CBSE · Class 12 · Chemistry · Chapter 4

The d- and f-Block Elements — Formula Sheet

Board Formulas

19 formulas

1.General Electronic Configuration (d-block)
$(n - 1) d^{1 - 10} n s^{1 - 2}$
$n$ : Principal quantum number of the outermost shell (4, 5 or 6)
Holds for the 3d, 4d and 5d series; Pd (4d¹⁰ 5s⁰) is the notable exception. The (n−1)d and ns energies are very close, so electrons shift between them easily.
2.Exceptional Configurations of Cr and Cu
$Cr : [Ar] 3 d^{5} 4 s^{1}, Cu : [Ar] 3 d^{10} 4 s^{1}$
Half-filled (d⁵) and completely filled (d¹⁰) sets are extra stable, so one 4s electron moves into 3d. Writing 3d⁴4s² or 3d⁹4s² is marked wrong.
3.Configuration of Transition Metal Ions★
$Fe : [Ar] 3 d^{6} 4 s^{2} \to Fe^{2 +} : [Ar] 3 d^{6}, Fe^{3 +} : [Ar] 3 d^{5}$
On ionisation the 4s electrons leave first, then 3d. Shortcut for the 3d series: M²⁺ = [Ar]3dⁿ with n = Z − 20, and M³⁺ has n = Z − 21.
4.Highest Oxidation State (Sc to Mn)
$Max. O.S. = (n - 1) d + n s electrons : Sc (+ 3), Ti (+ 4), V (+ 5), Cr (+ 6), Mn (+ 7)$
Up to Mn every 3d and 4s electron can be used. After Mn the d electrons pair up and high states become rare (Zn shows only +2). Oxygen stabilises high states better than fluorine: Mn's highest fluoride is MnF₄ but its highest oxide is Mn₂O₇.
5.Stability of d⁰, d⁵, d¹⁰ and Half-filled t₂g
$E_{Cr^{3 +} / Cr^{2 +}}^{\circ} = - 0.41 V, E_{Mn^{3 +} / Mn^{2 +}}^{\circ} = + 1.57 V$
Cr²⁺ and Mn³⁺ are both d⁴. Cr²⁺ is reducing (d⁴ → d³, half-filled t₂g); Mn³⁺ is oxidising (d⁴ → stable d⁵). Cu is the only 3d metal with positive E°(M²⁺/M) (+0.34 V), so it does not liberate H₂ from dilute acids.
6.Spin-only Magnetic Moment★
$μ = n (n + 2) BM$
$μ$ : Spin-only magnetic moment (Bohr magneton, BM) · $n$ : Number of unpaired electrons (no unit)
n is the number of UNPAIRED electrons, not the total number of d electrons. n ≥ 1 ⇒ paramagnetic; n = 0 ⇒ diamagnetic.
7.Spin-only Values to Memorise
$n = 1, 2, 3, 4, 5 \Rightarrow μ = 1.73, 2.83, 3.87, 4.90, 5.92 BM$
Reverse use: from a given μ, solve n(n+2) = μ² for a whole number n. Unpaired electrons for free ions: d¹–d⁵ → 1–5; d⁶, d⁷, d⁸, d⁹ → 4, 3, 2, 1. (NCERT Table 4.7 prints 2.84 for n = 2; √8 = 2.828 rounds to 2.83.)
8.Colour of Transition Metal Ions
$Coloured if d^{1} to d^{9}; colourless if d^{0} or d^{10} (Sc^{3 +}, Ti^{4 +}, Cu^{+}, Zn^{2 +})$
Colour comes from d–d transitions: an electron absorbs visible light to jump to a higher d level and the complementary colour is seen. MnO₄⁻ and Cr₂O₇²⁻ are d⁰ — their colour is due to charge transfer, not d–d transitions.
9.K₂Cr₂O₇ Preparation — Step 1: Fusion of Chromite★
$4 FeCr_{2} O_{4} + 8 Na_{2} CO_{3} + 7 O_{2} \to 8 Na_{2} CrO_{4} + 2 Fe_{2} O_{3} + 8 CO_{2}$
Chromite ore is fused with sodium carbonate in free access of air. Cr is oxidised from +3 to +6 (yellow sodium chromate).
10.K₂Cr₂O₇ Preparation — Steps 2 and 3
$2 Na_{2} CrO_{4} + 2 H^{+} \to Na_{2} Cr_{2} O_{7} + 2 Na^{+} + H_{2} O; Na_{2} Cr_{2} O_{7} + 2 KCl \to K_{2} Cr_{2} O_{7} + 2 NaCl$
