Electron Configurations of Transition Metal Ions

Introduction

Every crystal-field diagram starts with a d-electron count. A wrong count makes the subsequent spin state, stabilisation energy and magnetic moment wrong even if the diagram is drawn neatly. The chief trap is remembering neutral transition-metal configurations, then removing 3d electrons before 4s electrons when forming positive ions. For the first transition series, the 4s electrons are removed first.

Core explanation

The formal d count of a complex follows from the metal identity and oxidation state. Assign the usual charges to ligands: H₂O, NH₃, CO and ethylenediamine are neutral; Cl⁻, CN⁻ and OH⁻ each carry −1. The oxidation state x satisfies x plus the sum of ligand charges equals the complex-ion charge. Then write the electron configuration of the resulting metal ion, removing outer ns electrons before (n−1)d electrons. A useful shortcut for a d-block metal is group number minus formal oxidation state, although special cases and later transition series deserve direct configuration checks.

Neutral iron is commonly written [Ar]3d⁶4s². Fe²⁺ has lost the two 4s electrons and is [Ar]3d⁶, or d⁶. Fe³⁺ has lost one further 3d electron and is d⁵. The observed energy ordering of 4s and 3d depends on occupancy and ionisation; the simple neutral-atom Aufbau filling order is not an instruction for which electron leaves first. For [Fe(CN)₆]⁴⁻, six cyanide ligands contribute −6, so Fe is +2 and d⁶ regardless of whether the resulting complex is low spin.

For [Co(NH₃)₆]³⁺, ammonia is neutral, making Co +3. Cobalt is group 9, so Co³⁺ is d⁶. For [CuCl₄]²⁻, four chloride ions total −4 and the ion is −2; Cu must be +2. Neutral Cu is [Ar]3d¹⁰4s¹ and Cu²⁺ is d⁹. The apparent irregularity of neutral Cu is another reason to check rather than blindly write 4s² for every metal.

Formal d count is distinct from the detailed electron density of a covalent metal–ligand bond. A ligand can donate a pair to a metal and share electron density without changing the formal oxidation-state calculation used in elementary CFT. A change in ligand charge or complex charge may change oxidation state; a change in ligand field strength alone does not. The same Fe²⁺ d⁶ can be high spin with water and low spin with cyanide.

Some ligands are noninnocent, meaning their formal oxidation level can be ambiguous because ligand-centred redox chemistry occurs. For routine examination examples, the question normally supplies conventional ligand charges. For advanced real complexes, spectroscopic and structural evidence may be needed to decide whether a redox change occurred on metal or ligand.

Step-by-step reasoning

Write the complex charge outside brackets, list ligand charges explicitly, and solve x+Σq(ligands)=q(complex). Identify the metal’s neutral configuration or group number. Remove ns electrons first, then (n−1)d electrons to obtain dⁿ. Only after that choose octahedral, tetrahedral or square-planar splitting and distribute the d electrons.

Visual explanation

Draw a two-column bookkeeping table: ligand charges on the left and metal charge on the right. Under it show Fe: [Ar]3d⁶4s² → Fe²⁺: [Ar]3d⁶ → Fe³⁺: [Ar]3d⁵. Cross out 4s before crossing out 3d.

Real-world analogy

Imagine a building that was filled floor by floor but whose occupants leave by the currently easiest exit. The order of arrival does not force the reverse order of departure. Neutral-atom filling order and ionisation order likewise answer different questions, especially for 4s and 3d electrons.

Real-world example

Blue aqueous Cu²⁺ is usually represented by six water ligands around a d⁹ copper ion, often with a distorted geometry. Replacing water ligands with chloride can alter colour and geometry, but copper remains d⁹ if it remains in oxidation state +2.

Why?

Why is a ligand donation not counted as an extra d electron? The donated pair belongs to a metal–ligand bonding description, whereas dⁿ is a formal count for the metal ion before ligand-field orbital filling. Mixing the two conventions would double-count electrons and spoil magnetic predictions.

Common misconception

“Fe²⁺ is d⁴ because two electrons must be taken from 3d⁶.” The first two electrons lost from neutral iron are 4s electrons. Fe²⁺ is d⁶; Fe³⁺ is d⁵.

Worked example

Find the d count of [Cr(H₂O)₄Cl₂]⁺. Water contributes zero; two chlorides contribute −2. Thus x−2=+1 and chromium is +3. Neutral chromium is [Ar]3d⁵4s¹; remove 4s¹ and two 3d electrons to get [Ar]3d³. Therefore the metal is d³. The four waters do not add eight to the formal metal d count.

Quick check

1. What is the d count of [Co(NH₃)₆]³⁺? Answer: NH₃ is neutral, so Co is +3; group 9 minus 3 gives d⁶. 2. Which electrons leave neutral Fe first when Fe²⁺ forms? Answer: The two 4s electrons leave before any 3d electron.

Exam focus

Show the oxidation-state equation before announcing dⁿ. Remember unusual neutral Cr and Cu configurations, and never equate ligand-pair donation with an increase in formal metal d count.

Advanced insight

The formal dⁿ label is a powerful classification but not a direct measurement of precisely n electrons spatially confined to metal d orbitals. Covalency delocalises electron density, and noninnocent ligands can make alternative oxidation-state assignments chemically plausible.

Summary

Determine the complex charge and ligand charges, solve for the metal oxidation state, then remove ns electrons before (n−1)d electrons. This yields the formal d count needed for splitting diagrams; spin state comes later.

Practice questions

1. Find the d count of [Mn(CN)₆]⁴⁻. Answer: Six CN⁻ ligands total −6; Mn is +2 to give −4 overall. Manganese is group 7, hence Mn²⁺ is d⁵. 2. A compound changes from [Fe(H₂O)₆]²⁺ to [Fe(CN)₆]⁴⁻. Does Fe change formal oxidation state? Answer: No. Water is neutral, giving Fe²⁺ in the first ion. Six cyanides total −6 and the second ion is −4, again giving Fe²⁺. Both are formally d⁶. 3. Why can Cu²⁺ not be represented as 3d¹⁰4s⁻¹? Answer: Orbital occupations cannot be negative. Neutral Cu has 3d¹⁰4s¹; losing the 4s electron and one 3d electron gives Cu²⁺ as 3d⁹.