Why are Zn²⁺ salts colourless while Cu²⁺ salts are blue?

Zn²⁺ is 3d¹⁰ with no vacancy for a d–d transition, while Cu²⁺ is 3d⁹

Copper has a higher atomic mass

The intense purple colour of MnO₄⁻ is due to:

Unpaired electrons in the 4s orbital

Mercury is liquid at room temperature mainly because:

Its filled 5d¹⁰ shell leaves only weak metallic bonding from the s electrons

It has variable oxidation states

Cu⁺ disproportionates in aqueous solution because:

The large hydration enthalpy of Cu²⁺ outweighs the second ionisation enthalpy

Cu⁺ is 3d¹⁰ and therefore unstable

The large hydration enthalpy of Cu²⁺ outweighs the second ionisation enthalpy

Water oxidises Cu⁺ to Cu²⁺ directly

Actinoids show a much wider range of oxidation states than lanthanoids because:

5f orbitals are less buried than 4f and so are more available for bonding

They form more stable complexes

The highest oxidation state of manganese, +7, is possible because:

It has seven valence electrons available in 3d⁵4s²

It is the largest atom in the series

It has the lowest ionisation enthalpy in the series

Cerium commonly shows a +4 state, unusual among lanthanoids, because Ce⁴⁺:

Attains the empty 4f⁰ configuration of xenon

Has a half-filled 4f⁷ configuration

Attains the empty 4f⁰ configuration of xenon

The d- and f-Block Elements

Every distinctive property of the transition metals — variable oxidation state, colour, paramagnetism, catalytic activity, complex formation, alloy formation — comes from one structural fact: the (n−1)d and ns orbitals are so close in energy that d electrons are chemically available. In the s-block the d orbitals are empty and in the p-block they are irrelevant; only here do they participate. Hold that single sentence and the chapter organises itself, because each of those six properties is a different consequence of partly filled d orbitals. The exceptions — zinc's colourlessness, manganese's wide oxidation range, the near-identity of the lanthanoids — are equally derivable once you know what to look at.

Why the Transition Metals Behave As They Do

Six characteristic properties and the single cause behind each
Property	Cause
Variable oxidation states	ns and (n−1)d electrons are close in energy, so both can be lost
Coloured ions	d–d transitions absorbing visible light
Paramagnetism	Unpaired d electrons
Catalytic activity	Variable oxidation state plus surface adsorption of reactants
Complex formation	Small highly charged ions with vacant d orbitals to accept lone pairs
Alloy formation	Similar atomic radii let one metal replace another in the lattice

⚠️ Zn, Cd and Hg are not really transition metals

A transition element is defined by having a partly filled d subshell in its ground state or in a common oxidation state. Zinc is 3d¹⁰4s² and Zn²⁺ is 3d¹⁰ — completely full in both — so it is a d-block element but not a transition element. This single fact explains a cluster of exam answers at once: zinc compounds are colourless because a full d¹⁰ set has no vacancy for a d–d transition, zinc is diamagnetic because nothing is unpaired, zinc shows only the +2 state, and zinc is a poor catalyst. So the standard question "why is Zn²⁺ colourless while Cu²⁺ is blue" has a one-line answer: Cu²⁺ is 3d⁹ and has a vacancy, Zn²⁺ is 3d¹⁰ and does not. Sc³⁺ (3d⁰) and Ti⁴⁺ (3d⁰) are colourless for the mirror-image reason — no electron to promote.

The highest oxidation state peaks in the middle of a series, because that is where the maximum number of d electrons is available for bonding while the nucleus can still let them go. Manganese, with 3d⁵4s², reaches +7 in MnO₄⁻ and shows every state from +2 to +7; by iron the nuclear pull has grown enough that +6 is the practical limit, and by zinc only +2 remains. The most stable state in aqueous solution is +2 for most of the series, and Mn²⁺ (3d⁵) and Fe³⁺ (3d⁵) are unusually stable because a half-filled d shell carries extra exchange energy.

🎯 Why transition metals are hard, dense and high-melting — and why Zn, Cd and Hg are not

Metallic bonding is stronger when more unpaired electrons are available to delocalise, and a partly filled d shell supplies them in quantity — which is why tungsten melts at 3422 °C, the highest of all metals, and why chromium is hard enough to be a plating. Zinc, cadmium and mercury have full d¹⁰ shells, so only the two s electrons contribute; their metallic bonding is correspondingly feeble, and mercury is so weakly bound that it is liquid at room temperature, the only metal that is. So the "why is mercury liquid" question is answered by the same d¹⁰ fact that explains its colourlessness, not by anything peculiar to mercury as an element.

