$Nd ^{2+}=$ ___________
$Nd ^{2+}=$ ___________
Show Hint
- $4 f ^3$
- $4 f^4 6 s^2$
- $4 f ^2 6 s ^2$
- $4 f ^4$
The Correct Option is D
Approach Solution - 1
Nd(60)=[Xe]4f45d06s2
Nd2+=[Xe]4f45d05s0
So, the correct option is (D): 4f4
Approach Solution -2
\[ \text{Nd(60)} = [\text{Xe}] 4f^4 5d^0 6s^2 \]
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 \( 4f^4 \).Top JEE Main Chemistry Questions
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