Lanthanoids: Electronic Configuration and Oxidation States (Why Ce Shows +4)
Chemistry · D And F Block Elements · NEET
Lanthanoids (Ce to Lu) fill the 4f subshell. Their most common and stable oxidation state is +3 (the ion is 4f^n). Cerium also shows +4 because after losing one extra electron, Ce4+ reaches the very stable empty 4f^0 (noble-gas Xe) configuration. Memory hook: "empty, half, full are happy" — 4f^0 (Ce4+), 4f^7 (Eu2+, Gd3+, Tb4+), 4f^14 (Yb2+, Lu3+) are the stable ones NEET loves to ask.
Cerium reaches the stable empty 4f^0 (xenon) configuration in the +4 state, which is why it shows +4 in addition to the common +3. Ce^4+ then acts as an oxidising agent by accepting an electron back to 4f^1.
Your doubts, answered
What is the general electronic configuration of lanthanoids?
The general outer electronic configuration is [Xe] 4f^(1-14) 5d^(0-1) 6s^2. The 6s^2 part is common to all of them; the 4f subshell fills up as you move from Ce (Z=58) to Lu (Z=71). In simple words, the last electron goes into a 4f orbital, which is an inner shell. That is why they are called inner-transition elements. For NEET, remember: 6s^2 is fixed, only the 4f count changes.
Why is +3 the most common (and most stable) oxidation state of lanthanoids?
In the +3 ion (Ln^3+), the lanthanoid loses its two 6s electrons and one more electron (from 5d or 4f). This gives a clean 4f^n configuration for the ion, which is energetically stable across the whole series. Because of this, almost every lanthanoid shows +3 as its main oxidation state. NCERT says: 'the electronic configurations of all the tripositive ions are of the form 4f^n (n = 1 to 14).' So +3 is the default answer unless the question points to a special stable config.
Why does cerium (Ce) show the +4 oxidation state?
Cerium in +3 is Ce^3+ = [Xe] 4f^1. If it loses one more electron, it becomes Ce^4+ = [Xe] 4f^0, which is an empty 4f subshell (same as noble gas xenon). An empty subshell is extra stable, so Ce readily gives up that fourth electron. This is exactly why Ce^4+ works as an oxidising agent (it wants to grab an electron back to return to 4f^1). This is the 2026 NEET answer: Ce shows +4 to reach 4f^0.
Why does europium (Eu) show the +2 oxidation state?
Europium atom is [Xe] 4f^7 6s^2. If it loses only its two 6s electrons, it becomes Eu^2+ = [Xe] 4f^7, which is a stable half-filled 4f subshell. A half-filled shell is extra stable, so Eu stops at +2 instead of going all the way to +3. The same half-filled logic explains Tb showing +4 (Tb^4+ = 4f^7) and Yb showing +2 (Yb^2+ = 4f^14, fully filled).
What are the configurations of Ce^4+ and Yb^2+, and why are they special?
Ce^4+ = [Xe] 4f^0 (empty) and Yb^2+ = [Xe] 4f^14 (completely filled). Both have zero unpaired electrons, so both are diamagnetic — this was the answer to NEET 2024. Empty (4f^0), half-filled (4f^7), and fully-filled (4f^14) are the three 'happy' configurations that decide which unusual oxidation states appear.
Why do Gd and some lanthanoids have a 5d^1 electron in their configuration?
To keep a stable half-filled or empty 4f, the extra electron sometimes enters 5d instead of 4f. For example, Gd (Z=64) is [Xe] 4f^7 5d^1 6s^2, not 4f^8, so that the 4f^7 half-filled shell is preserved. Lanthanum (La, 4f^0 5d^1) and lutetium (Lu, 4f^14 5d^1) also carry a 5d^1 electron. NEET 2016 tested exactly this Eu–Gd–Tb pattern.
