Lanthanoid ions with 4f 7 configuration are : (A) Eu 2+ (B) Gd 3+ (C) Eu 3+ (D) Tb 3+ (E) Sm 2+ Choose the correct answer from the options given below : (1) (A) and (B) only (2) (A) and (D) only (3) (B) and (E) only (4) (B) and (C) only

Key Ideas Needed

1. Electronic Configuration of Lanthanoids
   The general filling order beyond xenon is
   $[Xe] \;6s^2 \;4f^{1-14}\;5d^{0-1}$ as we go from Ce ($4f^1$) to Lu ($4f^{14}$).

2. Oxidation State Effect
   • Remove electrons first from the outermost $6s$ level.
   • If more have to be removed, take from $5d$, then $4f$.
   Hence a $+2$ state often means $6s^2$ is gone; a $+3$ state removes $6s^2$ plus one more.

3. Half-Filled Stability
   A subshell exactly half-filled ($4f^7$) has extra exchange-energy stabilization. This makes Eu$^{2+}$ and Gd$^{3+}$ unusually common and stable.

Logical Chain to Solve

1. Write the neutral atom’s $4f$ count.
   • Eu (Z = 63) → $4f^7$
   • Gd (Z = 64) → $4f^7$
   • Sm (Z = 62) → $4f^6$
   • Tb (Z = 65) → $4f^9$

2. Subtract electrons according to the oxidation state:
   $\text{4f electrons in ion} = \text{4f in atom} - \text{any 4f removed}$

3. For +2 or +3 states, outer $6s$ (and $5d$ if necessary) are removed first, so 4f usually stays the same.

4. Check which ions end with $4f^7$.

5. Mark the matching option.

Imagine a Long Row of Lockers

1. Lockers = Orbitals: Think of the tiny spaces where electrons stay ($s$, $p$, $d$, $f$ orbitals) as lockers in a long hallway.
2. 4f Lockers: The lanthanoids open a brand-new hallway called 4f that can hold 14 students (electrons).
3. Half-Filled is Special: If exactly 7 lockers out of 14 are filled (half-filled), it is extra stable—like each student getting a full desk to themselves.
4. Who owns which lockers?
   • Europium (Eu) normally has $4f^7\,6s^2$.
   • Gadolinium (Gd) normally has $4f^7\,5d^1\,6s^2$.
5. Remove Keys = Form Ions: Making a $+2$ or $+3$ ion is like taking away 2 or 3 keys from the outer most lockers ($6s$ and $5d$ first).
   • Eu$^{2+}$ ➔ loses 2 keys from $6s^2$ → still $4f^7$.
   • Gd$^{3+}$ ➔ loses 2 keys from $6s^2$ and 1 from $5d^1$ → still $4f^7$.

So the ions with exactly seven 4f electrons are Eu$^{2+}$ and Gd$^{3+}$.

Step-by-Step Calculation

1. List Neutral Configurations

   \text{Sm (Z 62)} & : [Xe]\,4f^6\,6s^2 \\ \text{Eu (Z 63)} & : [Xe]\,4f^7\,6s^2 \\ \text{Gd (Z 64)} & : [Xe]\,4f^7\,5d^1\,6s^2 \\ \text{Tb (Z 65)} & : [Xe]\,4f^9\,6s^2 \end{aligned}$$

2. Form the Given Ions

   • Eu$^{2+}$: remove $6s^2$
   $[Xe]\,4f^7$

   • Gd$^{3+}$: remove $6s^2$ and $5d^1$
   $[Xe]\,4f^7$

   • Eu$^{3+}$: remove $6s^2$ and one 4f (or 5d if present). No 5d present, so 4f loses one
   $[Xe]\,4f^6$

   • Tb$^{3+}$: remove $6s^2$ and one 4f
   $[Xe]\,4f^8$

   • Sm$^{2+}$: remove $6s^2$
   $[Xe]\,4f^6$

3. Select Ions with $4f^7$

   $Eu^{2+},\;Gd^{3+}$

4. Match to Options

   Option (1) (A) and (B) only is correct.

Final Answer: Option 1.