The Compounds That Are Actually Examined

Potassium permanganate and dichromate
Point	KMnO₄	K₂Cr₂O₇
Oxidation state of the metal	Mn is +7	Cr is +6
Colour and its origin	Deep purple — charge transfer, not d–d, since Mn is d⁰	Orange — also charge transfer
Behaviour in alkali	Reduced only to green manganate, n-factor 1	Converts to yellow chromate CrO₄²⁻
Behaviour in acid	Reduced to Mn²⁺, n-factor 5	Reduced to Cr³⁺, n-factor 6
Use as a primary standard	No — it is not stable enough to weigh accurately	Yes — stable, non-hygroscopic, weighable

🎯 Why MnO₄⁻ is intensely coloured even though Mn has no d electrons

In MnO₄⁻ manganese is +7, so its configuration is 3d⁰ — there is no d electron to promote and therefore no possible d–d transition. Yet permanganate is one of the most intensely coloured ions known. The colour comes from ligand-to-metal charge transfer: an electron jumps from an oxygen p orbital into an empty metal d orbital, absorbing visible light in the process. Charge-transfer bands are far more intense than d–d bands, which is precisely why a permanganate solution is deeply coloured at concentrations where a copper sulphate solution is barely tinted. This is the answer the paper wants whenever a d⁰ or d¹⁰ species is nonetheless coloured.

Two further compound facts recur. Cu⁺ disproportionates in water — 2Cu⁺ → Cu²⁺ + Cu — because the very large hydration enthalpy of the doubly charged Cu²⁺ more than pays for the second ionisation, so copper(I) salts are stable only when insoluble (CuCl, CuI) or complexed. And the +2 state of the 3d series is stabilised across the board in water for the same hydration reason, which is why Fe²⁺ exists happily in solution while Fe⁴⁺ does not.

The Lanthanoids and Actinoids

The fourteen lanthanoids are so alike chemically that separating them was one of the hardest problems in nineteenth-century chemistry. The reason is that the differentiating electrons enter the deeply buried 4f orbitals, which barely affect the outer 5d/6s shell that does the bonding — so all of them behave as +3 ions of nearly the same size. The lanthanoid contraction across the series is a steady shrinkage caused by the poor screening of those 4f electrons, and its consequences reach far outside the f-block, making Hf the size of Zr and the 5d metals unusually dense.

Lanthanoids compared with actinoids
Feature	Lanthanoids (4f)	Actinoids (5f)
Common oxidation state	Almost always +3	+3 but also +4, +5, +6, +7
Reason for the difference	4f is buried and shielded	5f is less buried, so more available for bonding
Radioactivity	Only promethium	All of them
Contraction across the series	Lanthanoid contraction	Actinoid contraction — sharper still
Complex formation	Limited tendency	Greater tendency, from higher charge and smaller size

ℹ️ Ce⁴⁺ and Eu²⁺ are the two lanthanoid exceptions worth knowing

Cerium reaches +4 because losing the fourth electron leaves the empty-shell 4f⁰ configuration of xenon, and europium drops to +2 because that leaves the half-filled 4f⁷. Both are the familiar stability of empty, half-filled and full subshells reappearing in the f-block. Practically, this is why ceric ammonium sulphate is a useful oxidising agent in volumetric analysis — Ce⁴⁺ reverting to the stable Ce³⁺ is a clean one-electron change — and why europium is the odd one out in an otherwise uniform series.

