Question

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

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

Answer 1

The Correct Option is A

To determine the correct set of quantum numbers for the valence electron of rubidium (Rb), we need to understand the electronic configuration and the meaning of quantum numbers.

1. Electronic Configuration: Rubidium has an atomic number \( Z = 37 \). Its electronic configuration is: \[ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^1 \] The valence electron is the outermost electron, which is in the \( 5s \) orbital.
2. Quantum Numbers Explanation:
   • Principal Quantum Number \((n)\): This indicates the main energy level. For Rubidium's valence electron in the \( 5s \) orbital, \( n = 5 \).
   • Azimuthal Quantum Number \((l)\): This defines the shape of the orbital. For an \( s \)-orbital, \( l = 0 \).
   • Magnetic Quantum Number \((m_l)\): This represents the orientation of the orbital in space. For \( l = 0 \), \( m_l = 0 \) because there is only one orientation possible for an \( s \)-orbital.
   • Spin Quantum Number \((m_s)\): It can be \( +\frac{1}{2} \) or \( -\frac{1}{2} \). For simplicity, we usually consider \( m_s = +\frac{1}{2} \) for singly occupied orbitals unless specified otherwise.
3. Combining these, the correct set of quantum numbers for the valence electron in Rubidium is: \[ n = 5, \, l = 0, \, m_l = 0, \, m_s = +\frac{1}{2} \]

Thus, the correct answer is \(5,0,0,+\frac{1}{2}\).