The number of $$f$$ electrons in the ground state electronic configuration of Np (Z = 93) is _________. (Integer answer)

We have to find how many electrons of neptunium (atomic number $$Z=93$$) occupy any subshell of type $$f$$ in its ground-state electronic configuration. To do this we first write the complete configuration and then simply count the electrons that sit in all the $$f$$ subshells that appear.

The rule governing the order in which subshells are filled is called the Aufbau principle. It can be expressed through the $$(n+\ell)$$ rule:

For two subshells, the one with smaller }(n+\ell)\text{ is filled first. If }(n+\ell)\text{ is the same, the subshell with the smaller }n\text{ is filled first.

Using this filling order we proceed step by step, allotting a total of $$93$$ electrons.

We start with the lowest available subshells:

$$1s^2$$, $$\;2s^2$$, $$\;2p^6$$, $$\;3s^2$$, $$\;3p^6$$, $$\;4s^2$$, $$\;3d^{10}$$, $$\;4p^6$$, $$\;5s^2$$, $$\;4d^{10}$$, $$\;5p^6$$, $$\;6s^2$$, $$\;4f^{14}$$, $$\;5d^{10}$$, $$\;6p^6$$, $$\;7s^2$$

Up to this point the number of electrons already assigned is

$$2+2+6+2+6+2+10+6+2+10+6+2+14+10+6+2=86.$$

The first $$86$$ electrons complete the configuration of radon, so we write

$$\text{Neptunium}: [\mathrm{Rn}]\,\cdots$$

Now there are $$93-86 = 7$$ electrons still to be placed. According to the same $$(n+\ell)$$ rule the next subshells in line are $$5f$$, then $$6d$$.

Following experimental evidence for the actinide series, the most stable (ground-state) distribution for neptunium is

$$5f^4\,6d^1\,7s^2.$$

Combining this with the radon core gives the full ground-state configuration:

$$\boxed{[\mathrm{Rn}]\,5f^{4}\,6d^{1}\,7s^{2}}.$$

Now we count all electrons that reside in any $$f$$ subshell:

• The $$4f$$ subshell is completely filled inside the radon core, so it contains $$14$$ electrons.
• The $$5f$$ subshell just written contains $$4$$ electrons.

Adding these, the total number of $$f$$ electrons is

$$14 + 4 = 18.$$

Hence, the correct answer is Option 18.