Incorrect Statements Regarding Molecular Orbital Theory for Diatomic Molecules

To determine which of the given statements regarding Molecular Orbital (MO) theory are INCORRECT, we analyze each option individually:

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Analysis of Option (A):

• Molecule: $\text{Ne}_2$ (Total number of electrons = $20$)
• MO Electronic Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^2 (\pi^* 2p_y)^2 (\sigma^* 2p_z)^2$
• Bond Order Calculation: $\text{Bond Order (B.O.)} = \frac{N_b - N_a}{2} = \frac{10 - 10}{2} = 0$
• Thus, the bond order of $\text{Ne}_2$ is zero.
• Statement (A) is CORRECT.

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Analysis of Option (B):

• Molecule: $\text{F}_2$ (Total number of electrons = $18$)
• MO Electronic Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^2 (\pi^* 2p_y)^2$
• The Highest Occupied Molecular Orbital (HOMO) contains the last filled electrons, which reside in the $\pi^* 2p_x$ and $\pi^* 2p_y$ antibonding orbitals.
• Therefore, the HOMO of $\text{F}_2$ is of $\pi$-type symmetry ($\pi^*$), not $\sigma$-type.
• Statement (B) is INCORRECT.

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Analysis of Option (C):

• Molecules: $\text{O}_2$ ($16$ electrons) and $\text{O}_2^+$ ($15$ electrons)
• For $\text{O}_2$:
  • MO Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$
  • Number of bonding electrons ($N_b$) = $10$, antibonding electrons ($N_a$) = $6$
  • $\text{B.O. of O}_2 = \frac{10 - 6}{2} = 2.0$
• For $\text{O}_2^+$:
  • One electron is removed from an antibonding $\pi^*$ orbital.
  • Number of bonding electrons ($N_b$) = $10$, antibonding electrons ($N_a$) = $5$
  • $\text{B.O. of O}_2^+ = \frac{10 - 5}{2} = 2.5$

Since Bond Energy is directly proportional to Bond Order: $\text{Bond Energy of } \text{O}_2^+ > \text{Bond Energy of } \text{O}_2$

• Therefore, the statement claiming that the bond energy of $\text{O}_2^+$ is smaller than that of $\text{O}_2$ is wrong.
• Statement (C) is INCORRECT.

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Analysis of Option (D):

• Molecules: $\text{Li}_2$ ($6$ electrons) and $\text{B}_2$ ($10$ electrons)
• For $\text{Li}_2$:
  • MO Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2$
  • $\text{B.O.} = \frac{4 - 2}{2} = 1$
• For $\text{B}_2$:
  • MO Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^1 (\pi 2p_y)^1$
  • $\text{B.O.} = \frac{6 - 4}{2} = 1$

Although both molecules have a bond order of $1$, moving across the period from Lithium ($\text{Li}$) to Boron ($\text{B}$) increases the effective nuclear charge ($Z_{\text{eff}}$), resulting in a smaller atomic radius for Boron ($r_{\text{B}} < r_{\text{Li}}$). Consequently, the bond length of $\text{Li}_2$ ($\approx 267\text{ pm}$) is significantly larger than that of $\text{B}_2$ ($\approx 159\text{ pm}$).

• Statement (D) is CORRECT.

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Conclusion:

The incorrect statements are B and C.

Correct Answer: B, C