The chromate solution is acidified with H₂SO₄ to give orange sodium dichromate. KCl is then added: K₂Cr₂O₇ is less soluble than Na₂Cr₂O₇, so orange crystals separate.
11.Chromate–Dichromate Equilibrium
$2 CrO_{4}^{2 -} + 2 H^{+} ⇌ Cr_{2} O_{7}^{2 -} + H_{2} O$
Acid shifts it right (yellow → orange); alkali shifts it back: Cr₂O₇²⁻ + 2OH⁻ → 2CrO₄²⁻ + H₂O. Cr is +6 in both ions — this is not a redox change. Dichromate is two tetrahedra sharing one O (Cr–O–Cr angle 126°).
12.K₂Cr₂O₇ as Oxidising Agent (acidic medium)★
$Cr_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 Cr^{3 +} + 7 H_{2} O (E^{\circ} = 1.33 V)$
Orange → green (Cr³⁺). Six electrons per dichromate ion, so it oxidises 6 Fe²⁺, 6 I⁻ (to 3 I₂), 3 Sn²⁺ or 3 H₂S (to 3 S).
13.Dichromate Oxidising Fe²⁺
$Cr_{2} O_{7}^{2 -} + 14 H^{+} + 6 Fe^{2 +} \to 2 Cr^{3 +} + 6 Fe^{3 +} + 7 H_{2} O$
1 mol Cr₂O₇²⁻ oxidises 6 mol Fe²⁺ — the basis of estimating iron(II) by titration. Check: total charge is +24 on each side.
14.Preparation of KMnO₄ from Pyrolusite
$2 MnO_{2} + 4 KOH + O_{2} \to 2 K_{2} MnO_{4} + 2 H_{2} O; 3 MnO_{4}^{2 -} + 4 H^{+} \to 2 MnO_{4}^{-} + MnO_{2} + 2 H_{2} O$
MnO₂ is fused with KOH and an oxidising agent (air or KNO₃) to give dark green K₂MnO₄, which disproportionates in neutral or acidic solution to purple permanganate. Commercially, manganate is oxidised to permanganate electrolytically in alkaline solution.
15.KMnO₄ as Oxidising Agent — Acidic Medium★
$MnO_{4}^{-} + 8 H^{+} + 5 e^{-} \to Mn^{2 +} + 4 H_{2} O (E^{\circ} = 1.52 V)$
Purple → colourless; 5 electrons per MnO₄⁻. Example: 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 10CO₂ + 8H₂O. Acidify with dilute H₂SO₄, not HCl (HCl would be oxidised to Cl₂).
16.KMnO₄ — Neutral or Faintly Alkaline Medium
$MnO_{4}^{-} + 2 H_{2} O + 3 e^{-} \to MnO_{2} + 4 OH^{-}$
Only 3 electrons; brown MnO₂ forms. Example: 2MnO₄⁻ + H₂O + I⁻ → 2MnO₂ + 2OH⁻ + IO₃⁻ (iodide goes to iodate, not to I₂).
17.Lanthanoids — Configuration and Oxidation States
$[Xe] 4 f^{1 - 14} 5 d^{0 - 1} 6 s^{2}; common O.S. = + 3 (also Ce^{4 +}, Eu^{2 +}, Yb^{2 +})$
+3 is characteristic. Ce⁴⁺ (4f⁰) is a strong oxidant (E° Ce⁴⁺/Ce³⁺ = +1.74 V) and returns to +3; Eu²⁺ (4f⁷) and Yb²⁺ (4f¹⁴) are reductants — again empty, half-filled and full subshells explain them.
18.Lanthanoid Contraction★
$r (La^{3 +}) > r (Ce^{3 +}) > \dots > r (Lu^{3 +}) (poor shielding by 4 f electrons)$
Steady decrease in atomic and ionic radii from La to Lu. Key consequence: 4d and 5d elements of the same group have almost equal radii (Zr 160 pm, Hf 159 pm).
19.Actinoids — Configuration and Oxidation States
$[Rn] 5 f^{1 - 14} 6 d^{0 - 1} 7 s^{2}; O.S. mostly + 3, up to + 7 (Np)$
General pattern; thorium (6d² 7s², no 5f electron) is a known exception. Actinoids show more oxidation states than lanthanoids because 5f, 6d and 7s levels are close in energy (Th +4, Pa +5, U +6, Np +7). Actinoid contraction from element to element is greater than lanthanoid contraction because 5f electrons shield even more poorly. All actinoids are radioactive.