Question

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

• A. $5, 0, 0, +\frac{1}{2}$
• B. $5, 0, 1, +\frac{1}{2}$
• C. $5, 1, 0, +\frac{1}{2}$
• D. $5, 1, 1, +\frac{1}{2}$

Answer 1

Answer: (A)

$5, 0, 0, +\frac{1}{2}$

Answer 2

Rubidium (Rb) has the electron configuration: [Kr]5s1.
For the valence electron in the 5s orbital: - Principal quantum number, n = 5. - Azimuthal quantum number, l = 0 (since it is an s-orbital).
Magnetic quantum number, m = 0 (as m can range from –l to +l, and l = 0 allows only m = 0).
Spin quantum number, s = $+\frac{1}{2}$ (or $-\frac{1}{2}$ as it can have either spin).
Thus, the correct set of quantum numbers is (5, 0, 0, $+\frac{1}{2}$).

So, the correct answer is: 5, 0, 0, $+\frac{1}{2}$