Arrabge in decreasing order, the energy of 2s orbital in the following atoms H, Li, Na, K

Options

A. `E_(2s) (H) gt E_(2s) (Li) gt E_(2s) (Na) gt E_(2s) (K)`
B. `E_(2s) (H) gt E_(2s) (Na) gt E_(2s) (Li) gt E_(2s) (K)`
C. `E_(2s) (H) gt E_(2s) (Na) = E_(2s) (K) gt E_(2s) (Li)`
D. `E_(2s) (K) gt E_(2s) (Na) gt E_(2s) (Li) gt E_(2s) (H)`

Explanation

To arrange the energy of the 2s orbital in the atoms H (Hydrogen), Li (Lithium), Na (Sodium), and K (Potassium) in decreasing order, we need to consider the atomic structure and the effect of increasing atomic number on the energy levels of orbitals.

### Step-by-Step Solution:

1. **Identify the Atomic Numbers**:
- Hydrogen (H) has an atomic number of 1.
- Lithium (Li) has an atomic number of 3.
- Sodium (Na) has an atomic number of 11.
- Potassium (K) has an atomic number of 19.

2. **Understand the Concept of Orbital Energy**:
- In general, as the atomic number increases, the energy of the orbitals also increases. This is due to the increasing nuclear charge which affects the energy levels of the electrons.

3. **Determine the Energy Order**:
- The 2s orbital energy is influenced by the effective nuclear charge experienced by the electrons in that orbital.
- For hydrogen, with only one electron, the 2s orbital has the lowest energy.
- As we move to lithium, sodium, and potassium, the 2s orbital energy increases because of the increased electron shielding and the effective nuclear charge.

4. **Arrange the Atoms**:
- Based on the above reasoning, we can arrange the atoms in decreasing order of the energy of their 2s orbital:
- Hydrogen (H) has the highest energy for its 2s orbital.
- Lithium (Li) comes next.
- Sodium (Na) follows.
- Potassium (K) has the lowest energy for its 2s orbital.

5. **Final Order**:
- The final order of the energy of the 2s orbital from highest to lowest is:
- H > Li > Na > K

### Conclusion:
The decreasing order of the energy of the 2s orbital in the given atoms is:
**H > Li > Na > K**