Question 1

Why are the 4s electrons removed before the 3d electrons?

Accepted Answer

In a transition-metal atom, 3d and 4s energies are close, but the 4s electrons are removed first when a positive ion is formed. This is why Mn [Ar] 3d5 4s2 gives Mn2+ [Ar] 3d5 rather than a configuration with 3d4.

Question 2

Why are Mn2+ and Fe3+ especially associated with extra stability?

Accepted Answer

Both ions have a half-filled d5 configuration, with one electron in each of the five d orbitals. This arrangement is associated with the exchange energy involved in pairing electrons. Fe2+ is d6, so it does not have the same half-filled configuration.

Question 3

Is Mn in permanganate a bare Mn7+ ion?

Accepted Answer

No. Permanganate is the MnO4- ion, in which manganese has the formal oxidation state +7. The complete ion has a charge of -1, and its d0 configuration does not by itself explain the violet colour.

Question 4

Why do copper and zinc need special electron configurations?

Accepted Answer

Copper has the ground-state configuration [Ar] 3d10 4s1, giving Cu+ as d10 and Cu2+ as d9. Zinc loses its two 4s electrons to form the commonly observed Zn2+ ion with a d10 configuration.

Question 5

Does d0 automatically mean a species is colourless?

Accepted Answer

No. A d0 ion has no ordinary d-d transition, but ligand-to-metal charge transfer or other interactions can still produce colour. Permanganate with Mn(VII), a d0 formal configuration, is the standard example of a violet species.

Variable Oxidation States — Stripping 4s then 3d

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