The decreasing order of energy of the `3d,4s,3p,3s` orbital is :-

Options

A. `3dgt3sgt4sgt3p`
B. `3sgt4sgt3pgt3d`
C. `3dgt4sgt3pgt3s`
D. `4sgt3dgt3sgt3p`

Explanation

To determine the decreasing order of energy of the orbitals `3d`, `4s`, `3p`, and `3s`, we will use the n + l rule. This 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 larger the value of n + l, the higher the energy of the orbital. If two orbitals have the same n + l value, the one with the higher n value will have the higher energy.

### Step-by-Step Solution:

1. **Identify the quantum numbers for each orbital:**
- For `3d`:
- n = 3 (principal quantum number)
- l = 2 (for d orbitals)
- n + l = 3 + 2 = 5
- For `4s`:
- n = 4
- l = 0 (for s orbitals)
- n + l = 4 + 0 = 4
- For `3p`:
- n = 3
- l = 1 (for p orbitals)
- n + l = 3 + 1 = 4
- For `3s`:
- n = 3
- l = 0 (for s orbitals)
- n + l = 3 + 0 = 3

2. **List the n + l values:**
- `3d`: n + l = 5
- `4s`: n + l = 4
- `3p`: n + l = 4
- `3s`: n + l = 3

3. **Determine the order of energy based on n + l values:**
- The highest n + l value is for `3d` (5), followed by `4s` and `3p` (both 4), and then `3s` (3).
- Since `4s` and `3p` have the same n + l value, we compare their n values. `4s` has a higher n (4) than `3p` (3).

4. **Write the decreasing order of energy:**
- Therefore, the decreasing order of energy is:
- `3d` > `4s` > `3p` > `3s`

### Final Answer:
The decreasing order of energy of the orbitals is: **3d > 4s > 3p > 3s**