If nitrogen atoms had electronic configuration 1s7 It would have energy lower than that of the nornal ground state configuration `1s^(2) 2s^(2) 2p^(3)` because the electrons would be clear to the nucleus yet `1s^(7)` is not oberved because it violates ?

Options

A. Heisenberg uncertainty principle
B. Hund's rule
C. Pauli's exxlusion principal
D. Bohr's postulate of stationary orbital

Explanation

To solve the question, we need to analyze why the electronic configuration `1s^7` for nitrogen is not observed, despite the claim that it would have lower energy than the normal ground state configuration `1s^2 2s^2 2p^3`.

### Step-by-Step Solution:

1. **Understanding Electron Configuration**:
- The normal ground state configuration of nitrogen is `1s^2 2s^2 2p^3`, which means nitrogen has 7 electrons distributed among its orbitals.

2. **Hypothetical Configuration**:
- The question proposes a hypothetical configuration of `1s^7`. This means that all 7 electrons would be in the 1s orbital.

3. **Energy Consideration**:
- It is suggested that `1s^7` would have lower energy because the electrons would be closer to the nucleus. However, energy levels are not solely determined by proximity to the nucleus; electron-electron interactions and quantum mechanical principles also play a significant role.

4. **Pauli's Exclusion Principle**:
- According to Pauli's Exclusion Principle, no two electrons can have the same set of quantum numbers. In the case of the `1s` orbital, it can hold a maximum of 2 electrons with opposite spins. Therefore, having `1s^7` would violate this principle, as it would require more than 2 electrons to occupy the same orbital.

5. **Conclusion**:
- Since the configuration `1s^7` violates Pauli's Exclusion Principle, it cannot exist. Thus, despite the claim of lower energy, this configuration is not physically realizable.

### Final Answer:
The configuration `1s^7` is not observed because it violates **Pauli's Exclusion Principle**.

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