Question

$Nd ^{2+}=$ ___________

• A. $4 f ^3$
• B. $4 f^4 6 s^2$
• C. $4 f ^2 6 s ^2$
• D. $4 f ^4$

Hint:

The electron configuration for transition metal ions is derived by removing electrons from the highest energy orbitals, typically starting from the outermost \( s \)-orbital.

Answer 1

The Correct Option is D

Nd(60)=[Xe]4f⁴5d⁰6s² 
Nd²⁺=[Xe]4f⁴5d⁰5s⁰
So, the correct option is (D): 4f⁴

Answer 2

The electronic configuration of Nd (atomic number 60) is:

\[ \text{Nd(60)} = [\text{Xe}] 4f^4 5d^0 6s^2 \]

For \( \text{Nd}^{2+} \), two electrons are removed (typically from the 6s orbital first):

\[ \text{Nd}^{2+} = [\text{Xe}] 4f^4 5d^0 5s^0 \]

Thus, the correct answer is \( 4f^4 \).