The correct set of four quantum numbers for the valence electrons of rubidium atom ($$Z = 37$$) is:

We begin by recalling that any electron in an atom is completely described by four quantum numbers: the principal quantum number $$n$$, the azimuthal (orbital‐angular‐momentum) quantum number $$l$$, the magnetic quantum number $$m_l$$ and the spin quantum number $$m_s$$.

The principal quantum number $$n$$ tells us the main energy level (or shell) to which the electron belongs. The azimuthal quantum number $$l$$ takes integral values from $$0$$ to $$(n-1)$$ and designates the subshell: $$l = 0$$ corresponds to an $$s$$ subshell, $$l = 1$$ to a $$p$$ subshell, $$l = 2$$ to a $$d$$ subshell, and so on. For any fixed $$l$$, the magnetic quantum number $$m_l$$ can assume all integral values from $$-l$$ through $$0$$ to $$+l$$. Finally, the spin quantum number $$m_s$$ can be either $$+\dfrac{1}{2}$$ or $$-\dfrac{1}{2}$$, representing the two possible spin orientations.

Now, rubidium has atomic number $$Z = 37$$. To locate its valence electron, we must write the complete ground-state electronic configuration, building it up step by step according to the Aufbau (energy‐ordering) principle, the Pauli exclusion principle and Hund’s rule:

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

Up to krypton $$[Kr]$$, which accounts for $$36$$ electrons, we have filled all lower subshells. The $$37^{\text{th}}$$ electron of rubidium therefore enters the next available subshell, which is the $$5s$$ subshell, giving the outer-electronic configuration $$[Kr]\,5s^1$$. Hence the valence (outermost) electron of rubidium occupies a $$5s$$ orbital.

For this $$5s$$ electron:

• The principal quantum number is simply the shell number: $$n = 5$$.

• Because it is in an $$s$$ subshell, we use the correspondence $$s \rightarrow l = 0$$. So we have $$l = 0$$.

• For any $$s$$ orbital (where $$l = 0$$), the only possible magnetic quantum number is $$m_l = 0$$, because $$m_l$$ ranges from $$-l$$ to $$+l$$ and that gives only one value when $$l = 0$$.

• The spin quantum number can be either $$+\dfrac{1}{2}$$ or $$-\dfrac{1}{2}$$. Conventionally, in listing a single electron we choose the $$+\dfrac{1}{2}$$ value, although the other value would also represent an allowed state. Here, every option in the question specifies $$+\dfrac{1}{2}$$, so we select $$m_s = +\dfrac{1}{2}$$.

Collecting these results, the complete set of quantum numbers for the valence electron of rubidium is

$$n = 5,\; l = 0,\; m_l = 0,\; m_s = +\dfrac{1}{2}.$$

We now inspect the given options:

A. $$5,\;0,\;0,\;+\dfrac{1}{2}$$

B. $$5,\;1,\;0,\;+\dfrac{1}{2}$$

C. $$5,\;1,\;1,\;+\dfrac{1}{2}$$

D. $$5,\;0,\;1,\;+\dfrac{1}{2}$$

Only Option A matches the required set $$\left(5,\;0,\;0,\;+\dfrac{1}{2}\right)$$ exactly. The other options are incorrect because they assign either an inappropriate azimuthal quantum number $$l$$ or an impossible magnetic quantum number $$m_l$$ for the $$s$$ subshell.

Hence, the correct answer is Option A.