Which of the following options does not represent ground state electronic configuration of an atom? (i) 1s2 2s

NCERT Exemplar · Q2

Q. Which of the following options does not represent ground state electronic configuration of an atom?

(i) 1s2 2s2 2p6 3s2 3p6 3d8 4s2

(ii) 1s2 2s2 2p6 3s2 3p6 3d9 4s2

(iii) 1s2 2s2 2p6 3s2 3p6 3d10 4s1

(iv) 1s2 2s2 2p6 3s2 3p6 3d5 4s1

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Concept understanding — Stability of Oxidation States

Stability of Oxidation States – From Intuition to Precision

Imagine you're holding a ball on a hill. If you place it exactly at the top, it's balanced — but the slightest push sends it rolling down. That's an unstable position. If you place it in a small dip on the hillside, it stays put even if nudged — that's stable. Oxidation states work the same way: some are like the hilltop (easily changed), others like the dip (hard to change).

The Core Intuition

An oxidation state is just a number we assign to an atom to track how many electrons it has gained or lost compared to its neutral state. But atoms don't "want" to stay in arbitrary oxidation states — they want to reach a configuration that minimises their energy.

Stability here means: how reluctant is that oxidation state to change under normal conditions? A stable oxidation state resists being oxidised further or reduced further. An unstable one readily changes into something else.

The Precise Statement

Stability of an oxidation state refers to the tendency of an element to maintain that particular oxidation state under given conditions (temperature, pH, presence of other reagents). A stable oxidation state is one that does not easily undergo redox reactions — it is neither easily oxidised nor easily reduced.

This depends on three key factors:

Electronic configuration – Half-filled and fully-filled d or f subshells confer extra stability (e.g., $\text{Fe}^{3+}$ with $d^5$ is more stable than $\text{Fe}^{2+}$ with $d^6$ in some contexts).
Inert pair effect – Heavier p-block elements (like Tl, Pb, Bi) show lower oxidation states (e.g., +1 for Tl) as more stable than higher ones (+3 for Tl), because the s-electrons become reluctant to participate.
Disproportionation tendency – Some oxidation states are unstable because they spontaneously convert into two other states (e.g., $\text{Cu}^+$ in aqueous solution gives $\text{Cu}^{2+}$ and $\text{Cu}$ ).

Important

Stability is relative — it depends on the environment. $\text{Mn}^{2+}$ is stable in acidic solution but easily oxidised in alkaline medium. Always specify conditions when discussing stability.

Examples That Make It Concrete

Transition metals – $\text{Cr}^{3+}$ ( $d^3$ ) and $\text{Mn}^{2+}$ ( $d^5$ ) are exceptionally stable because half-filled/half-filled-like configurations have low energy. $\text{Cr}^{2+}$ ( $d^4$ ) is easily oxidised to $\text{Cr}^{3+}$ — it's unstable.

p-block elements – $\text{Pb}^{2+}$ is stable, $\text{Pb}^{4+}$ is a strong oxidising agent (unstable). $\text{Sn}^{2+}$ is a reducing agent (easily oxidised to $\text{Sn}^{4+}$ ), so $\text{Sn}^{4+}$ is more stable for tin.

Common pattern – For most elements, the most common oxidation state is the most stable one under standard conditions. But "most common" isn't always "most stable" — e.g., $\text{Fe}^{3+}$ is common but $\text{Fe}^{2+}$ is more stable in acidic solution.

Watch out

Do not confuse "stability" with "occurrence". $\text{Mn}^{7+}$ (as $\text{MnO}_4^-$ ) is common in the lab but is a powerful oxidising agent — it is not stable in the sense of resisting change. It readily accepts electrons.

How to Think About It in Exams

When asked "Explain the stability of oxidation states of [element]", follow this mental checklist:

Write the electronic configuration of the atom.
Write configurations for each possible oxidation state.
Look for half-filled, fully-filled, or inert pair effects.
Check if the state can disproportionate (common for +1 states of Cu, Au, and +3 states of Mn).
Mention the medium (acidic/alkaline) if relevant.

Tip

For d-block elements, remember: $d^0$ , $d^5$ , and $d^{10}$ are especially stable. For p-block, the inert pair effect makes lower oxidation states more stable as you go down the group.

The Bottom Line

Stability of an oxidation state is a measure of how strongly an atom holds onto that oxidation number — how hard it is to push it up or down. It's determined by electronic structure, the element's position in the periodic table, and the chemical environment. Master this, and you'll predict redox behaviour without memorising every reaction.

Stability of oxidation states among transition and inner-transition elements is discussed in the NCERT/CBSE Class 12 Chemistry chapter on d- and f-Block Elements, and ‘stability of oxidation states in transition elements’ is a frequently searched important-question topic for board exams, JEE Main and NEET. Predicting which oxidation state is most stable is a reasoning skill regularly tested in competitive-exam inorganic chemistry MCQs.

Why this formula?

Stability of Oxidation States: Why It Works

This concept explains why certain oxidation states of an element are more stable than others — and why some states are never observed at all.

The Core Idea: Energy Minimisation

An oxidation state is stable when the total energy of the system is at a minimum. This depends on three competing factors:

Ionisation energy (energy needed to remove electrons)
Lattice energy (for ionic compounds) or bond energy (for covalent compounds)
Electronic configuration (half-filled / fully-filled subshells)

There is no single formula for stability — instead, we use trends and principles that act as "formulae" for prediction.

Key Principle 1: Inert Pair Effect (for p-block elements)

Why it holds:

For heavier elements (e.g., Tl, Pb, Bi), the 6s² electrons are held very tightly due to poor shielding and relativistic effects. They resist removal.

