In multi-electron atom 4s-orbital is lower in energy than

Options

A. `3d - ` orbital
B. 3p- orbital
C. 2s - orbital
D. 2p - orbital

Answer

To determine which orbital has a higher energy than the 4s orbital in a multi-electron atom, we can use the n + l rule. Here’s how to solve the problem step by step:

### Step-by-Step Solution:

1. **Understand the n + l Rule**:
The n + l rule states that the energy of an orbital is determined by the sum of its principal quantum number (n) and its azimuthal quantum number (l). The lower the value of n + l, the lower the energy of the orbital.

2. **Identify the Values of n and l**:
- For the **4s orbital**:
- n = 4 (principal quantum number)
- l = 0 (for s orbitals)
- Therefore, n + l = 4 + 0 = 4.

- For the **3d orbital**:
- n = 3
- l = 2 (for d orbitals)
- Therefore, n + l = 3 + 2 = 5.

- For the **3p orbital**:
- n = 3
- l = 1 (for p orbitals)
- Therefore, n + l = 3 + 1 = 4.

- For the **2s orbital**:
- n = 2
- l = 0
- Therefore, n + l = 2 + 0 = 2.

- For the **2p orbital**:
- n = 2
- l = 1
- Therefore, n + l = 2 + 1 = 3.

3. **Calculate and Compare n + l Values**:
- 4s: n + l = 4
- 3d: n + l = 5
- 3p: n + l = 4
- 2s: n + l = 2
- 2p: n + l = 3

4. **Arrange the Orbitals by Energy**:
- The lowest n + l value is for 2s (2), followed by 2p (3), then 4s and 3p (both 4), and finally 3d (5).
- Since 3p has the same n + l value as 4s, we compare their n values. The 3p orbital has a lower n (3) than the 4s (4), so 3p is lower in energy than 4s.

5. **Conclusion**:
- The order of increasing energy is: 2s < 2p < 3p < 4s < 3d.
- Therefore, the 4s orbital is lower in energy than the 3d orbital.

### Final Answer:
The 4s orbital is lower in energy than the **3d orbital**.