`sigma2s, pi^(**) (2p_(x))` and `pi(2p_(x))` are gerade `MO` .

Answer

To determine whether the statement "sigma2s, pi^(**) (2p_(x)) and pi(2p_(x)) are gerade molecular orbitals" is correct, we need to analyze each of the mentioned molecular orbitals in terms of their nodal planes.

### Step-by-Step Solution:

1. **Understanding Molecular Orbitals**:
- Molecular orbitals (MOs) are formed from the combination of atomic orbitals. They can be classified as either "gerade" (g) or "ungerade" (u) based on their symmetry with respect to inversion through the center of the molecule.

2. **Analyzing Sigma 2s**:
- The sigma 2s molecular orbital is formed by the constructive interference of two 2s atomic orbitals.
- In this case, there are no nodal planes present.
- Since there are zero nodal planes, sigma 2s is classified as a gerade molecular orbital (g).

3. **Analyzing Pi Star 2p_x (π*)**:
- The pi star 2p_x molecular orbital is formed by the sidewise overlapping of two 2p_x atomic orbitals, which leads to destructive interference.
- This results in two nodal planes: one vertical and one horizontal.
- Since there are two nodal planes, pi star 2p_x is also classified as a gerade molecular orbital (g).

4. **Analyzing Pi 2p_x (π)**:
- The pi 2p_x molecular orbital is formed by the constructive interference of two 2p_x atomic orbitals.
- In this case, there is only one nodal plane, which is horizontal.
- Since there is one nodal plane, pi 2p_x is classified as an ungerade molecular orbital (u).

5. **Conclusion**:
- Based on the analysis:
- Sigma 2s is gerade (g).
- Pi star 2p_x is gerade (g).
- Pi 2p_x is ungerade (u).
- Therefore, the statement "sigma2s, pi^(**) (2p_(x)) and pi(2p_(x)) are gerade molecular orbitals" is **false**.

### Final Answer:
The statement is false because only sigma 2s and pi star 2p_x are gerade molecular orbitals, while pi 2p_x is ungerade.

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