d & f Block Elements — One-Shot Mind Map

What counts as a d-block element

Defd-Block elements

Elements in which the last electron enters the d-subshell.

Also called transition elements — but with a condition.

The condition A transition element must have a partially filled d-subshell (d¹–d⁹) either in the neutral atom or in any of its known oxidation states.

Exd-Block but NOT transition

Zn, Cd, Hg, Cn (group 12) have only d⁰ / d¹⁰ configurations in the atom and in all their known oxidation states.

So they sit in the d-block, yet they are not transition elements.

Exam trap "Which of the following is not a transition element?" → the answer is almost always Zn, Cd or Hg.

Atomic properties

1General electronic configuration

(n–1)d^1–10 ns^0–2

ns² is the normal case. ns¹ and ns⁰ are the exceptions you must memorise.

ns⁰ — only one element

Pd (Z = 46) → [Kr] 4d¹⁰ 5s⁰

ns¹ — nine elements

Cr, Cu · Nb, Mo, Ru, Rh, Ag · Pt, Au

2Atomic & ionic size

Size generally decreases left → right across a series (added d-electrons shield poorly, Z_eff rises).
For ions of the same charge, size also decreases left → right.
Near the end of the series the decrease flattens out and then reverses (electron–electron repulsion in the filled d-orbitals).

3d series — final order

Sc > Ti > V > Cr < Mn > Fe ≈ Co ≈ Ni < Cu < Zn

3Ionisation enthalpy

General trend: IE increases left → right, but the actual values are irregular.

3d series IE₁ — final (data-based) order

Sc < V < Cr < Ti < Mn < Ni < Cu < Co < Fe < Zn

Why irregular Every IE value is decided by the stability of the ion that is left behind — half-filled (d⁵) and fully-filled (d¹⁰) products are extra stable, so removing that electron costs more.

4Important IE comparisons

Each reversal is explained by the configuration of the resulting ion
Comparison	Order	Reason (configuration left behind)
Cr vs Mn	Cr < Mn (IE₁) Cr > Mn (IE₂)	Cr⁺ = 3d⁵4s⁰ (stable, easy to form) · Mn⁺ = 3d⁵4s¹, so the 2nd removal from Cr breaks a stable d⁵
Cu vs Zn	Cu < Zn (IE₁) Cu > Zn (IE₂)	Cu⁺ = 3d¹⁰4s⁰ (stable) · Zn⁺ = 3d¹⁰4s¹, so the 2nd removal from Cu breaks a stable d¹⁰
Mn vs Fe	Mn < Fe (IE₁ & IE₂) Mn > Fe (IE₃)	Mn²⁺ = 3d⁵ (half-filled, very stable) → third electron is hard to pull out; Fe²⁺ = 3d⁶ loses one easily to reach d⁵

Alloy formation & interstitial compounds

5Alloy formation

Alloys form between metals whose atomic radii are within about 15% of each other — that is why d-block metals alloy so freely with one another.

Must-know examples

Alloy	Composition
Brass	Cu + Zn
German silver	Cu + Zn + Ni
Bronze	Cu + Sn

6Interstitial compounds

Formed when small atoms — H, B, C, N — get trapped in the interstitial voids of a metallic lattice.

Properties

Non-stoichiometric — composition is not a fixed whole-number ratio (e.g. Fe₃H, TiC, VH_0.56).
Harder than the pure metal.
Higher melting point than the pure metal.
Conductance is retained — they still conduct like a metal.
Chemically inert.

Magnetic moment & melting point

7Magnetic moment

Unpaired electrons present → paramagnetic.
Unpaired electrons absent (n = 0) → diamagnetic.

Spin-only formula

μ = √[ n(n + 2) ] BM n = number of unpaired electrons

Memory hook Magnetic moment = "n point something" — for n = 1, μ = 1.73; n = 2 → 2.83; n = 3 → 3.87; n = 4 → 4.90; n = 5 → 5.92. The answer always begins with a digit close to n.