Key takeaways

Every transition-metal property traces to (n−1)d and ns being close in energy, so d electrons are available.
Zn, Cd and Hg are d-block but not transition metals — d¹⁰ explains their colourlessness and mercury being liquid.
The highest oxidation state peaks mid-series: Mn reaches +7, and by zinc only +2 survives.
Colour usually comes from d–d transitions, but MnO₄⁻ (d⁰) is coloured by ligand-to-metal charge transfer.
KMnO₄ has n-factor 5 in acid, 3 in neutral and 1 in base; K₂Cr₂O₇ has n-factor 6 and is a primary standard.
Cu⁺ disproportionates in water because Cu²⁺ hydration enthalpy pays for the second ionisation.
Lanthanoids are nearly identical because 4f is buried; actinoids show many states because 5f is more available.
Ce⁴⁺ (4f⁰) and Eu²⁺ (4f⁷) are the f-block exceptions, and both come from stable subshell configurations.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Why are Zn²⁺ salts colourless while Cu²⁺ salts are blue?
1. Zn²⁺ is larger than Cu²⁺
2. Zn²⁺ is 3d¹⁰ with no vacancy for a d–d transition, while Cu²⁺ is 3d⁹
3. Zinc is not a metal
4. Copper has a higher atomic mass
Show answer
Answer: B — Zn²⁺ is 3d¹⁰ with no vacancy for a d–d transition, while Cu²⁺ is 3d⁹
Colour in transition-metal ions normally arises from an electron being promoted between split d orbitals, which requires both an electron to move and a vacancy to move into. Zn²⁺ has a completely full 3d¹⁰ set so no such transition is possible, while Cu²⁺ is 3d⁹ and has exactly one vacancy. The same reasoning makes Sc³⁺ and Ti⁴⁺ colourless from the opposite end — d⁰, so no electron available to promote.
The intense purple colour of MnO₄⁻ is due to:
1. A d–d transition in Mn⁷⁺
2. Ligand-to-metal charge transfer
3. Unpaired electrons in the 4s orbital
4. The presence of potassium
Show answer
Answer: B — Ligand-to-metal charge transfer
Manganese in MnO₄⁻ is +7 and therefore 3d⁰, so there is no d electron to promote and a d–d transition is impossible. The colour instead comes from an electron jumping from an oxygen p orbital into an empty manganese d orbital. Charge-transfer bands are much more intense than d–d bands, which is why permanganate is deeply coloured at concentrations where copper sulphate is barely tinted.
Mercury is liquid at room temperature mainly because:
1. Its atoms are unusually large
2. Its filled 5d¹⁰ shell leaves only weak metallic bonding from the s electrons
3. It is radioactive
4. It has variable oxidation states
Show answer
Answer: B — Its filled 5d¹⁰ shell leaves only weak metallic bonding from the s electrons
Metallic bond strength depends on how many electrons are available to delocalise, and a full d¹⁰ shell contributes none, leaving only the two 6s electrons. That gives feeble bonding and an extremely low melting point. The contrast is tungsten, whose partly filled d shell gives the strongest metallic bonding of any element and a melting point of 3422 °C.
Which is used as a primary standard in volumetric analysis?
1. KMnO₄
2. K₂Cr₂O₇
3. NaOH
4. HCl
Show answer
Answer: B — K₂Cr₂O₇
Potassium dichromate is stable, non-hygroscopic and can be weighed accurately, so a solution of known concentration can be made up directly. KMnO₄ slowly decomposes and is contaminated by MnO₂, NaOH absorbs both moisture and carbon dioxide from the air, and concentrated HCl loses HBr gas — all three must therefore be standardised against something else rather than weighed out.
Cu⁺ disproportionates in aqueous solution because:
1. Cu⁺ is 3d¹⁰ and therefore unstable
2. The large hydration enthalpy of Cu²⁺ outweighs the second ionisation enthalpy
3. Copper is a noble metal
4. Water oxidises Cu⁺ to Cu²⁺ directly
Show answer
Answer: B — The large hydration enthalpy of Cu²⁺ outweighs the second ionisation enthalpy
Cu²⁺ is smaller and doubly charged, so its hydration enthalpy is very large — large enough to more than repay the energy cost of removing a second electron, which makes 2Cu⁺ → Cu²⁺ + Cu spontaneous in water. Note that 3d¹⁰ is normally a stable configuration, so the first option has the reasoning backwards, and copper(I) salts are in fact perfectly stable when insoluble, as CuCl and CuI are.
Actinoids show a much wider range of oxidation states than lanthanoids because:
1. They are all radioactive
2. 5f orbitals are less buried than 4f and so are more available for bonding
3. They have larger atomic radii
4. They form more stable complexes
Show answer
Answer: B — 5f orbitals are less buried than 4f and so are more available for bonding
Lanthanoid 4f orbitals lie deep inside the atom and are shielded from chemistry, so almost every lanthanoid behaves as +3 and nothing else. Actinoid 5f orbitals are more extended and closer in energy to 6d, so their electrons can take part in bonding, giving uranium states from +3 to +6 and neptunium up to +7. Radioactivity is a nuclear property and has no bearing on which electrons can bond.
The highest oxidation state of manganese, +7, is possible because:
1. It has seven valence electrons available in 3d⁵4s²
2. It is the largest atom in the series
3. It has the lowest ionisation enthalpy in the series
4. Its 4s orbital is empty
Show answer
Answer: A — It has seven valence electrons available in 3d⁵4s²
Manganese is 3d⁵4s², so all seven of those electrons can in principle be used in bonding, which is exactly the +7 of MnO₄⁻. This is also why the maximum oxidation state peaks in the middle of the series: earlier elements have fewer d electrons to offer, and later ones hold theirs too tightly as the nuclear charge builds up, so by zinc only +2 remains.
Cerium commonly shows a +4 state, unusual among lanthanoids, because Ce⁴⁺:
1. Has a half-filled 4f⁷ configuration
2. Attains the empty 4f⁰ configuration of xenon
3. Is radioactive
4. Is larger than Ce³⁺
Show answer
Answer: B — Attains the empty 4f⁰ configuration of xenon
Removing a fourth electron from cerium empties the 4f subshell entirely, giving the noble-gas configuration of xenon, and that stability is what makes +4 accessible. The half-filled 4f⁷ answer belongs to europium and gadolinium — Eu²⁺ is stabilised for exactly that reason. A cation is always smaller than its lower-charged form, so the last option is impossible.