Match the following species with their corresponding ground state electronic configuration. Atom / Ion (i) Cu

NCERT Exemplar · Q42

Q. Match the following species with their corresponding ground state electronic configuration. Atom / Ion

(i) Cu

(ii) Cu^2+

(iii) Zn^2+

(iv) Cr^3+
Electronic configuration

(a) 1s2 2s2 2p6 3s2 3p6 3d10

(b) 1s2 2s2 2p6 3s2 3p6 3d10 4s2

(d) 1s2 2s2 2p6 3s2 3p6 3d9

(e) 1s2 2s2 2p6 3s2 3p6 3d3

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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 …

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: …

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Transition metals and their ions fill $d$ orbitals in ways that favor stability (half-filled or fully-filled subshells). Copper adopts an exceptional $3d^{10}4s^1$ configuration; removing electrons to form cations strips the $4s$ first, then $3d$ . The matches are: (i)→(c), (ii)→(d), (iii)→(a), (iv)→(e).

Why electron configurations matter for transition elements

The aufbau principle tells us to fill orbitals in order of increasing energy: $4s$ before $3d$ for neutral atoms. But two wrinkles appear with the first-row transition metals. First, chromium and copper adopt exceptional ground-state configurations because half-filled ( $d^5$ ) and fully-filled ( $d^{10}$ ) subshells carry extra exchange-energy stabilization. Second, when we ionize a transition metal the $4s$ electrons are always removed before the $3d$ , even though $4s$ filled first. That is because once $3d$ electrons are present they shield the nucleus more effectively than $4s$ , making $4s$ higher in energy for the cation.

Step-by-step matching

1. Copper, Cu (atomic number 29)

A naïve aufbau filling would give $[\text{Ar}]\,3d^9\,4s^2$ . But promoting one $4s$ electron into the $3d$ subshell yields $[\text{Ar}]\,3d^{10}\,4s^1$ , which has a completely filled $d$ subshell. The extra exchange stabilization wins, so copper's ground state is

1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\,4s^1.

This is configuration (c).

2. Cupric ion, $\text{Cu}^{2+}$

Remove two electrons from neutral copper. The single $4s$ electron goes first, then one $3d$ electron:

[\text{Ar}]\,3d^{10}\,4s^1 \;\xrightarrow{-2e^-}\; [\text{Ar}]\,3d^9.

Written out in full,

1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^9.

This is configuration (d).

Watch out

A common mistake is to remove two $3d$ electrons because " $3d$ is higher energy." In the cation, $4s$ lies above $3d$ , so $4s$ empties first.

3. Zinc(II) ion, $\text{Zn}^{2+}$ (parent atom $Z=30$ )

Neutral zinc has the straightforward configuration $[\text{Ar}]\,3d^{10}\,4s^2$ —no exception, because $d^{10}$ is already fully filled. Removing two electrons strips both $4s$ electrons:

[\text{Ar}]\,3d^{10}\,4s^2 \;\xrightarrow{-2e^-}\; [\text{Ar}]\,3d^{10}.

Expanded,

1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}.

This is configuration (a).

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Why electron configurations matter for transition elements

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Q.Match the following species with their corresponding ground state electronic configuration.
Atom / Ion

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