Is it correct to say: Writing Fe²⁺ as [Ar]3d⁴4s² (removing 3d electrons before 4s).?

The 4s electrons are lost first: Fe²⁺ = [Ar]3d⁶ and Fe³⁺ = [Ar]3d⁵.

Is it correct to say: Putting the total number of d electrons into μ = √n(n+2).?

n is the number of UNPAIRED electrons. Fe²⁺ (3d⁶) has n = 4, so μ = 4.90 BM, not √48.

Is it correct to say: Writing Cr as [Ar]3d⁴4s² and Cu as [Ar]3d⁹4s².?

Cr = [Ar]3d⁵4s¹ and Cu = [Ar]3d¹⁰4s¹, because half-filled and completely filled d subshells are extra stable.

Is it correct to say: Calling the chromate ⇌ dichromate change a redox reaction.?

Cr is +6 in both CrO₄²⁻ and Cr₂O₇²⁻. Only the pH shifts the equilibrium.

Is it correct to say: Using 5 electrons for KMnO₄ in every medium.?

Acidic: MnO₄⁻ → Mn²⁺ (5 e⁻). Neutral or faintly alkaline: MnO₄⁻ → MnO₂ (3 e⁻).

Is it correct to say: Saying lanthanoid contraction makes 5d elements much smaller than 4d elements.?

It cancels the expected size increase, so 4d and 5d partners are almost the SAME size (Zr 160 pm, Hf 159 pm).

📄 Formula sheet (PDF)

The d- and f-Block Elements

Electronic configurations, oxidation states, spin-only magnetic moment, colour, K₂Cr₂O₇ and KMnO₄, lanthanoid contraction and actinoids — NCERT Class 12 Chemistry Ch 4

📐 19 formulas✏️ 4 examples🎯 6 practice⚖️ 7 marks🏫 CBSE📚 Class 12✓ 2025–26 syllabus

💡

Board Exam Tips

→Learn the spin-only values for n = 1 to 5 (1.73, 2.83, 3.87, 4.90, 5.92 BM). A magnetic-moment numerical needs only the ion's configuration and this table.
→For any first-row ion: write the atom's configuration, remove the 4s electrons first, then count unpaired 3d electrons using Hund's rule.
→Most 'explain why' questions come down to the extra stability of d⁰, d⁵ and d¹⁰ (and half-filled t₂g): Mn²⁺ is stable, Cr²⁺ is reducing, Mn³⁺ is oxidising, Zn is not a transition element.
→Write the balanced equations for making K₂Cr₂O₇ from chromite and KMnO₄ from pyrolusite (MnO₂), plus their acidic-medium half-reactions. Check that atoms AND charges balance.
→For lanthanoid contraction, give the cause (poor shielding by 4f electrons) and at least two consequences (Zr/Hf have almost equal radii; lanthanoids are hard to separate).

📐 Formulas(19)

General Electronic Configuration (d-block)

(n - 1) d^{1 - 10} n s^{1 - 2}

Symbol	Meaning
$n$	Principal quantum number of the outermost shell (4, 5 or 6)

Exceptional Configurations of Cr and Cu

Cr : [Ar] 3 d^{5} 4 s^{1}, Cu : [Ar] 3 d^{10} 4 s^{1}

Configuration of Transition Metal Ions★ Board fav

Fe : [Ar] 3 d^{6} 4 s^{2} \to Fe^{2 +} : [Ar] 3 d^{6}, Fe^{3 +} : [Ar] 3 d^{5}

Highest Oxidation State (Sc to Mn)

Max. O.S. = (n - 1) d + n s electrons : Sc (+ 3), Ti (+ 4), V (+ 5), Cr (+ 6), Mn (+ 7)

Stability of d⁰, d⁵, d¹⁰ and Half-filled t₂g

E_{Cr^{3 +} / Cr^{2 +}}^{\circ} = - 0.41 V, E_{Mn^{3 +} / Mn^{2 +}}^{\circ} = + 1.57 V

Spin-only Magnetic Moment★ Board fav

μ = n (n + 2) BM

Symbol	Meaning
$μ$	Spin-only magnetic moment (Bohr magneton, BM)
$n$	Number of unpaired electrons (no unit)

