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9476 / 2027
Transition Elements

Lesson 2 of 6

Physical properties and oxidation states

Explain the small physical trends and flexible electron accounting.

A-Level 9476 (2026-2027)

  • I can explain relatively small radius and first-ionisation-energy changes.Syllabus

    Syllabus 9476, 13(c). Explain relatively small radius and first-ionisation-energy changes.

  • I can compare transition-metal physical properties with calcium.Syllabus

    Syllabus 9476, 13(d). Compare transition-metal physical properties with calcium.

  • I can explain variable oxidation states.Syllabus

    Syllabus 9476, 13(e). Explain variable oxidation states.

  • I can suggest likely oxidation states from configuration.Syllabus

    Syllabus 9476, 13(f). Suggest likely oxidation states from configuration.

Why the physical trends are comparatively gentle

Increasing nuclear attraction is partly offset by added 3d shielding.

From scandium towards copper, nuclear charge rises while electrons are added mainly to 3d. These added electrons partly shield the outer electrons from the increasing nuclear charge, and the atoms do not acquire a new principal shell. The two changes largely offset one another: atomic radii and first ionisation energies vary relatively little compared with a full main-group period. Radii generally contract modestly and first ionisation energies show small, irregular changes; neither quantity is exactly constant.

Qualitative comparison with calcium
PropertyFirst-row transition metals compared with CaExplanation
Melting pointGenerally higher.More extensive participation of 3d and 4s electrons in metallic bonding, together with relatively small atoms, generally produces stronger bonding than in calcium. The precise trend is irregular.
DensityGenerally higher.Mass increases while atomic volumes remain relatively small. More mass packed into a similar or smaller volume gives a higher density.

Do not explain a high melting point using strong intermolecular forces: these are metallic structures. Do not explain density only by atomic mass; the volume occupied is also essential.

Worked exampleExplain a supplied trend

A data table shows only a small change in atomic radius across Sc to Cu, despite eight extra protons. What balances the increased nuclear attraction?

Try it first. Then show the worked stepsHide the worked steps
  1. The added electrons mainly enter the inner 3d subshell rather than a new outer shell.
  2. Their additional shielding partly offsets the increase in nuclear charge experienced by outer electrons.
  3. A small net increase in effective attraction can still cause a modest contraction.
Answer

The radius is relatively invariant because nuclear charge and 3d shielding increase together. This is not a claim of perfect cancellation.

Check your understanding

Why do atomic radii change only a little from Sc to Cu?

  1. Each element adds a new outer shell, which cancels the rising nuclear charge.
  2. Nuclear charge stays the same across the series, so the pull is unchanged.
  3. Extra electrons go into the inner 3d subshell and shield the outer 4s electrons.
Show the answer

Extra electrons go into the inner 3d subshell and shield the outer 4s electrons.

Nuclear charge rises across the series, but the added 3d electrons partly shield the 4s electrons. The two effects largely cancel, and no new shell is added.

Close 3d and 4s energies allow several oxidation states

Configuration suggests possibilities; the reaction environment decides which are stable.

Transition elements commonly have variable oxidation states because their 3d and 4s electrons have comparable energies and different numbers can participate in bonding. Fe commonly forms +2 and +3 compounds; copper commonly forms +1 and +2 compounds. Calcium, by contrast, usually forms +2: removing more electrons would disrupt a much more tightly bound noble-gas core.

A configuration can suggest likely oxidation states by accounting for 4s electrons and, where chemically feasible, additional 3d electrons. Early members can use more of these electrons in high oxidation states. This is a prediction of possibilities, not proof that every intermediate state is equally stable in water.

Worked exampleFrom configuration to likely states

Vanadium has [Ar]3d34s2. Explain why +2 and +5 are plausible oxidation states.

Try it first. Then show the worked stepsHide the worked steps
  1. Using the two 4s electrons gives a +2 state and a d3 metal centre.
  2. Using three 3d electrons as well gives the formal +5 state. Strong bonding to oxygen can stabilise this high state.
  3. The formula VO2+ has V at +5 because x + 2(-2) = +1.
Answer

Both +2 and +5 are plausible; V(III) and V(IV) also occur. VO2+ does not contain a free V5+ ion in water.

Watch out for this

A student says: "Mn is +7 in MnO4-, so the solution contains bare Mn7+ ions." What is wrong with this?

Show the answer

In MnO4-, Mn has oxidation number +7; this does not mean the ion contains a bare aqueous Mn7+ cation. Ligand exchange can change colour without changing oxidation number.

Check your understanding

Why does iron form both Fe2+ and Fe3+, while calcium forms only Ca2+?

  1. Iron has more protons than calcium, so it can afford to lose more electrons each time.
  2. Iron's 3d and 4s electrons are close in energy; calcium's next electron is in a tightly held inner shell.
  3. Calcium's 3d subshell is full, so it cannot lose any more electrons than two.
Show the answer

Iron's 3d and 4s electrons are close in energy; calcium's next electron is in a tightly held inner shell.

Iron can lose its two 4s electrons and then a 3d electron for little extra energy. Removing a third electron from calcium would break into its noble-gas core, which costs far too much energy.