8Melting point

More unpaired electrons → stronger interatomic (metallic) bonding → higher enthalpy of atomisation → higher melting point.

Group-wise orders

Groups	Order
3 and 12	3d > 4d > 5d
11	3d > 5d > 4d (Cu > Au > Ag)
4 to 10	3d < 4d < 5d

Two numbers to remember Highest m.p. among all d-block elements → W (tungsten). Highest in each series → group 6: Cr, Mo, W.

Variable oxidation states

9Why the OS varies

More electrons are available — the (n–1)d electrons take part in bonding in addition to the ns electrons.
The energy gap between (n–1)d and ns subshells is very small, so both sets are usable.

Do not exist Fe⁺⁵, Co⁺⁵, Sc⁺²

10Range across the 3d series

Number of oxidation states first increases up to d⁵, then decreases.

Sc (only +3) → … → Mn (+2 to +7, maximum) → … → Zn (only +2)

Mn shows the maximum number of oxidation states in the 3d series: +2, +3, +4, +5, +6, +7.

11Stability of higher oxidation states

Down a group in the d-block, the stability of the higher oxidation state increases.

Cr⁺⁶ < Mo⁺⁶ < W⁺⁶

Species	Behaviour in +6
Cr₂O₇^2–, CrO₃	Not stable in +6 → work as oxidising agents
MoO₃, WO₃	Stable in +6 → not oxidising agents

12Which ligands support high OS

Higher oxidation states of d-block elements are stable only with the most electronegative elements — O and F.

Oxygen stabilises even better than fluorine because it can form multiple (pπ–dπ) bonds with the metal.

Exist	Do not exist
MnF₂, MnF₃, MnF₄ Mn₂O₇, Mn₃O₄, MnO₃F, Mn₂O₃	MnF₅, MnF₆, MnF₇ steric hindrance — too many F around Mn

Note that Mn reaches +7 with oxygen (Mn₂O₇) but only +4 with fluorine.

13Complex formation tendency

d-Block metal ions form complexes readily because of their

small size,
high charge (high charge density),
and availability of vacant d-orbitals of suitable energy.

14Colour of aqueous ions

Colour arises from d–d transition: an electron jumps between split d-orbitals by absorbing visible light.

Colourless ions d⁰ and d¹⁰ ions are colourless — there is no d–d transition possible (no electron to promote, or no vacancy to promote into). Examples: Sc³⁺, Ti⁴⁺, Cu⁺, Zn²⁺.

Standard reduction potentials (E°)

15E°_M²⁺/M

In the 3d series all values are negative except copper.

E°_Cu²⁺/Cu = +0.34 V (positive)

Reason: the high sublimation energy and high IE₂ of copper are not compensated by its hydration enthalpy.

Consequence Cu cannot liberate H₂ gas from acids — the classic one-liner question from this chapter.

16E°_{M³⁺/M²⁺}

Read every value as: how badly does M²⁺ want to exist?
Couple	Value	Reason
Sc³⁺/Sc²⁺	Low (–ve)	Sc³⁺ has stable noble-gas configuration, so it resists being reduced
Zn³⁺/Zn²⁺	High (+ve)	Zn²⁺ is d¹⁰ — very stable, strongly favoured
Mn³⁺/Mn²⁺	High	Mn²⁺ is d⁵ half-filled — extra stable

17What the E° values tell you

Strong oxidising agents (M³⁺ → M²⁺)

Mn³⁺ · Co³⁺

Both are desperate to pick up an electron and drop to +2.

Strong reducing agents (M²⁺ → M³⁺)

Ti²⁺ · V²⁺ · Cr²⁺

Cr²⁺ gives up an electron to reach the stable d³ (t_2g³) configuration.

Oxides of the d-block

18Nature of the oxides

All MO-type oxides (metal in +2) are ionic — low oxidation state means ionic character.
As the oxidation state increases, the electronegativity of the metal increases and the oxide becomes more acidic.

