While filling up of electrons in the atomic orbitals, the 4s orbital is filled before the 3d orbital but rever

NCERT Exemplar · Q40

Q. While filling up of electrons in the atomic orbitals, the 4s orbital is filled before the 3d orbital but reverse happens during the ionisation of the atom. Explain why?

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Concept understanding — Transition Element Definition

Transition Elements: From Intuition to Definition

Imagine you're building a house with bricks. Most bricks are identical — you stack them in neat rows. But some bricks are special: they have extra slots on their sides where you can attach hooks, magnets, or other bricks. These special bricks can change the shape of the wall, conduct electricity, or even change colour when you heat them.

In the periodic table, transition elements are those special bricks. They are the metals that sit in the middle block — groups 3 to 12 — and they have a unique ability: they can use their inner electrons (not just the outermost ones) to form bonds, change oxidation states, and create colourful compounds.

The Intuition: Why "Transition"?

The word "transition" comes from the idea that these elements form a bridge between the highly reactive metals on the left (like sodium, magnesium) and the less reactive metals / non-metals on the right (like aluminium, silicon). Their properties are not extreme — they are in-between.

But the real reason they are special lies in their electron configuration.

The Precise Definition (IUPAC)

A transition element is an element whose atom has an incomplete d sub-shell, or which can give rise to cations with an incomplete d sub-shell.

Let's unpack that.

1. The "d" sub-shell

Electrons are arranged in shells (K, L, M, N...) and sub-shells (s, p, d, f). The d sub-shell can hold a maximum of 10 electrons. In transition elements, the d sub-shell is being filled — but not completely.

For example, consider Iron (Fe):

Atomic number 26
Electron configuration: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$
The 3d sub-shell has 6 electrons — it is incomplete (it can hold 10).

So iron is a transition element.

2. The "or" part — cations matter

Some elements have a complete d sub-shell in their neutral atom, but when they lose electrons to form positive ions (cations), the d sub-shell becomes incomplete.

Example: Zinc (Zn)

Neutral Zn: $[Ar] 3d^{10} 4s^2$ — the 3d sub-shell is full (10 electrons).
But Zn commonly forms $Zn^{2+}$ : $[Ar] 3d^{10}$ — still full.
So zinc is NOT a transition element by the IUPAC definition.

Example: Copper (Cu)

Neutral Cu: $[Ar] 3d^{10} 4s^1$ — 3d is full.
But $Cu^{2+}$ : $[Ar] 3d^9$ — now the 3d sub-shell is incomplete.
So copper IS a transition element.

Watch out

A common mistake: thinking that all elements in the d-block (groups 3–12) are transition elements. They are not. Zinc, cadmium, and mercury are d-block elements but NOT transition elements because their common cations have a full d sub-shell.

The "d-block" vs "Transition Elements"

d-block elements	Transition elements
Groups 3 to 12	Groups 3 to 11 (excluding Zn, Cd, Hg)
All have d electrons	Must have incomplete d sub-shell in atom or common cation

Why this formula?

Transition Element Definition: The "Why" Behind the Definition

The Core Definition

A transition element (IUPAC definition) is an element whose atom has an incomplete d-subshell in its ground state or can form stable ions with an incomplete d-subshell.

Key exam point: This definition covers both the neutral atom and its common ions.

Why This Definition? The Reasoning

1. The d-orbital filling pattern

In the periodic table, transition elements belong to the d-block (Groups 3–12). As we move across a period, electrons fill the $(n-1)d$ orbitals after the $ns$ orbital.

