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

The Correct Option is A

The question asks us to find the correct set of four quantum numbers for the valence electron of a rubidium atom, which has an atomic number \( Z = 37 \). Let's break this down step by step: 

1. First, identify the electron configuration of rubidium (Rb). For Rb: \(1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^6\,5s^1\).
2. The valence electron is the outermost electron, which in this case is in the 5s orbital (\(5s^1\)).
3. Quantum numbers describe the properties of the electron:
   • Principal quantum number (\(n\)): For the 5s orbital, \(n = 5\).
   • Azimuthal quantum number (\(l\)): For an s orbital, \(l = 0\).
   • Magnetic quantum number (\(m_l\)): For \(l = 0\), \(m_l = 0\).
   • Spin quantum number (\(m_s\)): The electron can have a spin of +1/2 or -1/2. Generally, the electron starts with +1/2, so \(m_s = +\frac{1}{2}\).
4. Therefore, the correct set of quantum numbers for the valence electron of a rubidium atom is \((5, 0, 0, +\frac{1}{2})\).

Now, let's rule out the incorrect options:

• \((5, 0, 1, +\frac{1}{2})\): This is incorrect because for \(l = 0\), \(m_l\) must be 0, not 1.
• \((5, 1, 0, +\frac{1}{2})\): This is incorrect because \(l = 1\) would indicate a p orbital, not s.
• \((5, 1, 1, +\frac{1}{2})\): This is also incorrect for the same reason as the previous option - it does not correspond to an s orbital.

Thus, the correct answer, considering the rubidium electron configuration and quantum number rules, is \((5, 0, 0, +\frac{1}{2})\).

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, \(+\frac{1}{2}\)