Exception to remember ScO does not exist.

Acid–base character vs oxidation state

Oxidation state	Nature of oxide
+1, +2, +3	Basic
+4	Amphoteric
+5, +6, +7, +8	Acidic

19Named exceptions

Cr₂O₃ (+3) and ZnO (+2) are amphoteric, not basic.
V₂O₅ (+5) is mainly acidic but also amphoteric.

V₂O₅ — proof of amphoteric character

V₂O₅ + acid → VO₂⁺

V₂O₅ + base → VO₄^3–

Potassium permanganate, KMnO₄

20Name the two ions correctly

Compound	Name	Ion	OS of Mn	Colour
K₂MnO₄	Potassium manganate	MnO₄^2–	+6	Dark green
KMnO₄	Potassium permanganate	MnO₄^–	+7	Dark purple / violet

Disproportionation in acid

3 MnO₄^2– + 4 H⁺ → 2 MnO₄^– + MnO₂ + 2 H₂O

Green (+6) splits into purple (+7) and brown (+4).

21Preparation from pyrolusite ore

Ore: MnO₂ (pyrolusite), Mn in +4.

MnO₂ — fused with KOH, air or KNO₃ (oxidising agent) → K₂MnO₄

K₂MnO₄ → KMnO₄, three routes

Reagent	Product
H⁺ (disproportionation)	KMnO₄ + MnO₂
Electrolytic oxidation	KMnO₄ — commercial method
O₃ (strong oxidant)	KMnO₄

Laboratory method

Mn²⁺ + S₂O₈^2– → MnO₄^– + SO₄^2–

Colourless Mn²⁺ → violet permanganate. S₂O₈^2– is peroxodisulphate, the oxidising agent (a peroxide).

22MnO₄^2– — manganate ion

Mn is +6 → d¹, so n = 1 → paramagnetic.
Colour: dark green (a genuine d–d transition, since d¹).
Shape: tetrahedral.
Average Mn–O bond order = 6/4 = 1.5.
All Mn–O bond lengths are equal.
Number of dπ–pπ bonds = 2.

23MnO₄^– — permanganate ion

Mn is +7 → d⁰, n = 0 → diamagnetic (shows only weak, temperature-dependent paramagnetism).
Colour: dark purple / violet — not from a d–d transition (there are no d-electrons). It comes from LMCT (ligand-to-metal charge transfer).
Shape: tetrahedral.
Average Mn–O bond order = 7/4 = 1.75.
All Mn–O bond lengths are equal.
Number of dπ–pπ bonds = 3.

Compare the two Bond order: MnO₄^– > MnO₄^2– → Bond length: MnO₄^– < MnO₄^2–. (Bond order ↑ ⇒ bond length ↓.)

24Reactions — thermal decomposition

2 KMnO₄ (s) —Δ→ K₂MnO₄ + MnO₂ + O₂

Mn goes from +7 to +6 and +4; oxygen is released.

Why KMnO₄ is such a strong oxidant

Mn sits in its highest oxidation state (+7), so it can only be reduced. It readily drops to Mn²⁺ to achieve the stable d⁵ configuration.

It works as a strong oxidising agent in acidic medium, and also in faintly basic / neutral medium.

25Oxidations in acidic medium

Purple MnO₄^– → colourless Mn²⁺

I^–	→	I₂
Cl^–	→	Cl₂
Br^–	→	Br₂
Fe²⁺	→	Fe³⁺
NO₂^–	→	NO₃^–
H₂S / S^2–	→	S
C₂O₄^2–	→	CO₂
Sn²⁺	→	Sn⁴⁺
SO₂ / SO₃^2–	→	SO₄^2–

26Oxidations in neutral / faintly basic medium

MnO₄^– → brown MnO₂ (Mn ends at +4, not +2)

I^–	→	IO₃^–
Fe²⁺	→	Fe³⁺
Sn²⁺	→	Sn⁴⁺
C₂O₄^2–	→	CO₂

High-yield contrast Iodide is the giveaway: in acidic medium I^– → I₂, but in neutral / faintly basic medium I^– → IO₃^– (+5).

