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Question

The electronic configuration for Neodymium is:

[Atomic Number for Neodymium 60]

$[\mathrm{Xe}] 4 \mathrm{f}^4 6 \mathrm{~s}^2$
$[\mathrm{Xe}] 5 \mathrm{f}^4 7 \mathrm{~s}^2$
$[\mathrm{Xe}] 4 \mathrm{f}^6 6 \mathrm{~s}^2$
$[\mathrm{Xe}] 4 \mathrm{f}^1 5 \mathrm{~d}^1 6 \mathrm{~s}^2$

Solution

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