Which of the following set of quantum numbers shows orbital of highest energy ?

Options

A. n = 4, l = 0, m = 0, `s=+1/2`
B. n = 2, l = 0, m = 0, `s=+1/2`
C. n = 3, l = 1, m = 1, `s=+1/2`
D. n = 3, l = 2, m = 1, `s=+1/2`

Answer

To determine which set of quantum numbers corresponds to the orbital of highest energy, we will follow these steps:

### Step-by-Step Solution:

1. **Understand Quantum Numbers**:
- Each set of quantum numbers consists of:
- \( n \) (principal quantum number)
- \( l \) (azimuthal quantum number)
- \( m \) (magnetic quantum number)
- \( s \) (spin quantum number)

2. **Calculate \( n + l \)**:
- The energy of an orbital is primarily determined by the sum of the principal quantum number \( n \) and the azimuthal quantum number \( l \) (i.e., \( n + l \)).
- If two orbitals have the same \( n + l \) value, the one with the higher \( n \) value has higher energy.

3. **Evaluate Each Set of Quantum Numbers**:
- **Option A**: \( n = 4, l = 0 \)
- \( n + l = 4 + 0 = 4 \)
- **Option B**: \( n = 2, l = 0 \)
- \( n + l = 2 + 0 = 2 \)
- **Option C**: \( n = 3, l = 1 \)
- \( n + l = 3 + 1 = 4 \)
- **Option D**: \( n = 3, l = 2 \)
- \( n + l = 3 + 2 = 5 \)

4. **Compare \( n + l \) Values**:
- From the calculations:
- Option A: \( n + l = 4 \)
- Option B: \( n + l = 2 \)
- Option C: \( n + l = 4 \)
- Option D: \( n + l = 5 \)

5. **Determine Highest Energy Orbital**:
- The highest \( n + l \) value is from Option D, which is 5. Therefore, the orbital with the quantum numbers \( n = 3, l = 2, m = 1, s = +\frac{1}{2} \) has the highest energy.

### Conclusion:
The set of quantum numbers that shows the orbital of highest energy is:
- **Option D**: \( n = 3, l = 2, m = 1, s = +\frac{1}{2} \).