The 71st electron of an element X with an atomic number of 71 enters the orbital:

🧠 The f-block Finale The element with $Z=71$ is Lutetium ($\ce{Lu}$). It marks the very last spot in the lanthanoid series. We need to determine exactly where that final 71st electron "lands" during the building process.

🗺️ The Aufbau Ladder

1. Noble Gas: Xenon ($Z=54$).
2. Filling Period 6:
   • $6s^2$: electrons 55 and 56.
   • $4f^{14}$: electrons 57 through 70. (The f-subshell is fully filled here!)
3. The Next Room: With $4f$ full, the 71st electron must now enter the 5d subshell. Result: Electron 71 is in the $5d$ orbital.

⚡ The Periodic Anchor Look at the Periodic Table row. Lutetium ($Z=71$) is often grouped under the "d-block" because it has a partially filled $5d$ orbital. If you remember that the lanthanoids end at $Z=71$ with a "d" electron, this question is a 5-second check.

⚠️ Common Traps Many assume that since Lutetium is a Lanthanoid, the last electron must be in $4f$. This is incorrect! The $4f$ subshell is completed with the 70th electron ($\ce{Yb}$). Think of $\ce{Lu}$ as the "bridge" back to the d-block.

$\boxed{\text{Answer: (a)}}$