Spin-only Values to Memorise

n = 1, 2, 3, 4, 5 \Rightarrow μ = 1.73, 2.83, 3.87, 4.90, 5.92 BM

Colour of Transition Metal Ions

Coloured if d^{1} to d^{9}; colourless if d^{0} or d^{10} (Sc^{3 +}, Ti^{4 +}, Cu^{+}, Zn^{2 +})

K₂Cr₂O₇ Preparation — Step 1: Fusion of Chromite★ Board fav

4 FeCr_{2} O_{4} + 8 Na_{2} CO_{3} + 7 O_{2} \to 8 Na_{2} CrO_{4} + 2 Fe_{2} O_{3} + 8 CO_{2}

K₂Cr₂O₇ Preparation — Steps 2 and 3

2 Na_{2} CrO_{4} + 2 H^{+} \to Na_{2} Cr_{2} O_{7} + 2 Na^{+} + H_{2} O; Na_{2} Cr_{2} O_{7} + 2 KCl \to K_{2} Cr_{2} O_{7} + 2 NaCl

Chromate–Dichromate Equilibrium

2 CrO_{4}^{2 -} + 2 H^{+} ⇌ Cr_{2} O_{7}^{2 -} + H_{2} O

K₂Cr₂O₇ as Oxidising Agent (acidic medium)★ Board fav

Cr_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 Cr^{3 +} + 7 H_{2} O (E^{\circ} = 1.33 V)

Dichromate Oxidising Fe²⁺

Cr_{2} O_{7}^{2 -} + 14 H^{+} + 6 Fe^{2 +} \to 2 Cr^{3 +} + 6 Fe^{3 +} + 7 H_{2} O

Preparation of KMnO₄ from Pyrolusite

2 MnO_{2} + 4 KOH + O_{2} \to 2 K_{2} MnO_{4} + 2 H_{2} O; 3 MnO_{4}^{2 -} + 4 H^{+} \to 2 MnO_{4}^{-} + MnO_{2} + 2 H_{2} O

KMnO₄ as Oxidising Agent — Acidic Medium★ Board fav

MnO_{4}^{-} + 8 H^{+} + 5 e^{-} \to Mn^{2 +} + 4 H_{2} O (E^{\circ} = 1.52 V)

KMnO₄ — Neutral or Faintly Alkaline Medium

MnO_{4}^{-} + 2 H_{2} O + 3 e^{-} \to MnO_{2} + 4 OH^{-}

Lanthanoids — Configuration and Oxidation States

[Xe] 4 f^{1 - 14} 5 d^{0 - 1} 6 s^{2}; common O.S. = + 3 (also Ce^{4 +}, Eu^{2 +}, Yb^{2 +})

Lanthanoid Contraction★ Board fav

r (La^{3 +}) > r (Ce^{3 +}) > \dots > r (Lu^{3 +}) (poor shielding by 4 f electrons)

Actinoids — Configuration and Oxidation States

[Rn] 5 f^{1 - 14} 6 d^{0 - 1} 7 s^{2}; O.S. mostly + 3, up to + 7 (Np)

✏️ Solved Examples

1Solved Exampleeasy4 steps

Calculate the spin-only magnetic moment of Cr³⁺ (Z = 24).

Ground-state configuration of Cr (half-filled 3d is favoured)

Cr : [Ar] 3 d^{5} 4 s^{1}

2Solved Exampleboard4 steps

Write the electronic configurations of Fe²⁺ (Z = 26), Cu²⁺ (Z = 29) and Zn²⁺ (Z = 30). Calculate the spin-only magnetic moment of each and state which ion is diamagnetic and colourless.

Fe²⁺: remove both 4s electrons; 3d⁶ has 4 unpaired electrons

Fe^{2 +} : [Ar] 3 d^{6}, n = 4, μ = 24 = 4.90 BM

3Solved Exampleboard5 steps

Write the balanced ionic equation for the oxidation of oxalate ions by acidified KMnO₄. What mass of KMnO₄ (M = 158 g mol⁻¹) is needed to oxidise 0.50 mol of oxalate ions?

