If the value of` (n + l)` is more than `3` and less than `6` , then what will be the possible number of orbitals ?

Options

A. (1) 6
B. (2) 9
C. (3) 10
D. (4) 13

Explanation

To solve the question regarding the possible number of orbitals when the value of \( n + l \) is more than 3 and less than 6, we will follow these steps:

### Step 1: Understand the Range of \( n + l \)
The problem states that \( n + l \) is greater than 3 and less than 6. Therefore, the possible integer values for \( n + l \) are:
- \( n + l = 4 \)
- \( n + l = 5 \)

### Step 2: Determine Possible Values of \( n \) and \( l \)
We know that:
- \( n \) (the principal quantum number) can take values starting from 1 and going to infinity.
- \( l \) (the azimuthal quantum number) can take values from 0 to \( n - 1 \).

Now, we will find combinations of \( n \) and \( l \) for both cases of \( n + l \).

#### Case 1: \( n + l = 4 \)
1. If \( n = 3 \), then \( l = 1 \) (3 + 1 = 4).
- This corresponds to the 3p subshell, which has **3 orbitals** (px, py, pz).

2. If \( n = 4 \), then \( l = 0 \) (4 + 0 = 4).
- This corresponds to the 4s subshell, which has **1 orbital**.

3. If \( n = 4 \), then \( l = 1 \) (4 + 1 = 5).
- This corresponds to the 4p subshell, which has **3 orbitals**.

#### Case 2: \( n + l = 5 \)
1. If \( n = 3 \), then \( l = 2 \) (3 + 2 = 5).
- This corresponds to the 3d subshell, which has **5 orbitals**.

2. If \( n = 4 \), then \( l = 1 \) (4 + 1 = 5).
- This corresponds to the 4p subshell, which has **3 orbitals**.

3. If \( n = 5 \), then \( l = 0 \) (5 + 0 = 5).
- This corresponds to the 5s subshell, which has **1 orbital**.

### Step 3: Count the Total Number of Orbitals
Now, we will sum up the number of orbitals from the valid combinations we found:

- From \( n + l = 4 \):
- 3p: 3 orbitals
- 4s: 1 orbital
- 4p: 3 orbitals

- From \( n + l = 5 \):
- 3d: 5 orbitals
- 4p: 3 orbitals
- 5s: 1 orbital

Now, let's count them:
- For \( n + l = 4 \): \( 3 + 1 + 3 = 7 \) orbitals
- For \( n + l = 5 \): \( 5 + 3 + 1 = 9 \) orbitals

### Step 4: Final Count
Adding both cases together:
- Total orbitals = \( 7 + 9 = 16 \)

### Conclusion
Thus, the total number of orbitals possible when \( n + l \) is more than 3 and less than 6 is **16**.

---