Result: Lower oxidation state (e.g., +1 for Tl, +2 for Pb) becomes more stable than the higher state (+3, +4).
Example: $\ce{Tl^3+}$ is a strong oxidising agent — it readily gains two electrons to become $\ce{Tl^+}$ .

Derivation logic:

The energy cost to remove the 6s² electrons is greater than the energy gained by forming additional bonds or lattice. So the system stays in the lower state.

Key Principle 2: Half-Filled / Fully-Filled Subshell Stability

Why it holds:

A half-filled ( $d^5$ , $f^7$ ) or fully-filled ( $d^{10}$ , $f^{14}$ ) subshell has extra exchange energy and symmetry — making it unusually stable.

Example: $\ce{Mn^{2+}}$ ( $d^5$ ) is more stable than $\ce{Mn^{3+}}$ ( $d^4$ ). $\ce{Fe^{3+}}$ ( $d^5$ ) is more stable than $\ce{Fe^{2+}}$ ( $d^6$ ).

Derivation logic:

The exchange energy (Hund's rule) is maximum for half-filled configurations. Removing an electron from a half-filled shell costs extra energy — so the half-filled state is favoured.

Key Principle 3: Lattice Energy / Hydration Energy Compensation

For transition metals, stability of a particular oxidation state in aqueous solution depends on:

\Delta G^\circ = \Delta H^\circ_{\text{hydration}} - \Delta H^\circ_{\text{ionisation}}

Why it holds:

Higher oxidation states have higher ionisation energy (harder to remove electrons).
But they also have higher hydration energy (smaller, more charged ions attract water more strongly).
The balance determines which state is stable.

Example:

$\ce{Cu^+}$ is unstable in water because its hydration energy is too low to compensate for the loss of the second electron.
$\ce{Cu^{2+}}$ is stable in water.

Key Principle 4: Disproportionation

Some oxidation states are unstable and spontaneously convert to two other states:

\ce{2Cu^+ -> Cu + Cu^{2+}}

Why it holds:

The free energy change $\Delta G^\circ$ for the reaction is negative. This happens when the intermediate oxidation state is less stable than the extremes.

Formula (for aqueous ions):

If $E^\circ_{\text{reduction}}$ for the higher state is more positive than for the lower state, disproportionation is spontaneous.

Summary Table: Why Each "Formula" Holds

Principle	Why it works	Key exam example
Inert pair effect	6s² electrons are too tightly bound	$\ce{Pb^{2+}}$ stable, $\ce{Pb^{4+}}$ oxidising
Half-filled stability	Extra exchange energy	$\ce{Mn^{2+}}$ > $\ce{Mn^{3+}}$
Hydration vs ionisation	Energy balance in solution	$\ce{Cu^{2+}}$ stable, $\ce{Cu^+}$ not
Disproportionation	$\Delta G < 0$ for intermediate state	$\ce{Cu^+}$ in water

Final Takeaway for Exams

Never memorise stability blindly. Always ask:

Is the electronic configuration special? (half-filled / inert pair)
Is the medium aqueous or solid? (hydration vs lattice)
Does the element belong to a heavier group? (inert pair effect)

The "formula" is really a balance of energies — and the reasoning is what gets you marks.

The ground-state configuration must obey the Aufbau principle, Hund’s rule, and the Pauli exclusion principle. The configuration that violates the Aufbau order (filling 4s before 3d) is option (B).

The question is about recognising which electron configuration is not a valid ground state. That means we need to check each option against the rules that govern how electrons fill orbitals in an atom at its lowest energy state.

The key idea is the Aufbau principle: electrons fill orbitals in order of increasing energy. For the 3d and 4s orbitals, the energy ordering is well-known: 4s fills before 3d for neutral atoms of the first transition series. But there’s a subtlety — once 3d starts filling, the 4s energy rises slightly above 3d, so in the written configuration we list 3d before 4s (by convention), but the filling order is still 4s first.

Let’s examine each option carefully.

Option (A): $1s^2 2s^2 2p^6 3s^2 3p^6 3d^8 4s^2$

Total electrons = $2+2+6+2+6+8+2 = 28$ . That’s nickel (Ni). The 4s is filled before the 3d, so the filling sequence is $4s^2$ then $3d^8$ . This is the correct ground state of Ni. No violation.
Option (B): $1s^2 2s^2 2p^6 3s^2 3p^6 3d^9 4s^2$

Total electrons = $2+2+6+2+6+9+2 = 29$ . That’s copper (Cu). But here’s the catch: for copper, the actual ground state is $3d^{10} 4s^1$ , not $3d^9 4s^2$ . Why? Because a half-filled or fully-filled d-subshell is extra stable. The energy cost of promoting one electron from 4s to 3d is more than compensated by the exchange energy gain of a full $3d^{10}$ configuration. So the configuration $3d^9 4s^2$ is not the ground state — it’s an excited state. This option is suspect.
Option (C): $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1$

Total electrons = 29 again — copper. This is the actual ground state of Cu, as just explained. Valid.
Option (D): $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$

Total electrons = 24. That’s chromium (Cr). Chromium’s ground state is famously $3d^5 4s^1$ , not $3d^4 4s^2$ , because a half-filled d-subshell ( $d^5$ ) is extra stable. So this is correct.

Watch out

A common mistake is to think that because 4s fills before 3d, the configuration $3d^9 4s^2$ should be fine. But the anomalous configurations of Cr and Cu (and a few others) override the simple Aufbau order due to extra stability from half-filled or fully-filled subshells.

So the only option that does not represent a ground state is (B).

✓Final answer

The configuration that does not represent a ground state is option (B) .

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Q.Which of the following options does not represent ground state electronic configuration of an atom?

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