Which set of quantum numbers is possible for the last electron of `Mg^(+)` ion

Options

A. n=3,l=2,m=0,s=+1/2
B. n=2,l=3,m=0,s=+1/2
C. n=1,l=0,m=0,s=+1/2
D. n=3,l=0,m=0,s=+1/2

Answer

To determine the possible set of quantum numbers for the last electron of the `Mg^(+)` ion, we will follow these steps:

### Step 1: Determine the electronic configuration of Mg and Mg^+
Magnesium (Mg) has an atomic number of 12, which means it has 12 electrons in its neutral state. The electronic configuration is:
\[ \text{Mg: } 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \]

When magnesium loses one electron to form the `Mg^(+)` ion, it will have 11 electrons. Therefore, the electronic configuration of `Mg^(+)` is:
\[ \text{Mg}^+: 1s^2 \, 2s^2 \, 2p^6 \, 3s^1 \]

### Step 2: Identify the last electron
The last electron that enters the configuration of `Mg^(+)` is in the 3s subshell.

### Step 3: Assign quantum numbers
The quantum numbers for the last electron in the 3s subshell are defined as follows:
- **Principal quantum number (n)**: This indicates the energy level of the electron. For the last electron in `Mg^(+)`, it is in the 3rd energy level, so \( n = 3 \).
- **Azimuthal quantum number (l)**: This indicates the subshell. For an s subshell, \( l = 0 \).
- **Magnetic quantum number (m)**: This indicates the orientation of the orbital. For an s orbital, there is only one orientation, so \( m = 0 \).
- **Spin quantum number (s)**: This indicates the spin of the electron. The spin can be either \( +\frac{1}{2} \) or \( -\frac{1}{2} \). We can choose \( s = +\frac{1}{2} \) for this case.

### Step 4: Compile the quantum numbers
Thus, the set of quantum numbers for the last electron of `Mg^(+)` is:
- \( n = 3 \)
- \( l = 0 \)
- \( m = 0 \)
- \( s = +\frac{1}{2} \)

### Final Answer
The possible set of quantum numbers for the last electron of `Mg^(+)` is:
\[ (n, l, m, s) = (3, 0, 0, +\frac{1}{2}) \]

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