Correct set of four quantum numbers for the valence (outermost) electron of rubidium `(Z = 37)` is

Options

A. `5,0,0,+1/2`
B. `5,1,0,+1/2`
C. `5,1,1,+1/2`
D. `6,0,0,+1/2`

Answer

To determine the correct set of four quantum numbers for the valence (outermost) electron of rubidium (Rb, atomic number 37), we can follow these steps:

### Step 1: Determine the Electron Configuration
Rubidium has an atomic number of 37. The electron configuration for rubidium can be derived by filling the orbitals according to the Aufbau principle. The electron configuration is:
- 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹
This shows that the outermost electron is in the 5s subshell.

### Step 2: Identify the Principal Quantum Number (n)
The principal quantum number (n) indicates the energy level of the electron. For the outermost electron in the 5s subshell:
- n = 5

### Step 3: Determine the Azimuthal Quantum Number (l)
The azimuthal quantum number (l) defines the subshell type:
- For an s subshell, l = 0 (since s corresponds to l = 0).

### Step 4: Determine the Magnetic Quantum Number (m_l)
The magnetic quantum number (m_l) can take values from -l to +l. Since l = 0 for the s subshell:
- m_l = 0

### Step 5: Determine the Spin Quantum Number (m_s)
The spin quantum number (m_s) can take values of +1/2 or -1/2. For the outermost electron, we can choose:
- m_s = +1/2 (this is a common choice for the first electron in an orbital).

### Step 6: Compile the Quantum Numbers
Now we can compile the quantum numbers for the valence electron in rubidium:
- n = 5
- l = 0
- m_l = 0
- m_s = +1/2

Thus, the correct set of quantum numbers for the valence electron of rubidium is:
**(n, l, m_l, m_s) = (5, 0, 0, +1/2)**

### Conclusion
The correct set of four quantum numbers for the valence (outermost) electron of rubidium is:
- **(5, 0, 0, +1/2)**

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