Draw An Outer Electron Box Diagram For A Cation

You’re staring at a periodic table, a pencil in hand, and a problem set asking for the outer electron box diagram of a cation. Plus, maybe it’s Sn⁴⁺. The neutral atom configuration is burned into your memory, but the ion? Which means maybe it’s Fe³⁺. That’s where the points get lost.

I’ve graded enough of these to know the pattern. That said, students write the configuration for the neutral atom, draw the boxes perfectly, and forget the single most important rule: **electrons leave the highest principal energy level first. Worth adding: ** Not the highest energy sublevel. The highest n value.

Let’s walk through how to do this right, why the exceptions trip everyone up, and how to check your work before you hand it in.

What Is an Outer Electron Box Diagram

An outer electron box diagram — sometimes called an orbital diagram — shows the distribution of electrons in the valence orbitals of an atom or ion. Each box represents one orbital. Arrows represent electrons: up for +½ spin, down for –½.

For a cation, you’re drawing the diagram after* electrons have been removed.

The “outer” part matters. You don’t draw the 1s, 2s, 2p core orbitals for a transition metal cation. You start at the highest occupied principal energy level (the valence shell) and include any partially filled sublevels from the level just below it if they’re part of the valence chemistry — typically (n-1)d for transition metals.

So for a first-row transition metal cation, you’re usually drawing the 4s and 3d boxes. For a main-group cation like Al³⁺, you’re drawing the 3s and 3p boxes (which end up empty).

The notation rules

One box per orbital. s = 1 box, p = 3 boxes, d = 5 boxes, f = 7 boxes.
Fill each box singly before pairing (Hund’s rule).
Pair with opposite spins (Pauli exclusion principle).
Label the sublevel (3d, 4s, etc.) to the left of the boxes.

That’s the mechanics. The chemistry — which* electrons go missing — is where it gets interesting.

Why It Matters

You might wonder why professors obsess over these diagrams. It’s not busywork.

The outer electron configuration dictates:

Magnetic behavior — paramagnetic vs. diamagnetic. Unpaired electrons = paramagnetic. This shows up in lab data, not just theory.
Color and spectroscopy — d-d transitions in transition metal complexes depend on how those d orbitals are split and occupied.
Reactivity and oxidation states — the stability of Fe²⁺ vs. Fe³⁺, or why Cu⁺ exists but Cu²⁺ is more common in aqueous solution, traces back to the electron count in those boxes.
Coordination geometry — crystal field theory and ligand field theory start with the d-electron count.

If you draw the neutral atom’s diagram for a cation question, you’re not just losing a point. You’re showing you don’t understand how ionization actually works.

How to Draw It: Step by Step

Let’s use a concrete example: Fe³⁺. This leads to iron, atomic number 26. Neutral configuration: [Ar] 4s² 3d⁶.

Step 1: Write the neutral ground-state configuration

Don’t skip this. Write it out. Noble gas core plus valence.

Fe: [Ar] 4s² 3d⁶

Step 2: Identify the highest principal quantum number (n)

Here, n = 4 (the 4s orbital). The 3d is n = 3.

This is the rule: Electrons are removed from the highest n first. Always.

So the two 4s electrons go before a single 3d electron.

Step 3: Remove electrons for the charge

Fe³⁺ means three electrons removed.

First two: from 4s² → 4s⁰
Third: from 3d⁶ → 3d⁵

Final configuration: [Ar] 3d⁵

Step 4: Draw the boxes for the remaining* valence orbitals

Since 4s is empty, you could* draw an empty 4s box. Most instructors don’t require it, but it doesn’t hurt. The valence orbitals now are just 3d.

Draw five boxes for 3d. Five electrons remain. Apply Hund’s rule: one electron in each box, all spins parallel (all up arrows).

3d:  ↑  ↑  ↑  ↑  ↑
      □  □  □  □  □

That’s it. Consider this: five unpaired electrons. Paramagnetic. High-spin d⁵ — the most stable d-count for a first-row transition metal ion.

Another example: Cu²⁺

Copper, Z = 29. Neutral: [Ar] 4s¹ 3d¹⁰ (exception: full d, half-filled s).

Remove two electrons. Highest n = 4. The 4s¹ electron goes first. Then one 3d electron.

Result: [Ar] 3d⁹

Draw five 3d boxes. That said, nine electrons. Four boxes full (up/down), one box with a single up arrow.

3d:  ↑↓  ↑↓  ↑↓  ↑↓  ↑
      □   □   □   □   □

One unpaired electron. Paramagnetic.

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Main-group example: Sn⁴⁺

Tin, Z = 50. Neutral: [Kr] 5s² 4d¹⁰ 5p².

Highest n = 5. Remove the 5s² and 5p² electrons — all four from n = 5.

