For a given principal level `n = 4` the energy of its subshells is of the order

Options

A. `s lt d lt f lt p`
B. `s lt p lt d lt f`
C. `d lt f lt p lt s`
D. `s lt p lt f lt d`

Explanation

To determine the energy order of the subshells for the principal level \( n = 4 \), we can use the \( n + l \) rule. This rule states that the energy of an orbital increases with increasing values of \( n + l \). Here’s how to approach the problem step by step:

### Step-by-Step Solution

1. **Identify the Subshells**: For \( n = 4 \), the possible subshells are:
- 4s
- 4p
- 4d
- 4f

2. **Determine the Values of \( l \)**:
- For the 4s subshell, \( l = 0 \)
- For the 4p subshell, \( l = 1 \)
- For the 4d subshell, \( l = 2 \)
- For the 4f subshell, \( l = 3 \)

3. **Calculate \( n + l \) for Each Subshell**:
- For 4s:
\[ n + l = 4 + 0 = 4 \]
- For 4p:
\[ n + l = 4 + 1 = 5 \]
- For 4d:
\[ n + l = 4 + 2 = 6 \]
- For 4f:
\[ n + l = 4 + 3 = 7 \]

4. **Order the Subshells by Energy**: Based on the \( n + l \) values calculated:
- 4s (4)
- 4p (5)
- 4d (6)
- 4f (7)

The order of increasing energy is:
\[ 4s < 4p < 4d < 4f \]

5. **Conclusion**: The subshells for \( n = 4 \) are ordered by energy as follows:
- 4s has the lowest energy,
- followed by 4p,
- then 4d,
- and finally, 4f has the highest energy.

### Final Answer
The energy of the subshells for \( n = 4 \) is ordered as:
\[ 4s < 4p < 4d < 4f \]