The three quantum numbers `n, l, m` corresponding to the valence electron in rubidium `(Z=37)` are

Answer

To find the three quantum numbers \( n, l, m \) corresponding to the valence electron in rubidium (Rb, atomic number 37), we can follow these steps:

### Step 1: Determine the Electronic Configuration
The electronic configuration of rubidium can be written as follows:
- Start filling from the lowest energy level:
- \( 1s^2 \)
- \( 2s^2 \)
- \( 2p^6 \)
- \( 3s^2 \)
- \( 3p^6 \)
- \( 4s^2 \)
- \( 3d^{10} \)
- \( 4p^6 \)
- \( 5s^1 \)

Thus, the complete electronic configuration of rubidium is:
\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^6 \, 5s^1 \]

### Step 2: Identify the Valence Electron
The valence electron of rubidium is the one in the outermost shell, which is the \( 5s \) orbital.

### Step 3: Determine the Principal Quantum Number \( n \)
The principal quantum number \( n \) indicates the energy level of the electron. For the \( 5s \) orbital, the principal quantum number is:
\[ n = 5 \]

### Step 4: Determine the Azimuthal Quantum Number \( l \)
The azimuthal quantum number \( l \) describes the shape of the orbital. For an \( s \) orbital, \( l \) is:
\[ l = 0 \]

### Step 5: Determine the Magnetic Quantum Number \( m \)
The magnetic quantum number \( m \) specifies the orientation of the orbital. For \( l = 0 \) (an \( s \) orbital), the only possible value for \( m \) is:
\[ m = 0 \]

### Final Answer
Thus, the three quantum numbers corresponding to the valence electron in rubidium are:
- \( n = 5 \)
- \( l = 0 \)
- \( m = 0 \)