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Question 43

Nd$$^{2+}$$ = ______

Solution

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Neodymium (Nd) has atomic number 60.

Electronic configuration of Nd:

$$ \text{Nd} = [\text{Xe}] \; 4f^4 \; 6s^2 $$

For Nd$$^{2+}$$, two electrons are removed. In lanthanides, the $$6s$$ electrons are removed first (as they are in the outermost shell):

$$ \text{Nd}^{2+} = [\text{Xe}] \; 4f^4 $$

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