Q.Although +3 is the characteristic oxidation state for lanthanoids but cerium also shows +4 oxidation state because ___________.

The key idea is that cerium’s +4 state is stabilised by achieving an empty $4f$ subshell.

Reasoning:

1. Lanthanoids typically lose three electrons to form $M^{3+}$, but cerium has the electronic configuration $[Xe]\,4f^1\,5d^1\,6s^2$.
2. Removing all four outer electrons ($4f^1, 5d^1, 6s^2$) gives $Ce^{4+}$, which has the configuration $[Xe]$ — a noble gas core with an empty $4f^0$ subshell. …

Why does cerium break the +3 rule?

Lanthanoids typically lose three electrons to form stable +3 ions. This is because after losing two 6s electrons and one 4f electron, the remaining 4f subshell is either half-filled or completely filled — both are especially stable arrangements. But cerium is special.

Cerium has the electronic configuration $\text{[Xe]} \, 4f^1 \, 5d^1 \, 6s^2$. If it loses all four of its outer electrons (the two 6s, the one 5d, and the one 4f), it becomes $\text{Ce}^{4+}$, which has the configuration $\text{[Xe]}$. That is the same as the noble gas xenon — a completely filled shell, and also the $f^0$ configuration (an empty 4f subshell). Both are extremely stable.

The question asks why cerium shows +4. Let’s examine each option carefully.

1. Option (i): "It has variable ionisation enthalpy"

   This is true for many elements, but it’s not the reason cerium shows +4. Variable ionisation enthalpy is a consequence of electronic structure, not the cause. Moreover, all lanthanoids have variable ionisation enthalpies, yet most don’t show +4. So this is too vague.

2. Option (ii): "It has a tendency to attain noble gas configuration"

   This is almost correct. $\text{Ce}^{4+}$ does have the noble gas configuration of xenon. But the phrase "noble gas configuration" here is a bit misleading — it’s not the only reason. The real driving force is the stability of the empty 4f subshell ($f^0$), which happens to coincide with the noble gas core. Option (iii) is more precise.

3. Option (iii): "It has a tendency to attain $f^0$ configuration"

   This is the exact reason. The $f^0$ configuration (empty 4f subshell) is exceptionally stable. Cerium can achieve it by losing four electrons. No other lanthanoid can do this as easily because they would have to lose more electrons to reach $f^0$, which is energetically too costly. This is the standard explanation in textbooks. …

Method: Stability via Electronic Configuration (f-orbital rule)

This method uses the principle that half-filled, fully-filled, or empty f-orbitals are exceptionally stable. For lanthanoids, the characteristic +3 state arises from losing two 6s electrons and one 4f electron. But if losing one more electron leads to a special stability, a +4 state becomes possible.

Steps:

1. Write the ground state electronic configuration of the neutral atom

   Cerium (Ce, atomic number 58):

   $[ \text{Xe} ] \, 4f^1 \, 5d^1 \, 6s^2$

   (Note: The 4f and 5d energies are close; the actual configuration is $4f^1 5d^1 6s^2$.)

2. Determine the configuration after losing three electrons (Ce³⁺)

   Ce → Ce³⁺ + 3e⁻:

   Lose 6s² and one 4f electron → $[ \text{Xe} ] \, 4f^1$

   This is the characteristic +3 state.

3. Determine the configuration after losing four electrons (Ce⁴⁺)

   Ce → Ce⁴⁺ + 4e⁻:

   Lose 6s², one 4f, and one 5d electron → $[ \text{Xe} ]$

   This is the noble gas configuration of xenon.

4. Apply the stability rule

   A noble gas configuration ($f^0$ in the 4f subshell) is exceptionally stable. Ce⁴⁺ achieves this, making the +4 state favourable despite the high ionization energy required. …

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

Mistake 1: Choosing (i) — “It has variable ionisation enthalpy”

Why students pick it:

They confuse the reason for variable oxidation states in transition metals (where successive ionisation enthalpies are close) with the reason in lanthanoids.

Why it’s wrong:

All lanthanoids have variable ionisation enthalpies — that alone doesn’t explain why only Ce (and Tb) show +4. The key is the stability of the resulting electronic configuration, not the energy cost alone.

How to avoid:

Always link the special oxidation state to the electronic configuration after electron loss. For Ce, losing 4 electrons gives $[Xe]\,4f^0$ — an empty, stable subshell. That’s the deciding factor, not just “variable IE”.

Mistake 2: Choosing (iv) — “It resembles $Pb^{4+}$”

Why students pick it:

They recall that Ce(IV) and Pb(IV) are both strong oxidising agents, and think similarity in behaviour implies the cause.

Why it’s wrong:

Resemblance is a consequence, not the cause. The question asks why Ce shows +4 — the reason is internal to Ce’s electronic structure, not its similarity to another element.

How to avoid:

Read the question stem carefully: “because ___________.” The blank must be filled with a fundamental reason (electronic configuration), not an observation (resemblance).

Mistake 3: Confusing $f^0$ with noble gas configuration

Why students pick (ii):

They think “noble gas configuration” means $[Xe]$ — which Ce(IV) does have. But that’s not the precise reason given in the options.

Why (iii) is better:

The standard textbook explanation for Ce(IV) is “tendency to attain $f^0$ configuration”. While $f^0$ is indeed a noble gas core ($[Xe]$), the exam specifically tests the $f$-subshell terminology. Option (iii) is the exact phrase used in NCERT.

How to avoid:

Memorise the exact phrasing:

• Ce(IV) → $4f^0$ (empty $f$ subshell)
• Eu(II), Yb(II) → $4f^7$ (half-filled $f$ subshell)
• Tb(IV) → $4f^7$ (half-filled $f$ subshell) …