⚠️ The NEET trap ✗ Cerium shows +4 because after losing one more electron it reaches the 4f^14 (fully-filled) configuration. ✓ Cerium shows +4 because Ce^4+ reaches [Xe] 4f^0 — an empty 4f subshell (noble-gas Xe config), not 4f^14. Ce^3+ is 4f^1, so removing one more electron gives 4f^0, not 4f^14. 🧠 Count from Ce first: Ce^3+ = 4f^1, take one more away = 4f^0 (empty), NOT full. Full-filled 4f^14 belongs to Yb^2+/Lu^3+, not cerium.
Real NEET questions
NEET 2026
Although the +3 oxidation state is most common in lanthanoids, cerium still shows the +4 oxidation state because:
A · After losing one more electron, it acquires 4f^14 electronic configuration.
B · Its nearest inert gas is Radon.
C · Its atomic number is 61.
D · After losing one more electron, it acquires 4f^0 electronic configuration. ✓
Solution: Ce^3+ is [Xe] 4f^1. Losing one more electron gives Ce^4+ = [Xe] 4f^0, the stable empty 4f (noble-gas Xe) configuration. This extra stability is why Ce readily shows +4. Option (a) is the trap: 4f^0 is empty, not 4f^14 (full).
NEET 2016 Phase 1
The electronic configurations of Eu (Atomic no. 63), Gd (Atomic No. 64) and Tb (Atomic No. 65) are:
A · [Xe] 4f^7 6s^2, [Xe] 4f^8 6s^2 and [Xe] 4f^8 5d^1 6s^2
B · [Xe] 4f^6 5d^1 6s^2, [Xe] 4f^7 5d^1 6s^2 and [Xe] 4f^9 6s^2
C · [Xe] 4f^6 5d^1 6s^2, [Xe] 4f^7 5d^1 6s^2 and [Xe] 4f^8 5d^1 6s^2
D · [Xe] 4f^7 6s^2, [Xe] 4f^7 5d^1 6s^2 and [Xe] 4f^9 6s^2 ✓
Solution: Eu (63): [Xe] 4f^7 6s^2, a stable half-filled 4f^7. Gd (64): the extra electron enters 5d (not 4f) to keep the half-filled 4f^7, giving [Xe] 4f^7 5d^1 6s^2. Tb (65): [Xe] 4f^9 6s^2. This matches option (d).
NEET 2024
The pair of lanthanoid ions which are diamagnetic is
A · Ce^3+ and Eu^2+
B · Gd^3+ and Eu^3+
C · Pm^3+ and Sm^3+
D · Ce^4+ and Yb^2+ ✓
Solution: Diamagnetic means zero unpaired electrons. Ce^4+ = [Xe] 4f^0 (empty) and Yb^2+ = [Xe] 4f^14 (fully filled), so both have all electrons paired. The other ions (Ce^3+ 4f^1, Eu^2+/Gd^3+ 4f^7, Eu^3+ 4f^6, Pm^3+ 4f^4, Sm^3+ 4f^5) all have unpaired f-electrons and are paramagnetic.
Solved D And F Block Elements NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the most common oxidation state of lanthanoids?
+3. Nearly every lanthanoid shows +3 as its main and most stable state, because the Ln^3+ ion has a clean 4f^n configuration.
Which lanthanoids show +4 oxidation state?
Mainly Ce (Ce^4+ = 4f^0) and Tb (Tb^4+ = 4f^7). These reach stable empty or half-filled 4f, so +4 appears in addition to +3.
Which lanthanoids show +2 oxidation state?
Mainly Eu (Eu^2+ = 4f^7, half-filled) and Yb (Yb^2+ = 4f^14, fully filled). Sm can also show +2. They stop early to keep a stable 4f shell.
Why is Ce^4+ used as an oxidising agent?
Ce^4+ (4f^0) easily accepts one electron to go back to the stable Ce^3+ (4f^1). Because it grabs electrons, it acts as an oxidising agent and is used in cerimetric titrations.
Is Ce^4+ diamagnetic or paramagnetic?
Diamagnetic. Ce^4+ = [Xe] 4f^0 has no unpaired electrons, so it is diamagnetic (NEET 2024 tested this with the Ce^4+/Yb^2+ pair).
