Which of the following representations of excited states of atoms is impossible?

Options

A. (a) `1s^(1)2s^(1)`
B. (b) `[Ne]3s^(2)3p^(3)4s^(1)`
C. (c) `[Ne]3s^(2)3p^(6)4s^(1)3d^(6)`
D. (d) `1s^(2)2s^(2)2p^(7)3s^(2)`

Answer

To determine which representation of excited states of atoms is impossible, we need to understand the rules governing the quantum numbers and the maximum number of electrons that can occupy each subshell.

### Step-by-Step Solution:

1. **Understanding Quantum Numbers**:
- The principal quantum number (n) indicates the energy level or shell.
- The angular momentum quantum number (l) indicates the subshell type:
- For s subshell, l = 0 (can hold 2 electrons)
- For p subshell, l = 1 (can hold 6 electrons)
- For d subshell, l = 2 (can hold 10 electrons)
- For f subshell, l = 3 (can hold 14 electrons)
- The magnetic quantum number (m) indicates the orientation of the orbital and can take values from -l to +l.

2. **Analyzing Each Option**:
- **Option 1**: 1s², 2s¹
- 1s can hold 2 electrons, and 2s can hold 2 electrons. This configuration is possible.
- **Option 2**: 2s², 3p³, 4s¹
- 2s can hold 2 electrons, 3p can hold 6 electrons, and 4s can hold 2 electrons. This configuration is also possible.
- **Option 3**: 3s², 3p⁶, 4s¹, 3d⁶
- 3s can hold 2 electrons, 3p can hold 6 electrons, 4s can hold 2 electrons, and 3d can hold 10 electrons. This configuration is also possible.
- **Option 4**: 2p⁷
- The p subshell can only hold a maximum of 6 electrons. Therefore, having 2p⁷ is impossible.

3. **Conclusion**:
- The impossible representation of excited states of atoms is **2p⁷** since the p subshell can only accommodate a maximum of 6 electrons.

### Final Answer:
The impossible representation of excited states of atoms is **2p⁷**.

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