Arrange the following orbitals in decreasing order of energy.
A. `n=3,l=0,m=0`
B. `n=4,1=0,m=0`
C. `n=1,l=1,m=0`
D. `n=3,l=2,m=1`
The correct option for the order is :

Options

A. `D gt B gt C gt A`
B. `B gt D gt C gt A`
C. ` A gt C gt B gt D`
D. `D gt B gt A gt C`

Answer

To arrange the given orbitals in decreasing order of energy, we will apply the N + L rule. According to this rule, the energy of an orbital increases with increasing values of the sum of the principal quantum number (n) and the azimuthal quantum number (l). If two orbitals have the same N + L value, the one with the higher n value will have higher energy.

Let's analyze each of the given orbitals:

1. **Orbital A: `n=3, l=0, m=0`**
- N + L = 3 + 0 = 3

2. **Orbital B: `n=4, l=0, m=0`**
- N + L = 4 + 0 = 4

3. **Orbital C: `n=1, l=1, m=0`**
- N + L = 1 + 1 = 2

4. **Orbital D: `n=3, l=2, m=1`**
- N + L = 3 + 2 = 5

Now, we will list the N + L values for each orbital:

- A: N + L = 3
- B: N + L = 4
- C: N + L = 2
- D: N + L = 5

Next, we will arrange them in decreasing order of energy based on their N + L values:

1. Orbital D (N + L = 5)
2. Orbital B (N + L = 4)
3. Orbital A (N + L = 3)
4. Orbital C (N + L = 2)

Thus, the correct order of energy from highest to lowest is: **D > B > A > C**.

### Final Answer:
The correct option for the order is: **D, B, A, C**.