For which of the following options m=0 for all orbitals ?

Options

A. `2s ,2p_(x) , 3d _(xy)`
B. `3s,2p_(z) ,3d_(z^2)`
C. `2s,2p_(z) , 3d _(X^2 -y^2)`
D. `3s , 3p_(x) , 3d _(yz)`

Answer

To solve the question of which option has \( m = 0 \) for all orbitals, we need to analyze each given option based on the magnetic quantum number \( m \).

### Step-by-Step Solution:

1. **Understanding the Magnetic Quantum Number (m)**:
- The magnetic quantum number \( m \) indicates the orientation of the orbital in space.
- The possible values of \( m \) depend on the azimuthal quantum number \( l \) and range from \( -l \) to \( +l \).

2. **Analyzing Each Option**:
- **Option A: 2px, 3dxy**
- For the 2px orbital, \( l = 1 \) (p-orbital), so \( m \) can be -1, 0, or +1. Here, \( m \) is not 0.
- For the 3dxy orbital, \( l = 2 \) (d-orbital), so \( m \) can be -2, -1, 0, +1, or +2. Here, \( m \) is also not 0.
- Conclusion: **Not valid**.

- **Option B: 3s, 2pz, 3dz²**
- For the 3s orbital, \( l = 0 \), so \( m = 0\).
- For the 2pz orbital, \( l = 1 \), and \( m = 0\) is possible.
- For the 3dz² orbital, \( l = 2 \), and \( m = 0\) is also possible.
- Conclusion: **Valid**.

- **Option C: 2pz, 3dx²-y²**
- For the 2pz orbital, \( l = 1 \), and \( m = 0\) is possible.
- For the 3dx²-y² orbital, \( l = 2 \), and \( m \) can be -2, -1, 0, +1, or +2. Here, \( m \) is not 0.
- Conclusion: **Not valid**.

- **Option D: 3px, 3dyz**
- For the 3px orbital, \( l = 1 \), so \( m \) can be -1, 0, or +1. Here, \( m \) is not 0.
- For the 3dyz orbital, \( l = 2 \), and \( m \) can be -2, -1, 0, +1, or +2. Here, \( m \) is also not 0.
- Conclusion: **Not valid**.

3. **Final Conclusion**:
- The only option where \( m = 0 \) for all orbitals is **Option B: 3s, 2pz, 3dz²**.

### Answer:
**Option B: 3s, 2pz, 3dz²** is the correct answer.