Which of the following statement is not correct from the point of view of molecular orbital ?

Options

A. `Be_(2)` is not ,a stable molecule
B. `He_(2)` is not stable but `He_(2)^(+)` is expected to exist
C. Bond strength of `N_(2)` is maximum amongst th homonuclear diatomic molecules
D. The order of energies of molecular orbitals in `F_(2)` molecules is `pi_(2px)=pi_(2py) lt sigma_(2pz)`

Answer

To determine which statement is not correct from the point of view of molecular orbital theory, we will analyze each statement provided in the question step by step.

### Step 1: Analyze the first statement
**Statement:** BE2 is not a stable molecule.

**Explanation:**
- Beryllium (Be) has an atomic number of 4, meaning it has 4 electrons. For 2 Be atoms, the total number of electrons is 8.
- The molecular orbital configuration for Be2 is:
- σ1s², σ*1s², σ2s², σ*2s²
- The bond order is calculated as:
\[ \text{Bond Order} = \frac{(\text{Number of electrons in bonding MOs}) - (\text{Number of electrons in antibonding MOs})}{2} \]
Here, we have 4 electrons in bonding orbitals (σ1s², σ2s²) and 4 electrons in antibonding orbitals (σ*1s², σ*2s²).
\[ \text{Bond Order} = \frac{4 - 4}{2} = 0 \]
- A bond order of 0 indicates that Be2 is not stable. This statement is correct.

### Step 2: Analyze the second statement
**Statement:** HE2 is not stable but HE2+ is expected to exist.

**Explanation:**
- Helium (He) has an atomic number of 2, meaning it has 2 electrons. For 2 He atoms, the total number of electrons is 4.
- The molecular orbital configuration for He2 is:
- σ1s², σ*1s²
- The bond order is:
\[ \text{Bond Order} = \frac{(2) - (2)}{2} = 0 \]
- Thus, He2 is not stable.
- For He2+, one electron is removed, leaving 3 electrons:
- The configuration becomes σ1s², σ*1s¹.
- The bond order for He2+ is:
\[ \text{Bond Order} = \frac{(2) - (1)}{2} = \frac{1}{2} = 0.5 \]
- A bond order of 0.5 indicates that He2+ can exist. This statement is also correct.

### Step 3: Analyze the third statement
**Statement:** The bond strength of N2 is maximum among the homonuclear diatomic molecules.

**Explanation:**
- For nitrogen (N), with an atomic number of 7, the total number of electrons for N2 is 14.
- The molecular orbital configuration for N2 is:
- σ1s², σ*1s², σ2s², σ*2s², π2p_x², π2p_y², σ2p_z²
- The bond order is:
\[ \text{Bond Order} = \frac{(10) - (4)}{2} = 3 \]
- N2 indeed has the highest bond order (and thus bond strength) among homonuclear diatomic molecules. This statement is correct.

### Step 4: Analyze the fourth statement
**Statement:** The order of energy of molecular orbitals in F2 is 2Px, 2Py, and then 2Pz.

**Explanation:**
- For fluorine (F), with an atomic number of 9, the total number of electrons for F2 is 18.
- The molecular orbital configuration for F2 is:
- σ1s², σ*1s², σ2s², σ*2s², σ2p_z², π2p_x², π2p_y²
- In F2, due to the mixing of 2s and 2p orbitals, the order of energy levels is:
- σ2p_z has lower energy than π2p_x and π2p_y.
- Therefore, the correct order is:
- σ2p_z < π2p_x = π2p_y
- The statement given is incorrect as it suggests that 2Pz is of higher energy than 2Px and 2Py.

### Conclusion
The statement that is not correct from the point of view of molecular orbital theory is the fourth statement regarding the order of energy of molecular orbitals in F2.

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