Reduction half (acidic medium), multiplied by 2

2 MnO_{4}^{-} + 16 H^{+} + 10 e^{-} \to 2 Mn^{2 +} + 8 H_{2} O

4Solved ExampleHOTS4 steps

An ion Mnˣ⁺ (Z = 25) has a spin-only magnetic moment of 4.90 BM. Find x, and predict whether this ion acts as an oxidising or a reducing agent.

Find the number of unpaired electrons from μ

n (n + 2) = (4.90)^{2} = 24 \Rightarrow n = 4

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Writing Fe²⁺ as [Ar]3d⁴4s² (removing 3d electrons before 4s).
✓The 4s electrons are lost first: Fe²⁺ = [Ar]3d⁶ and Fe³⁺ = [Ar]3d⁵.
2
Putting the total number of d electrons into μ = √n(n+2).
✓n is the number of UNPAIRED electrons. Fe²⁺ (3d⁶) has n = 4, so μ = 4.90 BM, not √48.
3
Writing Cr as [Ar]3d⁴4s² and Cu as [Ar]3d⁹4s².
✓Cr = [Ar]3d⁵4s¹ and Cu = [Ar]3d¹⁰4s¹, because half-filled and completely filled d subshells are extra stable.
4
Calling the chromate ⇌ dichromate change a redox reaction.
✓Cr is +6 in both CrO₄²⁻ and Cr₂O₇²⁻. Only the pH shifts the equilibrium.
5
Using 5 electrons for KMnO₄ in every medium.
✓Acidic: MnO₄⁻ → Mn²⁺ (5 e⁻). Neutral or faintly alkaline: MnO₄⁻ → MnO₂ (3 e⁻).
6
Saying lanthanoid contraction makes 5d elements much smaller than 4d elements.
✓It cancels the expected size increase, so 4d and 5d partners are almost the SAME size (Zr 160 pm, Hf 159 pm).

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Calculate the spin-only magnetic moment of Ti³⁺ (Z = 22).
Q2
Calculate the spin-only magnetic moment of Ni²⁺ (Z = 28).
Q3
Zinc is a d-block element. Why is it not regarded as a transition element?
Q4
Complete and balance: Cr₂O₇²⁻ + H⁺ + I⁻ →
Q5
How many moles of Fe²⁺ are oxidised by 1 mol of MnO₄⁻ in acidic medium?
Q6
Zr and Hf have almost identical atomic radii. What is the cause?

📝 Notes

The d- and f-Block Elements

Transition (d-block) elements fill the (n−1)d orbitals; inner-transition (f-block) elements fill the 4f (lanthanoids) or 5f (actinoids) orbitals. Almost every property in this chapter comes from how many d or f electrons are present and how they are arranged.

Configuration first, everything else follows

Write the atom's configuration (remember Cr = 3d⁵4s¹ and Cu = 3d¹⁰4s¹), then remove the 4s electrons before any 3d electrons to get the ion. From the ion's configuration you can read off:

Unpaired electrons, giving the magnetic moment $\mu = \sqrt{n(n+2)}$ BM.
Colour: d¹–d⁹ ions are usually coloured (d–d transitions); d⁰ and d¹⁰ ions are colourless.
Stability: d⁰, d⁵ and d¹⁰ (and the half-filled t₂g of d³) are favoured, which explains why Mn²⁺ and Fe³⁺ are stable, Mn³⁺ is oxidising and Cr²⁺ is reducing.

Dichromate and permanganate

Both are strong oxidising agents in acid. Learn the electron count: Cr₂O₇²⁻ takes 6 e⁻ (→ 2Cr³⁺) and MnO₄⁻ takes 5 e⁻ (→ Mn²⁺), or only 3 e⁻ (→ MnO₂) in neutral solution. To balance any redox equation, multiply the two half-reactions so the electrons cancel, then check atoms and total charge on both sides. Preparation equations — chromite → chromate → dichromate, and MnO₂ → manganate → permanganate — are worth learning exactly.

f-block in brief

Lanthanoids show mainly +3; Ce⁴⁺, Eu²⁺ and Yb²⁺ are the exceptions you are expected to explain. Lanthanoid contraction — the steady fall in size from La to Lu caused by poor 4f shielding — makes Zr and Hf almost the same size and the lanthanoids hard to separate. Actinoids show a wider range of oxidation states (up to +7), are all radioactive, and their contraction is larger because 5f shielding is even poorer.