Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT statement(s) is (are)

To determine the incorrect statement(s) regarding the molecular orbital (MO) energy levels of the given homonuclear diatomic molecules and ions, we analyze each statement individually using Molecular Orbital Theory.

1. Analysis of Option A (Bond order of Ne₂ is zero):
Neon (Ne) has an atomic number of 10. The diatomic molecule Ne₂ has a total of 20 electrons.
The molecular orbital electronic configuration for Ne₂ (having more than 14 electrons) is:
$\sigma 1s^2 {\sigma }^{*}1s^2 \sigma 2s^2 {\sigma }^{*}2s^2 \sigma 2p_z{}^2 \pi 2p_x{}^2=\pi 2p_y{}^2 {\pi }^{*}2p_x{}^2={\pi }^{*}2p_y{}^2 {\sigma }^{*}2p_z{}^2$
The number of bonding electrons (N_b) is 10, and the number of antibonding electrons (N_a) is 10.
The bond order (BO) is given by:
$\text{Bond Order}=\frac{N_b-N_a}{2}=\frac{10-102}{2}=0$
Thus, this statement is correct.

2. Analysis of Option B (The highest occupied molecular orbital (HOMO) of F₂ is σ-type):
Fluorine (F) has an atomic number of 9. F₂ contains 18 electrons.
The molecular orbital electronic configuration of F₂ is:
$\sigma 1s^2 {\sigma }^{*}1s^2 \sigma 2s^2 {\sigma }^{*}2s^2 \sigma 2p_z{}^2 \pi 2p_x{}^2=\pi 2p_y{}^2 {\pi }^{*}2p_x{}^2={\pi }^{*}2p_y{}^2$
The highest energy orbital containing electrons (HOMO) is the antibonding π* orbital (specifically, ${\pi }^{*}2p_x$ and ${\pi }^{*}2p_y$). This is a π-type orbital, not a σ-type orbital.
Therefore, this statement is incorrect.

3. Analysis of Option C (Bond energy of O₂⁺ is smaller than the bond energy of O₂):
Oxygen (O) has 8 electrons. O₂ has 16 electrons, and O₂⁺ has 15 electrons.
The configuration of O₂ is:
$\sigma 1s^2 {\sigma }^{*}1s^2 \sigma 2s^2 {\sigma }^{*}2s^2 \sigma 2p_z{}^2 \pi 2p_x{}^2=\pi 2p_y{}^2 {\pi }^{*}2p_x{}^1={\pi }^{*}2p_y{}^1$
$\text{Bond Order of}{\text{O}}_2=\frac{10-6}{2}=2$
For O₂⁺, one electron is removed from the antibonding ${\pi }^{*}$ orbital:
$\text{Bond Order of}{\text{O}}_2^{+}=\frac{10-5}{2}=2.5$
Since bond order is directly proportional to bond energy, a higher bond order for O₂⁺ (2.5) means it has a larger bond energy than O₂ (2.0). Thus, the bond energy of O₂⁺ is greater than that of O₂.
Therefore, this statement is incorrect.

4. Analysis of Option D (Bond length of Li₂ is larger than the bond length of B₂):
Li₂ has 6 electrons (configuration: $\sigma 1s^2 {\sigma }^{*}1s^2 \sigma 2s^2$) with a Bond Order of 1.
B₂ has 10 electrons (configuration: $\sigma 1s^2 {\sigma }^{*}1s^2 \sigma 2s^2 {\sigma }^{*}2s^2 \pi 2p_x{}^1=\pi 2p_y{}^1$) with a Bond Order of 1.
Since both have a bond order of 1, we look at the atomic sizes. Boron is smaller in size than lithium due to a higher effective nuclear charge across the period. Therefore, the bond length of Li₂ is larger than that of B₂. This statement is correct.

Hence, the incorrect statements are:
- The highest occupied molecular orbital (HOMO) of F₂ is σ-type.
- Bond energy of O₂⁺ is smaller than the bond energy of O₂.