Potassium dichromate, K₂Cr₂O₇

27Chromate ⇌ dichromate

2 CrO₄^2– + 2 H⁺ (pH < 7) ⇌ Cr₂O₇^2– + H₂O reverse with OH^–, pH > 7

	CrO₄^2–	Cr₂O₇^2–
Name	Chromate ion	Dichromate ion
Colour	Yellow	Orange
Stable in	Basic medium	Acidic medium
OS of Cr	+6	+6

Not a redox reaction The interconversion is pH dependent only — the oxidation state of Cr stays +6 on both sides.

28Preparation from chromite ore

Ore: FeCr₂O₄ — a mixed oxide = FeO (+2) + Cr₂O₃ (+3).

4 FeCr₂O₄ + 7 O₂ (air) —oxidation→ 2 Fe₂O₃ (basic) + 8 CrO₃ (acidic)

CrO₃ + Na₂CO₃ (basic) → Na₂CrO₄ yellow solution of sodium chromate

Na₂CrO₄ —H₂SO₄ (H⁺)→ Na₂Cr₂O₇ orange solution

Na₂Cr₂O₇ + 2 KCl → K₂Cr₂O₇ orange crystals

Only the acidic oxide CrO₃ reacts with the basic Na₂CO₃; Fe₂O₃ (basic) does not.

29Structures

CrO₄^2– — chromate

Tetrahedral.
All Cr–O bond lengths equal.
Average Cr–O bond order = 6/4 = 1.5.
2 dπ–pπ bonds.

Cr₂O₇^2– — dichromate

Two tetrahedral units joined by a Cr–O–Cr bridge.
4 dπ–pπ bonds.
All Cr–O bond lengths are not equal.
Terminal Cr–O < bridging Cr–O bond length.

Magnetism Both ions are diamagnetic — Cr is +6, i.e. d⁰, so n = 0.

30Two "Imp Note" facts

Na₂Cr₂O₇ is more soluble than K₂Cr₂O₇.
K₂Cr₂O₇ is preferred over Na₂Cr₂O₇ as a primary standard in volumetric analysis.

Reason Na₂Cr₂O₇ is hygroscopic — it absorbs moisture from the atmosphere, so its mass is unreliable. That makes it a secondary standard, not a primary one.

31Oxidising reactions

Orange Cr₂O₇^2– → green Cr³⁺ (in acidic medium)

Fe²⁺	→	Fe³⁺
Sn²⁺	→	Sn⁴⁺
S^2–	→	S
SO₂ / SO₃^2–	→	SO₄^2–
Cl^–	✗	Cl₂

Why Cl^– is not oxidised Because Cl₂ is the stronger oxidant: Cl₂ > Cr₂O₇^2–. Dichromate simply is not powerful enough to take an electron from chloride.

32Thermal decomposition

4 K₂Cr₂O₇ (s) —Δ→ 4 K₂CrO₄ + 2 Cr₂O₃ + 3 O₂

Cr goes from +6 to +6 (in chromate) and +3 (in Cr₂O₃).

33KMnO₄ vs K₂Cr₂O₇ — the comparison question

Both are oxidising agents because in both, the metal sits in its highest oxidation state.

Oxidising power: MnO₄^– > Cl₂ > Cr₂O₇^2–

	MnO₄^– (+7)	Cr₂O₇^2– (+6)
Reduced to	Mn²⁺ — colourless	Cr³⁺ — green
Configuration reached	d⁵, half-filled & stable	d³ (t_2g³), half-filled t_2g & stable
Oxidises Cl^–?	Yes	No

f-Block: the two inner-transition series

LnLanthanoids

n = 6 · Group 3
Atomic numbers 58 to 71
Ce to Lu
4f subshell is progressively filled

AnActinoids

n = 7 · Group 3
Atomic numbers 90 to 103
Th to Lr
5f subshell is progressively filled

Lanthanoids in detail

34Electronic configuration

6s² is common to all; the occupancy of 4f varies. General form: [Xe] 4f^1–14 5d^0–1 6s².