For example, in Period 4:

Scandium (Sc): $[Ar]\,3d^1\,4s^2$ — has one d-electron → transition element
Zinc (Zn): $[Ar]\,3d^{10}\,4s^2$ — d-subshell is full → not a transition element

2. The "incomplete d-subshell" condition

The definition focuses on incompleteness because:

A full d-subshell ( $d^{10}$ ) is exceptionally stable (like a noble gas configuration for d-orbitals)
Elements with $d^{10}$ configurations do not show the characteristic properties of transition metals (variable oxidation states, coloured compounds, catalytic activity, paramagnetism)

3. Why include ions?

Consider Zinc (Zn):

Ground state: $[Ar]\,3d^{10}\,4s^2$ — d-subshell is full → not a transition element
Common ion: $\text{Zn}^{2+}$ : $[Ar]\,3d^{10}$ — still full → still not a transition element

Now consider Copper (Cu):

Ground state: $[Ar]\,3d^{10}\,4s^1$ — d-subshell is full → by atom definition alone, not a transition element
But $\text{Cu}^{2+}$ : $[Ar]\,3d^9$ — incomplete d-subshell → is a transition element

Therefore: The definition must include ions to correctly classify elements like Cu, which form stable ions with incomplete d-subshells.

The "Formula" — A Decision Tree

The definition can be expressed as a logical condition:

\text{Transition element} \iff \left( \text{Atom has } d^{1-9} \right) \lor \left( \text{Stable ion has } d^{1-9} \right)

Where:

$d^{1-9}$ means incomplete d-subshell (1 to 9 electrons)
$d^{0}$ or $d^{10}$ means complete (empty or full) → not a transition element

Common Exam Exceptions …

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The 4s orbital has lower energy than 3d for neutral atoms (so it fills first), but during ionisation, electrons are removed from the orbital with the highest principal quantum number $n$ (the outermost shell), which is 4s — not because 4s is higher in energy than 3d at that stage, but because it is farther from the nucleus and less tightly bound once the 3d is occupied.

Why the filling order and removal order are opposite

This is one of the most common points of confusion in transition element chemistry. The key lies in understanding that orbital energy is not fixed — it changes as the nuclear charge and electron configuration change.

1. The energy crossover for neutral atoms

For a neutral atom, the 4s orbital has a lower energy than the 3d orbital. This happens because:

The 4s electron has a small but significant probability of being found very close to the nucleus (it has a "penetrating" shape).
This penetration means the 4s electron experiences a higher effective nuclear charge ( $Z_{\text{eff}}$ ) than a 3d electron, which is more shielded by inner electrons.
Lower energy means it fills first — so in the Aufbau principle, we write $4s$ before $3d$ .

Z \leq 20

2. What happens once 3d starts filling

As electrons enter the 3d subshell, something crucial changes. The 3d orbitals are more compact and lie closer to the nucleus than the 4s orbital. Once the 3d subshell is partially or fully occupied:

The 3d electrons are inside the 4s orbital (radially).
The 4s electron now experiences greater shielding from the 3d electrons.
The energy of the 4s orbital rises relative to the 3d.

For atoms of transition elements (e.g., Sc, Ti, Fe), the 3d orbital actually becomes lower in energy than the 4s. But the 4s was already filled first because the energy ordering was different at the start.

Watch out

A common mistake is to think that 4s is always lower in energy than 3d. This is only true for neutral atoms with empty or nearly empty 3d subshells. Once 3d has electrons, the ordering reverses.

3. Ionisation: the rule of outermost electrons

When an atom loses an electron (ionisation), the electron removed is not necessarily the one that was filled last. The rule is:

Electrons are removed from the orbital with the highest principal quantum number $n$ — i.e., the outermost shell.

For a transition metal atom like iron ( $[\text{Ar}]\,3d^6 4s^2$ ):

The 4s orbital has $n=4$ , the 3d has $n=3$ .
The 4s electrons are farther from the nucleus on average.
Even though 4s was filled first, it is more loosely held because it is in a higher shell.

So during ionisation, the 4s electrons leave first — not because they are higher in energy (they may actually be lower in energy than 3d for some ions), but because they are farther out and experience less effective nuclear attraction.

Tip

Think of it like a building: the 4s is the top floor (higher $n$ ), the 3d is a lower floor. Even if the top floor was built first, it's still the easiest to leave from.

