Electron Configurations of Transition Metals

Learning Objectives (OCR Spec 5.3.1)

By the end of this lesson you should be able to:

• Write the full and shortened ([Ar] notation) electron configurations for the first-row d-block elements
• Explain the anomalous configurations of Cr and Cu
• Write electron configurations for the ions formed by these elements
• State and apply the rule that 4s electrons are removed first
• Relate electron configuration to observed oxidation states

The First Row: Ti to Cu

The first-row d-block elements are Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn (Z = 21 to 30). Their ground-state electron configurations should, according to the Aufbau principle, fill 4s completely before starting 3d. For most of the row this is exactly what happens.

Standard Configurations

Filling 4s then 3d gives:

Z	Element	Full configuration	Short ([Ar]) form
21	Sc	1s2 2s2 2p6 3s2 3p6 3d1 4s2	[Ar] 3d1 4s2
22	Ti	1s2 2s2 2p6 3s2 3p6 3d2 4s2	[Ar] 3d2 4s2
23	V	1s2 2s2 2p6 3s2 3p6 3d3 4s2	[Ar] 3d3 4s2
24	Cr	1s2 2s2 2p6 3s2 3p6 3d5 4s1	[Ar] 3d5 4s1
25	Mn	1s2 2s2 2p6 3s2 3p6 3d5 4s2	[Ar] 3d5 4s2
26	Fe	1s2 2s2 2p6 3s2 3p6 3d6 4s2	[Ar] 3d6 4s2
27	Co	1s2 2s2 2p6 3s2 3p6 3d7 4s2	[Ar] 3d7 4s2
28	Ni	1s2 2s2 2p6 3s2 3p6 3d8 4s2	[Ar] 3d8 4s2
29	Cu	1s2 2s2 2p6 3s2 3p6 3d10 4s1	[Ar] 3d10 4s1
30	Zn	1s2 2s2 2p6 3s2 3p6 3d10 4s2	[Ar] 3d10 4s2

The two anomalies are Cr and Cu. Both "steal" an electron from 4s and put it into 3d to give a particularly stable arrangement.

Why Are Cr and Cu Anomalous?

For Cr, the expected configuration would be [Ar] 3d4 4s2. Instead we observe [Ar] 3d5 4s1. The reason is that a half-filled 3d sub-shell (d5) is especially stable because:

• Each of the five d-orbitals contains exactly one electron (all parallel spin, Hund's rule maximum)
• Exchange energy is maximised when all five spins are aligned
• The energy gained by promoting one 4s electron into 3d to achieve d5 outweighs the small energy cost

For Cu, the expected configuration would be [Ar] 3d9 4s2. Instead we observe [Ar] 3d10 4s1. The reason is that a completely filled 3d sub-shell (d10) is especially stable because:

• All five d-orbitals are full and paired
• The energy of a full sub-shell is slightly lower than a 9/10 filled one
• Again, the promotion of a 4s electron to 3d pays for itself

In exam answers, write: "Chromium adopts [Ar] 3d5 4s1 because a half-filled 3d sub-shell is particularly stable. Copper adopts [Ar] 3d10 4s1 because a fully filled 3d sub-shell is particularly stable."

The Rule for Ions: Remove 4s First

This is the single most important and most counter-intuitive rule in d-block chemistry:

When forming a positive ion, the 4s electrons are removed before the 3d electrons.

This is true even though during atom-building, the 4s was filled before 3d. The reason is that once electrons are placed in 3d, the nuclear charge pulls 3d below 4s in energy. So the outermost - and most easily removed - electrons are in 4s.

Worked examples:

• Fe (Z = 26): atom is [Ar] 3d6 4s2. Fe2+ is [Ar] 3d6 (remove both 4s). Fe3+ is [Ar] 3d5 (remove both 4s + one 3d).
• Cu (Z = 29): atom is [Ar] 3d10 4s1. Cu+ is [Ar] 3d10 (remove 4s). Cu2+ is [Ar] 3d9 (remove 4s + one 3d).
• Mn (Z = 25): atom is [Ar] 3d5 4s2. Mn2+ is [Ar] 3d5. Mn4+ is [Ar] 3d3. Mn7+ is [Ar] (all 3d + 4s removed).
• Cr (Z = 24): atom is [Ar] 3d5 4s1. Cr2+ is [Ar] 3d4 (remove 4s + one 3d). Cr3+ is [Ar] 3d3. Cr6+ is [Ar] (all removed).

Notice the quirk with Cr: because 4s was already 4s1 in the atom, removing two electrons to make Cr2+ requires taking the 4s electron AND one 3d electron.

Why the Stability of Cr3+ and Fe3+?

Fe3+ [Ar] 3d5 is particularly stable because it has the half-filled d5 configuration - this is why iron readily forms +3 in air (rust is Fe2O3.xH2O; iron(III) chloride FeCl3 is much more common than iron(II) chloride at room temperature in dry conditions).