Chapter 9 Test Review

Question

What d-block elements are exceptions to the configuration rules? Write the electron configuration for those elements.

Answer 1

The two your course almost certainly wants are chromium and copper.

Cr (24): [Ar] 4s¹ 3d⁵  not [Ar] 4s² 3d⁴
Cu (29): [Ar] 4s¹ 3d¹⁰  not [Ar] 4s² 3d⁹

In both, one electron is promoted from 4s to 3d, leaving 4s singly occupied.

Why it happens

Two things make the swap cheap and worthwhile:

1. 4s and 3d are very close in energy in this region of the table, so moving an electron between them costs almost nothing.
2. Half-filled (d⁵) and completely filled (d¹⁰) d subshells are unusually stable. With one electron in each of the five d orbitals, or two in each, the electron distribution is spherically symmetric and the number of favourable same-spin exchange interactions is maximised.

Cr reaches d⁵ and Cu reaches d¹⁰ by borrowing a single 4s electron, and the stability gained outweighs the promotion cost.

If you need the full list — there are more than two

Period 5 (where the exceptions are far more common):
Nb (41): [Kr] 5s¹ 4d⁴
Mo (42): [Kr] 5s¹ 4d⁵
Ru (44): [Kr] 5s¹ 4d⁷
Rh (45): [Kr] 5s¹ 4d⁸
Pd (46): [Kr] 4d¹⁰ — the strangest of all, with no 5s electrons
Ag (47): [Kr] 5s¹ 4d¹⁰

Period 6:
Pt (78): [Xe] 6s¹ 4f¹⁴ 5d⁹
Au (79): [Xe] 6s¹ 4f¹⁴ 5d¹⁰

Notice that Nb, Ru, and Rh do not end at d⁵ or d¹⁰. That is the clue that the half-filled/filled story is not the whole explanation.

An honest caveat about the standard explanation

"Half-filled shells are extra stable" is what textbooks say and what your exam will want, but it is a simplification. The deeper cause is that 3d orbitals are compact, so putting two electrons in the same d orbital carries a large repulsion penalty; spreading them out and accepting a lone 4s electron is often the lower-energy arrangement. That is why exceptions cluster in period 5, where the orbital energies sit even closer together, and why Pd abandons its s electrons entirely.

Two practical notes: these are gas-phase, ground-state, neutral-atom configurations — the picture changes in compounds. And when a transition metal forms a cation, the s electrons leave first regardless: Cu⁺ is [Ar] 3d¹⁰, and Fe²⁺ is [Ar] 3d⁶, not [Ar] 4s² 3d⁴. That rule catches more students than the exceptions themselves do.