1. Identify the element and its series

Neodymium has atomic number $Z=60$. It is a lanthanoid element, so its electrons are filled after the noble gas core $[\mathrm{Xe}]$.

Since xenon has atomic number $54$, the remaining electrons after $[\mathrm{Xe}]$ are:

$60-54=6$

So, 6 electrons must be placed beyond the $[\mathrm{Xe}]$ core.

2. General filling pattern in lanthanoids

For lanthanoids, the usual filling is:

$[\mathrm{Xe}]\,4f^n 5d^{0\text{ or }1} 6s^2$

Neodymium is one of the early lanthanoids, and its ground-state configuration is known to be:

$[\mathrm{Xe}]\,4f^4 6s^2$

3. Check the options

• Option A: $[\mathrm{Xe}]\,4f^4 6s^2$

  • This matches the known configuration of neodymium.
  • Correct

• Option B: $[\mathrm{Xe}]\,5f^4 7s^2$

  • $5f$ and $7s$ belong to the actinoid region, not lanthanoids.
  • Incorrect

• Option C: $[\mathrm{Xe}]\,4f^6 6s^2$

  • This gives $54+6+2=62$ electrons, corresponding to samarium, not neodymium.
  • Incorrect

• Option D: $[\mathrm{Xe}]\,4f^1 5d^1 6s^2$

  • This corresponds to cerium-like filling, not neodymium.
  • Incorrect

4. Final answer

Therefore, the electronic configuration of neodymium is:

$[\mathrm{Xe}]\,4f^4 6s^2$

So, the correct option is A.