The 71$$^{st}$$ electron of an element X with an atomic number of 71 enters the orbital:

We have to decide where the very last (71st) electron of the element with atomic number $$Z = 71$$ will go. For this we apply the Aufbau (building-up) principle, which states:

Rule stated first: “Electrons occupy orbitals in the increasing order of $$n+\ell$$. When two orbitals have the same $$n+\ell$$ value, the one with the lower principal quantum number $$n$$ is filled first.”

Listing the subshells in the sequence dictated by this rule gives the familiar order

$$ 1s \;\rightarrow\; 2s \;\rightarrow\; 2p \;\rightarrow\; 3s \;\rightarrow\; 3p \;\rightarrow\; 4s \;\rightarrow\; 3d \;\rightarrow\; 4p \;\rightarrow\; 5s \;\rightarrow\; 4d \;\rightarrow\; 5p \;\rightarrow\; 6s \;\rightarrow\; 4f \;\rightarrow\; 5d \;\rightarrow\; 6p \;\rightarrow\; 7s \;\ldots $$

Now we successively place electrons until we reach the 71st one, keeping track of the capacity of each subshell ($$s:2,\; p:6,\; d:10,\; f:14$$):

$$ \begin{aligned} 1s^2 &\quad (\text{total }2)\\ 2s^2 &\quad (\text{total }4)\\ 2p^6 &\quad (\text{total }10)\\ 3s^2 &\quad (\text{total }12)\\ 3p^6 &\quad (\text{total }18)\\ 4s^2 &\quad (\text{total }20)\\ 3d^{10} &\quad (\text{total }30)\\ 4p^6 &\quad (\text{total }36)\\ 5s^2 &\quad (\text{total }38)\\ 4d^{10} &\quad (\text{total }48)\\ 5p^6 &\quad (\text{total }54)\; \Bigl(\text{configuration of }^{54}\text{Xe}\Bigr)\\[4pt] 6s^2 &\quad (\text{total }56)\\ 4f^{14} &\quad (\text{total }70) \end{aligned} $$

At this stage we have accommodated 70 electrons. We still need to place one more—the 71st electron.

The next subshell in the Aufbau list after $$4f$$ is $$5d$$ because $$5d$$ has $$n+\ell = 5+2 = 7$$, which is the same as $$6p$$ ( $$6+1 = 7$$ ), but its principal quantum number $$n = 5$$ is smaller than that of $$6p$$ ($$n = 6$$). Therefore $$5d$$ is lower in energy than $$6p$$ and is filled first.

So the 71st electron must enter the $$5d$$ subshell, giving the outer portion of the configuration

$$\boxed{6s^2\,4f^{14}\,5d^1}$$

Hence, the correct answer is Option A.