Question

Choose the correct metal/ ion from the brackets which -----------------------------
(A) has chemical reactivity similar to that of the first few members of the Lanthanoids (Zn, Ca, Fe, Cu).
(B) has stable \( 4f^7 \) electronic configuration, but acts as a strong reducing agent and converts to \( M{3+} \) state. \(Eu^{2+}, Ce^{2+}, Pr^{2+}, Dy^{2+}\)
(C) is a colorless ion (Tm^{3+}, Lu^{3+}, Gd^{3+}, Sm^{3+}).
(D) shows stable +2 oxidation state and is diamagnetic (Ce, Sm, Ho, Yb).

• A. A : Cu B : Dy$^${2+} C : Sm$^${3+} D : Ho
• B. A : Zn B : Ce$^${2+} C : Gd$^${3+} D : Sm
• C. A : Fe B : Pr$^${2+} C : Tm$^${3+} D : Ce
• D. A : Ca B : Eu$^${2+} C : Lu$^${3+} D : Yb

Hint:

Remember key lanthanoid factsHalf-filled and fully filled \( f \)-orbitals give extra stability and influence color and oxidation states.

Answer 1

The Correct Option is D

Step 1: Analyze part (A).
Lanthanoids show chemical similarity with alkaline earth metals due to similar ionic radii.
Thus, calcium (Ca) shows similar reactivity.

Step 2: Analyze part (B).
Europium \( Eu^{2+} \) has a stable half-filled \( 4f^7 \) configuration but easily oxidizes to \( Eu^{3+} \), acting as a strong reducing agent.

Step 3: Analyze part (C).
Colorless ions have no unpaired electrons or symmetric electron distribution.
\( Lu^{3+} \) has a fully filled \( 4f^{14} \) configuration, making it colorless.

Step 4: Analyze part (D).
Ytterbium (Yb) forms stable \( Yb^{2+} \) with \( 4f^{14} \) configuration, which is fully filled and hence diamagnetic.

Step 5: Conclusion.
Thus, correct matching is:
\[ A : Ca,\quad B : Eu^{2+},\quad C : Lu^{3+},\quad D : Yb \]
Therefore:
\[ \boxed{A : Ca,\; B : Eu^{2+},\; C : Lu^{3+},\; D : Yb} \]