When the value of (n + l) is not more than 3 which of the following subshell is not possible to exist?

Options

A. 2s
B. 3s
C. 3p
D. 2p

Explanation

To determine which subshell is not possible when the value of (n + l) is not more than 3, we will analyze the values of the principal quantum number (n) and the azimuthal quantum number (l).

### Step-by-Step Solution:

1. **Understanding Quantum Numbers**:
- The principal quantum number (n) indicates the main energy level or shell.
- The azimuthal quantum number (l) indicates the subshell and can take values from 0 to (n - 1).
- The subshells correspond to values of l:
- l = 0 → s subshell
- l = 1 → p subshell
- l = 2 → d subshell
- l = 3 → f subshell

2. **Setting Up the Condition**:
- We need to find subshells for which (n + l) ≤ 3.

3. **Case Analysis**:
- **Case 1**: If (n + l) = 1
- The only possibility is n = 1, l = 0 → 1s subshell.
- **Case 2**: If (n + l) = 2
- Possible combinations:
- n = 2, l = 0 → 2s subshell.
- n = 1, l = 1 → Not possible since l can only be 0 when n = 1.
- **Case 3**: If (n + l) = 3
- Possible combinations:
- n = 3, l = 0 → 3s subshell.
- n = 2, l = 1 → 2p subshell.
- n = 1, l = 2 → Not possible since l can only be 0 or 1 when n = 1.

4. **Identifying Possible Subshells**:
- From the above cases, the possible subshells are:
- 1s (n=1, l=0)
- 2s (n=2, l=0)
- 2p (n=2, l=1)
- 3s (n=3, l=0)

5. **Finding the Impossible Subshell**:
- The subshells that cannot exist under the condition (n + l) ≤ 3 are:
- 3p (n=3, l=1) → This gives (3 + 1) = 4, which is greater than 3.

### Conclusion:
The subshell that is not possible to exist when (n + l) is not more than 3 is **3p**.