For Rb(37) which of the following set of quantum numbers are correct for valence electron?

To find the correct set of quantum numbers for the valence electron of Rubidium (Rb), we first need to determine its electronic configuration.

Rubidium has an atomic number of Z = 37.
The noble gas configuration preceding Rubidium is Krypton (Kr, Z = 36).
Therefore, the ground-state electronic configuration of Rubidium (Rb) is:
[Kr] 5s¹

The valence electron of Rubidium is the single electron in the outermost shell, which resides in the 5s orbital.

Now, let's determine the four quantum numbers ($n$, $l$, $m$ or $m_l$, $s$ or $m_s$) for this 5s¹ electron:

1. Principal Quantum Number ($n$): Represents the main energy level or shell. Since the valence electron is in the 5s orbital, $n=5$.

2. Azimuthal (Orbital Angular Momentum) Quantum Number ($l$): Defines the shape of the orbital (s = 0, p = 1, d = 2, f = 3). For an s orbital, $l=0$.

3. Magnetic Quantum Number ($m_l$): Specifies the orientation of the orbital in space, ranging from $-l$ to $+l$. For $l=0$, the only possible value is $m_l=0$.

4. Spin Quantum Number ($m_s$): Describes the spin orientation of the electron, which can be either $+1/2$ (spin-up) or $-1/2$ (spin-down). For a single electron in an orbital, it is conventionally written as $+1/2$.

Combining these values, the set of quantum numbers ($n$, $l$, $m_l$, $m_s$) for the valence electron of Rubidium is:
5, 0, 0, +(1/2)