The electronic configuration for Neodymium is: (Atomic Number for Neodymium 60)

Question

Accepted Answer

Given: Neodymium (Nd) has atomic number $60$ .

Find: The correct electronic configuration of Nd.

Neodymium is a lanthanide, so we use the xenon core and then place the remaining electrons in the appropriate subshells.

Using the noble gas core:

[Xe] = 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6

After $[Xe]$ , the remaining electrons occupy the $4f$ and $6s$ orbitals. Therefore:

[Xe] \, 4f^4 \, 6s^2

This matches option A. Therefore, the correct electronic configuration is $[Xe] \, 4f^4 \, 6s^2$ .

Given: We must identify the correct condensed electronic configuration of Neodymium.

Find: Which option matches the lanthanide configuration for atomic number $60$ .

Check the options against the lanthanide filling pattern:

$[Xe] \, 4f^4 \, 6s^2$ matches the derived configuration.
$[Xe] \, 5f^4 \, 7s^2$ is incorrect because $5f$ filling belongs to actinides, not neodymium.
$[Xe] \, 4f^6 \, 6s^2$ has too many electrons in $4f$ for Nd.
$[Xe] \, 4f^5 \, 5d^1 \, 6s^2$ does not match the stated solution conclusion for this question.

Hence, the correct option is A.

Answer