Forming Transition-Metal Cations

Introduction

Forming a transition-metal cation is an electron-counting task with a common trap. Neutral first-row metal configurations often show 4s electrons as well as 3d electrons. On ionisation, the 4s electrons are generally removed first, even though introductory Aufbau filling lists 4s before 3d. The occupancies and energy ordering change as the species changes.

Core explanation

Fe is [Ar]3d⁶4s². Fe²⁺ has lost two electrons and is [Ar]3d⁶, while Fe³⁺ loses one additional 3d electron and is [Ar]3d⁵. The check is total electron count: Fe has 26 electrons; Fe²⁺ has 24 = 18 + 6, and Fe³⁺ has 23 = 18 + 5. Writing Fe²⁺ as [Ar]3d⁴4s² would preserve 4s at the expense of d and contradict the standard ionic configurations used for this chemistry.

Mn is [Ar]3d⁵4s², so Mn²⁺ is [Ar]3d⁵. The half-filled d⁵ ion can have unpaired electrons, but exact magnetic behaviour in a complex depends on ligand-field splitting and other interactions. Co is [Ar]3d⁷4s², giving Co²⁺ d⁷ and Co³⁺ d⁶. A higher oxidation state does not always mean “all d electrons gone”; count the electrons removed from the actual atom configuration. For Cr, the neutral exception [Ar]3d⁵4s¹ means Cr²⁺ is [Ar]3d⁴ after removing one 4s and one 3d electron, while Cr³⁺ is [Ar]3d³.

Cu illustrates another exception. Neutral Cu is [Ar]3d¹⁰4s¹. Cu⁺ loses its one 4s electron and is d¹⁰; Cu²⁺ loses one further 3d electron and is d⁹. These distinct counts help explain why Cu⁺ and Cu²⁺ often have different magnetic behaviour and coordination chemistry. Zn is [Ar]3d¹⁰4s² and Zn²⁺ remains d¹⁰, a filled d case. Sc is [Ar]3d¹4s² and Sc³⁺ is d⁰.

Why does the filling-versus-removal rule differ? In a neutral atom, relative energies of 4s and 3d are close and depend on electron occupation. Once the d subshell is occupied and the atom is ionised, 4s electrons are more spatially extended and are removed preferentially in the standard first-row cation accounting. “4s always lower than 3d” is not a permanent energy law across atoms and ions. Electron configurations are state-specific descriptions, not an irreversible history of electron arrival.

In a coordination complex, a metal oxidation state is assigned formally by ligand charges and overall complex charge. For [Fe(H₂O)₆]²⁺, water is neutral, so Fe is formally +2 and d⁶. For [Fe(CN)₆]³⁻, six CN⁻ ligands total −6; the complex is −3, so Fe is +3 and formally d⁵. The ligand field then splits the d orbitals and influences spin and colour. Do not infer the full orbital occupancy from the d count alone.

Oxidation state and actual metal partial charge differ when metal–ligand bonding is covalent. Formal electron counting remains useful and consistent for introductory chemistry. It also helps balance redox reactions: Fe²⁺ → Fe³⁺ + e⁻ removes one electron and reduces formal d count from d⁶ to d⁵. The accounting must match net charge and electron conservation.

Step-by-step reasoning

1. Write the neutral atom's correct configuration, including Cr or Cu exceptions. 2. Subtract electrons equal to the positive ionic charge. 3. Remove outer 4s before 3d in the simple first-row sequence. 4. Check the total electron count against atomic number minus charge. 5. For complexes, assign oxidation state from ligand charges before giving dⁿ.

Visual explanation

Draw three tracks: Fe d⁶s² → Fe²⁺ d⁶ → Fe³⁺ d⁵; Cr d⁵s¹ → Cr²⁺ d⁴ → Cr³⁺ d³; Cu d¹⁰s¹ → Cu⁺ d¹⁰ → Cu²⁺ d⁹. Highlight the first electron leaving from s in each track.

Real-world analogy

Seats filled first in a theatre are not necessarily the last seats people leave when the room empties. Changes in crowding and exits alter the best route; neutral-atom filling order and ion electron removal likewise describe different states.

Real-world example

Iron(II) and iron(III) compounds are used in redox and coordination chemistry. Their d⁶ and d⁵ counts set up later magnetic and colour analysis, although ligand identity determines the detailed spectrum.

Why?

Why is Cr²⁺ d⁴ rather than d⁵? Neutral Cr has only one 4s electron. Forming +2 removes that 4s electron and then one 3d electron, leaving 3d⁴.

Common misconception

“To form Fe²⁺, remove two 3d electrons because 3d was written last.” Printed term order is not the ionisation sequence. The standard Fe²⁺ configuration is 3d⁶ with 4s empty.

Worked example

Find the metal oxidation state and d count in [Co(NH₃)₆]³⁺. All six NH₃ ligands are neutral, so Co is formally +3. Neutral Co is [Ar]3d⁷4s²; remove two 4s and one 3d electron to give Co³⁺ d⁶. This count is the starting point for ligand-field analysis, not a completed prediction of spin state.

Quick check

1. What configuration follows from forming Cu⁺ from neutral [Ar]3d¹⁰4s¹ copper? Answer: [Ar]3d¹⁰ after loss of the 4s electron.

Exam focus

Show neutral configuration first, then remove electrons explicitly and verify totals. Assign ligand charges before a complex d count. Distinguish formal dⁿ from the detailed field-split occupancy.

Advanced insight

The simple “4s first out” rule works well for elementary first-row ions, but real many-electron atomic states and highly charged ions can require spectroscopic configuration data. Orbital energies shift with occupancy rather than obeying one fixed universal ladder.

Summary

First-row metal cations generally lose 4s electrons before 3d. Neutral exceptions matter: Cr²⁺ is d⁴ and Cu⁺ d¹⁰. In complexes, formal oxidation state from ligand charge gives a useful d count that precedes spin and colour analysis.

Practice questions

1. What is Fe³⁺'s simple configuration beyond [Ar]? Answer: 3d⁵. 2. What is Mn²⁺'s simple d count? Answer: d⁵. 3. Why is Cu²⁺ d⁹ rather than d⁸? Answer: Neutral Cu is 3d¹⁰4s¹; remove 4s then one 3d electron. 4. What d count does Co³⁺ have in [Co(NH₃)₆]³⁺? Answer: d⁶, because NH₃ is neutral and Co is formally +3.