5d⁰ is normal · 5d¹ is exceptional.
Energy order: 6s < 4f < 5d.

Fill-it-yourself trick Fill up to 56 first — [Xe] = 54, plus 6s² = 56, which is common to all. Then put every remaining electron into 4f. Only Ce, Gd, Lu break the rule with a 5d¹.

Mnemonic for the 5d¹ three Ce · Gd · Lu → "Cinema — Gadar — Lut gaye"

35The 4f configurations

Z	Element	Configuration
58	Ce	[Xe] 4f¹ 5d¹ 6s²
59	Pr	[Xe] 4f³ 5d⁰ 6s²
60	Nd	[Xe] 4f⁴ 5d⁰ 6s²
61	Pm	[Xe] 4f⁵ 5d⁰ 6s²
62	Sm	[Xe] 4f⁶ 5d⁰ 6s²
63	Eu	[Xe] 4f⁷ 5d⁰ 6s²
64	Gd	[Xe] 4f⁷ 5d¹ 6s²
65	Tb	[Xe] 4f⁹ 5d⁰ 6s²
66	Dy	[Xe] 4f¹⁰ 5d⁰ 6s²
67	Ho	[Xe] 4f¹¹ 5d⁰ 6s²
68	Er	[Xe] 4f¹² 5d⁰ 6s²
69	Tm	[Xe] 4f¹³ 5d⁰ 6s²
70	Yb	[Xe] 4f¹⁴ 5d⁰ 6s²
71	Lu	[Xe] 4f¹⁴ 5d¹ 6s²

Pattern to note f⁰, f², f⁸ do not appear in this series; f⁷ and f¹⁴ each appear twice (Eu & Gd, Yb & Lu).

36Size & lanthanoid contraction

Size decreases left → right because electrons enter the 4f subshell, which shields poorly: shielding (σ) falls, Z_eff rises, and the atom contracts. This steady shrink is the lanthanoid contraction.

Atomic size order

Eu > La > Ce > Pr > Nd > Pm > Sm < Eu > Gd > Tb > Dy > Ho > Er > Tm > Yb > Lu

Why Eu is the odd one out Eu has a 4f⁷ half-filled configuration → loose packing, weak metal–metal bonding → larger radius. It is the largest in the series.

Ionic size (M³⁺) — perfectly regular

La³⁺ > Ce³⁺ > Pr³⁺ > Nd³⁺ > Pm³⁺ > Sm³⁺ > Eu³⁺ > Gd³⁺ > Tb³⁺ > Dy³⁺ > Ho³⁺ > Er³⁺ > Tm³⁺ > Yb³⁺ > Lu³⁺

37Consequence: basicity of the hydroxides

Going from La(OH)₃ to Lu(OH)₃:

Property	Trend
Size of M³⁺	Decreases
Electronegativity of M	Increases
Acidic character of the OH^–/O^2–	Increases
Basic character of M(OH)₃	Decreases

La(OH)₃ is the most basic; Lu(OH)₃ the least.

38Oxidation states

Common oxidation state: +3.
Most stable oxidation state: +3.
Some elements also show +2 and +4, whenever that gives a vacant (f⁰), half-filled (f⁷) or fully-filled (f¹⁴) f-subshell.

Ion	Configuration	Why it is stable
Ce⁴⁺	4f⁰	Inert-gas / vacant f
Tb⁴⁺	4f⁷	Half-filled
Eu²⁺	4f⁷	Half-filled
Yb²⁺	4f¹⁴	Fully filled

39Oxidants and reductants

+4 → +3 : oxidising agents

Ce⁴⁺ → Ce³⁺ E° = +1.74 V

Tb⁴⁺ → Tb³⁺

They gain an electron to reach the very stable +3 state, so they work as oxidants.

