⁸ (n = 2).
Ions with the same d count have the same moment: $\mathrm{V^{2+}}$ and $\mathrm{Cr^{3+}}$ (d³), $\mathrm{Mn^{2+}}$ and $\mathrm{Fe^{3+}}$ (d⁵).
Working backwards: solve $n(n+2) = \mu^{2}$ and take the positive whole number.
'Nearest integer' answers: 1.73 → 2, 2.83 → 3, 3.87 → 4, 4.90 → 5, 5.92 → 6.
Count the free ion unless a strong-field complex is named; ligands that pair electrons are on the Coordination Compounds pages.

Spin-only magnetic moment

\mu = \sqrt{n(n+2)}\ \text{BM}

Worked example

Find the spin-only magnetic moments of

\mathrm{Co^{2+}}

(Z = 27) and

\mathrm{Ti^{3+}}

(Z = 22) as free ions.

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 3 Apr 2025 · Q29Moderate

Example 1 · The d- and f-Block Elements · Magnetic Moment and Colour

The metal ions that have the calculated spin only magnetic moment value of 4.9 B.M. are A.

Cr^{2 +}

Fe^{2 +}

Fe^{3 +}

Co^{2 +}

Mn^{3 +}

Choose the correct answer from the options given below.

Past d⁵ the electrons pair up

\mathrm{Cu^{2+}}

is d⁹ with ONE unpaired electron, not nine, and

\mathrm{Ni^{2+}}

is d⁸ with two. Use

10 - x

for d⁶ to d¹⁰.

Take the 4s electrons off first

\mathrm{Mn^{2+}}

is 3d⁵ with five unpaired electrons (5.92 BM). Writing it as 3d³4s² gives three and a wrong answer of 3.87 BM, which is usually one of the options.

Drill 15 more on the spin-only magnetic moment

Concept 2 of 2: Colours of the aqueous 3d ions

Water molecules around an ion split its d orbitals into two sets. Visible light can lift an electron from the lower set to the upper one, and we see the colour that is left. That needs at least one d electron and a gap in the upper set, so d⁰ and d¹⁰ ions are colourless. A few d⁰ ions such as permanganate are coloured another way: light moves an electron from oxygen to the metal, a charge-transfer transition.

Definition

Colourless: d⁰ ( $\mathrm{Sc^{3+}}$ , $\mathrm{Ti^{4+}}$ ) and d¹⁰ ( $\mathrm{Zn^{2+}}$ , $\mathrm{Cu^{+}}$ , $\mathrm{Cd^{2+}}$ ).
Coloured: d¹ to d⁹ in water, by d–d transitions.
Charge transfer: $\mathrm{KMnO_4}$ (purple), $\mathrm{K_2Cr_2O_7}$ (orange) and $\mathrm{K_2CrO_4}$ (yellow) are d⁰ but coloured. They are also diamagnetic.
Colour needs the ligands: anhydrous $\mathrm{CuSO_4}$ is white, while $\mathrm{CuSO_4 \cdot 5H_2O}$ is blue.
'Paramagnetic and coloured' both follow from unpaired d electrons, so a pair qualifies only if both ions are d¹ to d⁹.

Ion	d configuration	Unpaired electrons (free ion)	Colour in water
$\mathrm{Sc^{3+}}$	3d⁰	0	Colourless
$\mathrm{Ti^{4+}}$	3d⁰	0	Colourless
$\mathrm{Ti^{3+}}$	3d¹	1	Purple
$\mathrm{V^{4+}}$	3d¹	1	Blue
$\mathrm{V^{3+}}$	3d²	2	Green
$\mathrm{V^{2+}}$	3d³	3	Violet
$\mathrm{Cr^{3+}}$	3d³	3	Violet
$\mathrm{Mn^{3+}}$	3d⁴	4	Violet V²⁺, Cr³⁺ and Mn³⁺ are all violet.
$\mathrm{Cr^{2+}}$	3d⁴	4	Blue
$\mathrm{Mn^{2+}}$	3d⁵	5	Pink
$\mathrm{Fe^{3+}}$	3d⁵	5	Yellow
$\mathrm{Fe^{2+}}$	3d⁶	4	Green
$\mathrm{Co^{3+}}$	3d⁶	4	Blue
$\mathrm{Co^{2+}}$	3d⁷	3	Pink
$\mathrm{Ni^{2+}}$	3d⁸	2	Green
$\mathrm{Cu^{2+}}$	3d⁹	1	Blue
$\mathrm{Zn^{2+}}$	3d¹⁰	0	Colourless

Same colour does not mean same d count: Fe²⁺ (d⁶), Ni²⁺ (d⁸) and V³⁺ (d²) are all green.

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 23 January 2025 · Q41Moderate

Example 2 · The d- and f-Block Elements · Magnetic Moment and Colour

The correct set of ions (aqueous solution) with same colour from the following is :

Intensely coloured is not paramagnetic

Permanganate is deep purple but Mn(VII) is d⁰, so

\mathrm{MnO_4^{-}}

is diamagnetic. Its colour comes from charge transfer, not from d electrons. The same holds for dichromate and chromate.

Copper is colourless as Cu⁺

\mathrm{Cu^{2+}}

(d⁹) is blue, but

\mathrm{Cu^{+}}

(d¹⁰) is colourless. Always count the d electrons of the ION, not of the element.

Drill 8 more on colours of the aqueous 3d ions

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 (1)

The spin-only magnetic moment
Spin-only magnetic moment
$\mu = \sqrt{n(n+2)}\ \text{BM}$

Reference tables (1)

Colours of the aqueous 3d ions17 rows

Ion	d configuration	Unpaired electrons (free ion)	Colour in water
$\mathrm{Sc^{3+}}$	3d⁰	0	Colourless
$\mathrm{Ti^{4+}}$	3d⁰	0	Colourless
$\mathrm{Ti^{3+}}$	3d¹	1	Purple
$\mathrm{V^{4+}}$	3d¹	1	Blue
$\mathrm{V^{3+}}$	3d²	2	Green
$\mathrm{V^{2+}}$	3d³	3	Violet
$\mathrm{Cr^{3+}}$	3d³	3	Violet
$\mathrm{Mn^{3+}}$	3d⁴	4	Violet V²⁺, Cr³⁺ and Mn³⁺ are all violet.
$\mathrm{Cr^{2+}}$	3d⁴	4	Blue
$\mathrm{Mn^{2+}}$	3d⁵	5	Pink
$\mathrm{Fe^{3+}}$	3d⁵	5	Yellow
$\mathrm{Fe^{2+}}$	3d⁶	4	Green
$\mathrm{Co^{3+}}$	3d⁶	4	Blue
$\mathrm{Co^{2+}}$	3d⁷	3	Pink
$\mathrm{Ni^{2+}}$	3d⁸	2	Green
$\mathrm{Cu^{2+}}$	3d⁹	1	Blue
$\mathrm{Zn^{2+}}$	3d¹⁰	0	Colourless

Same colour does not mean same d count: Fe²⁺ (d⁶), Ni²⁺ (d⁸) and V³⁺ (d²) are all green.

Watch out for (4)

Past d⁵ the electrons pair up
→ The spin-only magnetic moment
Take the 4s electrons off first
→ The spin-only magnetic moment
Intensely coloured is not paramagnetic
→ Colours of the aqueous 3d ions
Copper is colourless as Cu⁺
→ Colours of the aqueous 3d ions

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