The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) is:
- \( 5, 0, 0, +\frac{1}{2} \)
- \( 5, 0, 1, +\frac{1}{2} \)
- \( 5, 1, 0, +\frac{1}{2} \)
- \( 5, 1, 1, +\frac{1}{2} \)
The Correct Option is A
Solution and Explanation
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, \(+\frac{1}{2}\)
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