Match species given in Column I with the electronic configuration given in Column II. Column I (i) Cr (ii) Fe²

NCERT Exemplar · Q47

Q. Match species given in Column I with the electronic configuration given in Column II. Column I

(i) Cr

(ii) Fe^2+

(iii) Ni^2+

(iv) Cu
Column II

(a) [Ar]3d8 4s0

(b) [Ar]3d10 4s1

(d) [Ar] 3d5 4s1

(e) [Ar]3d6 4s2

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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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To match the species with their electronic configurations, we apply the Aufbau principle, Hund's rule, and Pauli exclusion principle, remembering exceptions for Cr and Cu, and the rule for electron removal from ions. The correct matching is (i)-(d), (ii)-(c), (iii)-(a), (iv)-(b).

The electronic configuration of an atom or ion describes the arrangement of electrons in its atomic orbitals. For transition metals, understanding these configurations is crucial because they dictate chemical properties, oxidation states, and magnetic behavior.

The key concepts to remember are:

Aufbau Principle: Electrons fill atomic orbitals in increasing order of energy. For transition metals, the 4s orbital is generally filled before the 3d orbital.
Hund's Rule of Maximum Multiplicity: Orbitals of equal energy are first singly occupied with parallel spins before any orbital is doubly occupied.
Pauli Exclusion Principle: No two electrons in an atom can have the same set of four quantum numbers. This means an orbital can hold a maximum of two electrons, and they must have opposite spins.
Stability of Half-filled and Fully-filled Orbitals: Half-filled ( $d^5$ ) and fully-filled ( $d^{10}$ ) d-orbitals are exceptionally stable due to symmetry and exchange energy. This leads to exceptions in the Aufbau principle for elements like Chromium (Cr) and Copper (Cu).
Electron Removal for Ions: When forming cations from transition metals, electrons are removed first from the outermost shell (the 4s orbital) before electrons are removed from the 3d orbital, even though the 3d orbital was filled later. This is because the 4s orbital becomes higher in energy than the 3d orbital once the atom is ionized.

Let's determine the electronic configuration for each species:

Species (i): Cr (Chromium)
- Chromium has an atomic number $Z=24$ .
- Following the Aufbau principle, the expected configuration would be $[Ar]3d^4 4s^2$ .
- However, due to the extra stability associated with a half-filled d-subshell ( $3d^5$ ), one electron from the 4s orbital shifts to the 3d orbital.
- Therefore, the actual electronic configuration of Cr is $[Ar]3d^5 4s^1$ .
- This matches option (d) in Column II.
Species (ii): Fe $^{2+}$ (Iron(II) ion)
- Iron has an atomic number $Z=26$ .
- The electronic configuration of a neutral Fe atom is $[Ar]3d^6 4s^2$ .
- To form the Fe $^{2+}$ ion, two electrons must be removed. According to the rule for transition metal ions, electrons are removed from the outermost shell first, which is the 4s orbital.
- Removing two electrons from the $4s^2$ orbital leaves us with $4s^0$ .
- Therefore, the electronic configuration of Fe $^{2+}$ is $[Ar]3d^6 4s^0$ .
- This matches option (c) in Column II.
Species (iii): Ni $^{2+}$ (Nickel(II) ion) …

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Q.Match species given in Column I with the electronic configuration given in Column II.
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