The possible value of `l` and `m` for the last electron in the `Cl^(- )ion` are :

Options

A. `1` and `2`
B. `2` and `+1`
C. `3` and `-1`
D. `1` and `-1`

Explanation

To determine the possible values of the angular momentum quantum number (l) and the magnetic quantum number (m) for the last electron in the chloride ion (Cl⁻), we can follow these steps:

### Step 1: Determine the Electron Configuration of Cl⁻
The chloride ion (Cl⁻) has gained one electron compared to the neutral chlorine atom. Chlorine has an atomic number of 17, which means it has 17 electrons. When it gains one electron to become Cl⁻, it has a total of 18 electrons.

The electron configuration for Cl⁻ is:
- 1s² 2s² 2p⁶ 3s² 3p⁶

### Step 2: Identify the Last Electron
The last electron added to the chloride ion enters the 3p subshell. Thus, the last electron is in the 3p orbital.

### Step 3: Determine the Values of l
The angular momentum quantum number (l) corresponds to the subshell:
- For s subshell, l = 0
- For p subshell, l = 1
- For d subshell, l = 2
- For f subshell, l = 3

Since the last electron is in the p subshell, we have:
- l = 1

### Step 4: Determine the Values of m
The magnetic quantum number (m) can take values from -l to +l, including zero. Therefore, for l = 1:
- Possible values of m are: -1, 0, +1

### Conclusion
The possible values of l and m for the last electron in the Cl⁻ ion are:
- l = 1
- m = -1, 0, +1

### Final Answer
- l = 1
- m = -1, 0, +1