Topic 2 of 6
Physical properties and oxidation states
Explain the small physical trends and flexible electron accounting.
A-Level 9476 (2026-2027)
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.
| Property | First-row transition metals compared with Ca | Explanation |
|---|---|---|
| Melting point | Generally 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. |
| Density | Generally 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 example
Explain 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?
- The added electrons mainly enter the inner 3d subshell rather than a new outer shell.
- Their additional shielding partly offsets the increase in nuclear charge experienced by outer electrons.
- A small net increase in effective attraction can still cause a modest contraction.
The radius is relatively invariant because nuclear charge and 3d shielding increase together. This is not a claim of perfect cancellation.
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 example
From configuration to likely states
Vanadium has [Ar]3d34s2. Explain why +2 and +5 are plausible oxidation states.
- Using the two 4s electrons gives a +2 state and a d3 metal centre.
- Using three 3d electrons as well gives the formal +5 state. Strong bonding to oxygen can stabilise this high state.
- The formula VO2+ has V at +5 because x + 2(-2) = +1.
Both +2 and +5 are plausible; V(III) and V(IV) also occur. VO2+ does not contain a free V5+ ion in water.