Q.Match the property given in Column I with the element given in Column II.
Column I (Property):

Concept: Lanthanoid Oxidation States & Electronic Configurations

The stability of +2, +3, and +4 states in lanthanoids depends on achieving half-filled ($4f^7$) or fully-filled ($4f^{14}$) subshells. Promethium (Pm) is the only radioactive lanthanoid.

Reasoning steps:

1. +4 oxidation state – Ce (Cerium) loses 4 electrons to attain stable $4f^0$ configuration.
2. +2 oxidation state – Eu (Europium) gains stability as $4f^7$ in +2 state.
3. Radioactive – Pm (Promethium) has no stable isotopes. …

Why This Works: The Logic of Lanthanoid Properties

The lanthanoids (elements 57–71) are famous for their similar chemistry, but a few stand out due to special electronic configurations. The +3 oxidation state is the most common for all lanthanoids, but deviations to +2 or +4 happen when doing so leads to a half-filled ($4f^7$) or fully-filled ($4f^{14}$) subshell — these are exceptionally stable. Radioactivity is rare among lanthanoids, with promethium being the only one that has no stable isotopes. Let’s match each property step by step.

1. Property (i): Lanthanoid which shows +4 oxidation state

   A +4 state means the element loses four electrons. For a lanthanoid, this is favourable only if the resulting $4f$ configuration is stable. Cerium (Ce, atomic number 58) has the ground state $[Xe]4f^1 5d^1 6s^2$. In the +4 state, it loses all four outer electrons, leaving $[Xe]$ — an empty $4f$ subshell, which is not particularly stable. But Ce also commonly shows +4 because the $4f^1$ electron is easily removed, and the resulting Ce$^{4+}$ ion is isoelectronic with xenon. The real stability comes from the fact that Ce$^{4+}$ has no $4f$ electrons, and the next element, Pr, also shows +4 but less commonly. Ce is the classic example.

   Match: (i) → (b) Ce

2. Property (ii): Lanthanoid which can show +2 oxidation state

   A +2 state is rare among lanthanoids because losing only two electrons leaves a $4f^n$ configuration that is often not particularly stable. However, europium (Eu, atomic number 63) has the ground state $[Xe]4f^7 6s^2$. In the +2 state, it loses the two $6s$ electrons, giving $4f^7$ — a half-filled subshell, which is exceptionally stable. Similarly, ytterbium (Yb) gives $4f^{14}$ in +2 state, but Eu is the most famous example.

   Match: (ii) → (d) Eu

3. Property (iii): Radioactive lanthanoid

   All lanthanoids except promethium (Pm, atomic number 61) have at least one stable isotope. Pm has no stable isotopes; all its isotopes are radioactive, with the longest half-life being about 17.7 years for $^{145}$Pm. This makes it the only radioactive lanthanoid in the list.

   Match: (iii) → (a) Pm

4. Property (iv): Lanthanoid which has $4f^7$ electronic configuration in +3 oxidation state …

Method: Electronic Configuration & Oxidation State Mapping

This is a concept-application matching problem. The method is:

Step 1: Recall the general electronic configuration of lanthanoids

All lanthanoids have the outer configuration:

$[Xe]\,4f^{1-14}\,5d^{0-1}\,6s^2$

Step 2: For each property, identify the element by its stable +3 configuration and known stable oxidation states

(i) Lanthanoid showing +4 oxidation state

• Why: Ce ($Z=58$) has $4f^1$ in +3 state. Losing one more electron gives $4f^0$ — a stable empty shell.
• Answer: (b) Ce

(ii) Lanthanoid showing +2 oxidation state

• Why: Eu ($Z=63$) has $4f^7$ in +3 state. Gaining one electron gives $4f^7$ — a stable half-filled shell.
• Answer: (d) Eu

(iii) Radioactive lanthanoid

• Why: Pm ($Z=61$) has no stable isotopes — all are radioactive.
• Answer: (a) Pm

(iv) $4f^7$ configuration in +3 oxidation state

• Why: Gd ($Z=64$) has $4f^7$ in +3 state — half-filled stability.
• Answer: (e) Gd

(v) $4f^{14}$ configuration in +3 oxidation state …

Here are the most common mistakes students make on this Alloy Composition / Lanthanoid Properties matching question, along with how to avoid each.

Mistake 1: Confusing Ce (Cerium) with Eu (Europium) for +4 oxidation state

Why it happens:

Students often remember that lanthanoids show +3 as the most stable state, but they forget which specific elements deviate. Ce can lose all 4 valence electrons (4f¹5d¹6s² → 4f⁰) to show +4, while Eu prefers +2 due to its half-filled 4f⁷ stability.

How to avoid:

• Memorise the exceptions: Ce (+4) and Tb (+4) are the main +4 lanthanoids.
• Use electronic configuration logic:
  • Ce: [Xe] 4f¹5d¹6s² → losing 4 electrons gives stable noble gas configuration (4f⁰).
  • Eu: [Xe] 4f⁷6s² → losing 2 electrons gives stable half-filled 4f⁷.
• Trick: “Ce for +4, Eu for +2” — think of Ce as “Can be +4”.

Correct match: (i) → (b) Ce

Mistake 2: Forgetting that Eu shows +2, not Gd

Why it happens:

Gd has 4f⁷ in +3 state, so students assume it can also show +2. But Gd’s +3 state is already half-filled and very stable; losing one more electron to get +4 is harder. Eu, on the other hand, achieves half-filled 4f⁷ by losing only 2 electrons.

How to avoid:

• Remember the stability rule: Half-filled (4f⁷) and fully-filled (4f¹⁴) subshells are extra stable.
  • Eu → +2 gives 4f⁷ (half-filled).
  • Yb → +2 gives 4f¹⁴ (fully-filled).
• Do not confuse: Gd in +3 already has 4f⁷, so it does not need to show +2.

Correct match: (ii) → (d) Eu

Mistake 3: Not knowing that Pm (Promethium) is the only radioactive lanthanoid

Why it happens:

Students think all lanthanoids are stable, or they confuse Pm with other elements like U or Th (which are actinoids). Pm has no stable isotopes — all are radioactive.

How to avoid:

• Memorise this fact: Among lanthanoids (La to Lu), only Pm is radioactive.
• Link to periodic table: Pm is the only lanthanoid with atomic number 61 — all its isotopes decay.
• Trick: “Pm = Promethium = Problem (radioactive)”.

Correct match: (iii) → (a) Pm

Mistake 4: Mixing up Gd (4f⁷) and Lu (4f¹⁴) in +3 state

Why it happens:

Both are special configurations, but students often swap them. Gd has 4f⁷ (half-filled) and Lu has 4f¹⁴ (fully-filled) in +3 state.

How to avoid:

• Write the +3 configurations:
  • Gd³⁺: [Xe] 4f⁷ (half-filled)
  • Lu³⁺: [Xe] 4f¹⁴ (fully-filled)
• Use atomic number logic:
  • Gd (Z=64) → after losing 3 electrons, 4f⁷ remains.
  • Lu (Z=71) → after losing 3 electrons, 4f¹⁴ remains.
• Trick: “Gd = Good half (7), Lu = Lucky full (14)”. …