Question

The compound(s) having [Xe]4f\(^1\) configuration is(are)
(Given the atomic numbers Ce:58, Lu:71, Pr:59 and Nd:60)

• A. Na\(_3\)[Ce(NO\(_3\))\(_6\)]
• B. Na\(_3\)[LuCl\(_6\)]
• C. PrO\(_2\)
• D. Nd(NR\(_2\))\(_3\) (R = SiMe\(_3\))

Hint:

Lanthanide \(3+\) oxidation state usually reflects \(4f^n\) of the element (Ce\(^{3+}\,{\to}\,4f^1\); Nd\(^{3+}\,{\to}\,4f^3\)). Higher oxidation states remove 5d/6s first, then 4f (Pr\(^{4+}\,{\to}\,4f^1\)).

Answer 1

The Correct Option is A, C

Step 1: Determine oxidation states.
(A) \([{\rm Ce(NO_3)_6}]^{3-}\Rightarrow {\rm Ce}^{3+}\).
(B) \([{\rm LuCl_6}]^{3-}\Rightarrow {\rm Lu}^{3+}\).
(C) \({\rm PrO_2}\Rightarrow {\rm Pr}^{4+}\) (since O is \(-2\)).
(D) \({\rm Nd(NR_2)_3}\Rightarrow {\rm Nd}^{3+}\).
Step 2: Write 4f electron counts.
Ce: \([{\rm Xe}]4f^1 5d^1 6s^2 \Rightarrow {\rm Ce}^{3+}:[{\rm Xe}]4f^1\).
Lu: \([{\rm Xe}]4f^{14}5d^1 6s^2 \Rightarrow {\rm Lu}^{3+}:[{\rm Xe}]4f^{14}\).
Pr: \([{\rm Xe}]4f^3 6s^2 \Rightarrow {\rm Pr}^{4+}:[{\rm Xe}]4f^1\).
Nd: \([{\rm Xe}]4f^4 6s^2 \Rightarrow {\rm Nd}^{3+}:[{\rm Xe}]4f^3\).
Step 3: Pick compounds with 4f\(^1\).
Ce\(^{3+}\) and Pr\(^{4+}\) give \([{\rm Xe}]4f^1\) \(\Rightarrow\) (A) and (C).