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

We need to find the four quantum numbers for the valence electron of rubidium ($$Z = 37$$).

Determine the electronic configuration of Rb.

Rubidium has 37 electrons. Its electronic configuration is:

$$Rb: 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \, 4s^2 \, 4p^6 \, 5s^1$$

Or equivalently: $$[Kr] \, 5s^1$$

The valence (outermost) electron is in the $$5s$$ orbital.

Assign the quantum numbers.

For an electron in the $$5s$$ orbital:

- Principal quantum number ($$n$$): $$n = 5$$ (the electron is in the 5th shell)

- Azimuthal quantum number ($$l$$): $$l = 0$$ (for an s-orbital, $$l$$ is always 0)

- Magnetic quantum number ($$m_l$$): $$m_l = 0$$ (for $$l = 0$$, the only possible value is 0)

- Spin quantum number ($$m_s$$): $$m_s = +\frac{1}{2}$$ (for the first electron in an orbital, by convention)

The set of quantum numbers is: $$5, 0, 0, +\frac{1}{2}$$.

The correct answer is Option (1): $$5, 0, 0, +\frac{1}{2}$$.