+2 → +3 : reducing agents

Eu²⁺ → Eu³⁺

Yb²⁺ → Yb³⁺

They lose an electron to reach +3, so they work as reductants.

40General characteristics

Silvery white metals.
Tarnish rapidly in air.
Hardness increases left → right; Sm is steel-hard.
Melting point up to 1623 K — the maximum in the series.
Good conductors of heat and electricity.
Almost all M³⁺ ions are coloured in both the solid and aqueous state — except La³⁺ and Lu³⁺, which are colourless.
f⁰ and f¹⁴ species are diamagnetic (n = 0): La³⁺, Lu³⁺, Ce⁴⁺, Yb²⁺.

41Ionisation enthalpy

IE₁ ≈ 600 kJ mol^–1 and IE₂ ≈ 1200 kJ mol^–1 — comparable to calcium.

Abnormally low IE₃ La, Gd, Lu — because after losing the third electron they land on empty (f⁰), half-filled (f⁷) and fully-filled (f¹⁴) f-orbitals respectively.

Atom	M³⁺ ends as
La [Xe] 4f⁰ 5d¹ 6s²	La³⁺ [Xe] 4f⁰
Gd [Xe] 4f⁷ 5d¹ 6s²	Gd³⁺ [Xe] 4f⁷
Lu [Xe] 4f¹⁴ 5d¹ 6s²	Lu³⁺ [Xe] 4f¹⁴

42Chemical reactivity

Earlier members behave much like calcium.
As atomic number increases they start to behave like aluminium.

Ln³⁺(aq) + 3e^– → Ln(s) E° = –2.2 to –2.4 V

Ln is the common symbol used for all lanthanoids.

43Reactions of Ln

Reagent	Product
N₂, Δ	LnN
Acids	H₂ gas released
H₂O	Ln(OH)₃ + H₂
Halogens (X₂)	LnX₃
C, Δ	LnC₂ / Ln₃C / Ln₂C₃ — carbides
S	Ln₂S₃
O₂	Ln₂O₃

Every product follows from Ln → Ln³⁺ + 3e^–, so the formulas fall out of the +3 state.

Actinoids

44Basic nature

All actinoids are radioactive.
Earlier members have relatively long half-lives.
Later members are extremely short-lived — from about a day down to 3 minutes for Lr (Z = 103).

45Electronic configuration

The 5f subshell is progressively filled and 7s² is common to all: [Rn] 5f^1–14 6d^0–1 7s².

Fill-it-yourself trick Fill up to 88 first — [Rn] = 86, plus 7s² = 88. Then put every remaining electron into 5f.

6d⁰ is normal; 6d¹ or 6d² is exceptional.
Exceptions: Pa (91), U (92), Np (93), Cm (96), Lr (103) → 6d¹.
Special case: Th (90) → 5f⁰ 6d².

Same pattern as the lanthanoids f¹, f⁵, f⁸ are skipped; f⁷ and f¹⁴ repeat.

46Size & actinoid contraction

Size generally decreases left → right in both M and M³⁺ — the actinoid contraction — caused by the poor shielding of 5f electrons.

Z_eff = Z – σ shielding ↓ ⇒ Z_eff ↑ ⇒ size ↓

Shielding order

s > p > d > f and 4f > 5f

Which contraction is stronger? Actinoid contraction > Lanthanoid contraction, because 5f electrons shield even more poorly than 4f electrons.

47Oxidation states & other points

Common oxidation state: +3, but actinoids show a much wider range because 7s, 5f and 6d are of comparable energy.
Maximum oxidation state: +7, shown by Np and Pu.
Ionisation enthalpies are generally lower than those of the lanthanoids.
Sizes are generally greater than the corresponding lanthanoids.
Highly reactive, especially in powdered form.
With concentrated HNO₃ they undergo passivation — a protective oxide layer forms on the metal surface.