4. Experimental evidence

This is not just theory — it is confirmed by:

Ionisation energies: The first ionisation energy of transition metals always removes a 4s electron. …

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Method: Aufbau Principle vs. Ionisation – The (n + ℓ) Rule & Stability of Half-Filled/Full Subshells

Step 1 – Recall the Aufbau Principle for Ground State

Electrons fill orbitals in order of increasing (n + ℓ).
If two orbitals have the same (n + ℓ), the one with lower n fills first.

Orbital	n	ℓ	n + ℓ
4s	4	0	4
3d	3	2	5

Since 4s has lower (n + ℓ) = 4, it fills before 3d.

Step 2 – Why Ionisation Reverses the Order

Once 4s is filled, the 3d orbital has lower energy than 4s (due to greater nuclear penetration and shielding effects).
During ionisation, electrons are removed from the highest energy orbital first — which is now 4s, not 3d.

Key point: The 4s orbital is higher in energy than 3d after both are occupied.

Step 3 – The Stability Factor

For transition metals (e.g., Sc, Ti, V), the 3d electrons are closer to the nucleus and more tightly bound.
The 4s electron is more loosely held, so it is lost first during ionisation. …

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Here are the common mistakes students make on this question, along with how to avoid each one.

Mistake 1: Confusing "Filling" with "Stability"

The Error: Students often say "4s is filled first because it is lower in energy than 3d." Then, when asked why 4s electrons are lost first during ionization, they incorrectly say "because 4s is higher in energy now." This sounds contradictory and leads to confusion.
Why it’s wrong: The energy order is not static. The relative energies of 4s and 3d change depending on whether the atom is neutral or ionized.
How to Avoid: Understand the n+l rule (Aufbau principle) for neutral atoms:
- For 4s: $n=4, l=0 \Rightarrow n+l = 4$
- For 3d: $n=3, l=2 \Rightarrow n+l = 5$
- Lower $n+l$ means lower energy. So in a neutral atom, 4s fills first.
- However, once electrons are present in 3d, the 3d orbital experiences greater nuclear attraction (poor shielding by 4s). During ionization, the 4s electron is farther from the nucleus and is shielded by the 3d electrons, making it easier to remove.

Mistake 2: Forgetting the "Penetration" Effect

The Error: Students memorize "4s is filled first, lost first" without understanding why the energy order flips.
Why it’s wrong: The 4s orbital has a small region of high electron density very close to the nucleus (it penetrates the inner shells). This makes it lower in energy than 3d in the neutral atom. But once 3d is occupied, the 3d electrons are closer to the nucleus on average and are less shielded.
How to Avoid: Visualize the radial probability distribution. The 4s orbital has a "spike" near the nucleus, giving it lower energy initially. After filling, the 3d orbital contracts and becomes more stable.

Mistake 3: Writing the Wrong Electronic Configuration

The Error: Students write the configuration of $\ce{Fe}$ as $[\ce{Ar}] 3d^6 4s^2$ (correct) but then write $\ce{Fe^{2+}}$ as $[\ce{Ar}] 3d^4 4s^2$ (incorrect).
Why it’s wrong: During ionization, electrons are removed from the outermost shell (highest $n$ value), not from the orbital that was filled last. For transition elements, the 4s orbital has $n=4$ , so it is removed first.
How to Avoid: Always remember the rule for writing ions:
- Neutral atom: Write according to Aufbau (4s before 3d).
- Ion: Remove electrons from the orbital with the highest principal quantum number ( $n$ ) first.
- For $\ce{Fe}$ : $[\ce{Ar}] 3d^6 4s^2$ . Remove two 4s electrons → $\ce{Fe^{2+}} = [\ce{Ar}] 3d^6$ .

Mistake 4: Mixing Up "Transition Element" Definition

The Error: Students think that because 4s is filled first, the valence electrons are in 4s, so transition elements must have incomplete 4s subshell. …

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Q.While filling up of electrons in the atomic orbitals, the 4s orbital is filled before the 3d orbital but reverse happens during the ionisation of the atom. Explain why?

Why the filling order and removal order are opposite

1. The energy crossover for neutral atoms

2. What happens once 3d starts filling

3. Ionisation: the rule of outermost electrons

4. Experimental evidence

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