Result: [Kr] 4d¹⁰

The outer electron box diagram shows a filled 4d sublevel (ten electrons, five paired boxes) and empty* 5s and 5p boxes if you choose to draw them. Diamagnetic.

Common Mistakes / What Most People Get Wrong

1. Removing from the highest energy* sublevel instead of highest n

This is the big one. In the neutral atom, 4s fills before 3d because 4s is lower in energy for the neutral atom*. But once the atom is ionized, the energy ordering flips. The 3d orbitals drop below 4s. Ionization removes the outermost electrons — highest n — which are the 4s electrons.

I’ve seen students write Fe³⁺ as [Ar] 4s² 3d³. Also, wrong. Worth adding: the 4s electrons are physically farther out. On top of that, they kept the 4s² and pulled three from 3d. They leave first.

2. Forgetting the exceptions (Cr, Cu, Mo, Ag, Au…)

Neutral Cr is [Ar] 4s¹ 3d⁵, not 4s² 3d⁴. Neutral Cu is [Ar] 4s¹ 3d¹⁰.

If you start with the wrong neutral configuration, your cation is wrong before you even draw a box.

Cr³⁺? Start with Cr: 4s¹ 3d⁵. Remove three electrons. In practice, the 4s¹ goes. Highest n = 4. Then two from 3d⁵ → 3d³.

Result: [Ar] 3d³. Three unpaired electrons.

If you started with the “textbook” 4s² 3d⁴, you’d remove 4s² and

If you began with the “textbook” configuration 4s² 3d⁴, the first two electrons are taken from the 4s subshell, leaving it empty, and the third electron must be removed from the 3d set. Day to day, the resulting ion is therefore [Ar] 3d³. In a box diagram the five 3d orbitals are each represented by a square; three of those squares contain a single up‑arrow, all with the same spin direction. The ion possesses three unpaired electrons, making it paramagnetic and illustrating a high‑spin d³ configuration, which is the most stable arrangement for a first‑row transition‑metal cation with three d electrons.

Additional pitfalls to watch for

Assuming the 4s orbital stays populated – After ionization the 4s level drops in energy below the 3d set, but it remains the outermost (highest n) subshell. As a result, any electrons that were originally in 4s are the first to leave, regardless of the altered energy ordering. Keeping a 4s electron in a cation is a frequent slip.
Misidentifying the “outermost” shell – For main‑group elements the highest principal quantum number (n) is the guide. In transition‑metal ions, however, the (n − 1)d subshell can become the effective valence shell once the ns electrons have been removed. Forgetting this shift leads to incorrect electron counts, as seen when a student writes Fe³⁺ as [Ar] 4s² 3d⁵ instead of the correct [Ar] 3d⁵.
Overlooking subshell exceptions – Elements such as Cr, Cu, Mo, Ag, and Au possess atypical neutral configurations (e.g., Cr = 4s¹ 3d⁵, Cu = 4s¹ 3d¹⁰). Starting from the wrong neutral arrangement propagates the error directly into the cation’s electron count. Always verify the ground‑state configuration before beginning the removal process.
Incorrect application of Hund’s rule after electron loss – Once the appropriate electrons have been removed, the remaining electrons must be distributed according to Hund’s rule within the partially filled subshell. A common error is to pair electrons prematurely or to place opposite‑spin electrons in the same orbital when the subshell is not yet half‑filled.
Neglecting to consider the total electron count – After removal, the sum of electrons in the diagram must equal the charge of the ion. As an example, a Cu²⁺ ion (neutral Cu = 29 electrons) loses two electrons, leaving 27 electrons in the diagram; the 3d⁹ configuration accounts for 9 + 18 = 27, confirming consistency.

Another illustrative case: Ti³⁺

Neutral titanium (Z = 22) has the configuration [Ar] 4s² 3d². Removing three electrons means taking both 4s electrons first and then one 3d electron, yielding [Ar] 3d¹. The box diagram for 3d shows a single up‑arrow in one orbital, indicating one unpaired electron and a paramagnetic Ti³⁺ ion.

Summary

The essential workflow for constructing an orbital box diagram of a transition‑metal cation is:

Write the correct ground‑state electron configuration, observing all subshell exceptions.
Identify the subshell(s) with the highest principal quantum number; these are the outermost electrons and are removed first.
Subtract the required number of electrons, starting with the outermost subshell and proceeding inward.
Sketch the remaining valence orbitals, placing electrons according to Hund’s rule and the Pauli exclusion principle.
Verify that the total electron count matches the ion’s charge and interpret the magnetic behavior from the number of unpaired electrons.

By adhering to these steps, the diagram accurately reflects the electronic structure of any transition‑metal cation, clarifies its magnetic properties, and avoids the most common sources of error.