The user wants me to identify the incorrect statements about the molecular orbital (MO) energy levels of homonuclear diatomic molecules.

Let's analyze each statement based on Molecular Orbital Theory.

The general order of MO energy levels for homonuclear diatomic molecules is:

1. For $Li_2$ to $N_2$ (due to s-p mixing): $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$
2. For $O_2$, $F_2$, $Ne_2$ (s-p mixing is less significant): $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$

The bond order (B.O.) is calculated as: $B.O. = \frac{1}{2} (N_b - N_a)$ where $N_b$ is the number of electrons in bonding molecular orbitals and $N_a$ is the number of electrons in anti-bonding molecular orbitals.

Analysis of Statement A:

A: Bond order of $Ne_2$ is zero.

1. Neon (Ne) has an atomic number of 10. A $Ne_2$ molecule has $2 \times 10 = 20$ electrons.
2. The electronic configuration of $Ne_2$ is: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^2 (\pi^* 2p_y)^2 (\sigma^* 2p_z)^2$
3. Number of bonding electrons ($N_b$) = $2+2+2+2+2 = 10$.
4. Number of anti-bonding electrons ($N_a$) = $2+2+2+2+2 = 10$.
5. Bond order = $\frac{1}{2} (10 - 10) = 0$.
6. A bond order of zero indicates that the molecule is not stable.
7. Thus, the statement is CORRECT.

Analysis of Statement B:

B: The highest occupied molecular orbital (HOMO) of $F_2$ is $\sigma$-type.

1. Fluorine (F) has an atomic number of 9. An $F_2$ molecule has $2 \times 9 = 18$ electrons.
2. The electronic configuration of $F_2$ is: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^2 (\pi^* 2p_y)^2$
3. The highest occupied molecular orbitals (HOMO) are the degenerate $\pi^* 2p_x$ and $\pi^* 2p_y$ orbitals.
4. These are $\pi$-type orbitals, not $\sigma$-type.
5. Thus, the statement is INCORRECT.

Analysis of Statement C:

C: Bond energy of $O_2^{+}$ is smaller than the bond energy of $O_2$.

1. Bond energy is directly proportional to bond order. Higher bond order implies higher bond energy.
2. For $O_2$ (16 electrons): The electronic configuration is ...$(\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$. $N_b = 10$, $N_a = 6$. Bond order = $\frac{1}{2}(10-6) = 2.0$.
3. For $O_2^{+}$ (15 electrons): One electron is removed from the highest energy orbital of $O_2$, which is an anti-bonding $\pi^*$ orbital. The configuration is ...$(\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1$. $N_b = 10$, $N_a = 5$. Bond order = $\frac{1}{2}(10-5) = 2.5$.
4. Since the bond order of $O_2^{+}$ (2.5) is greater than the bond order of $O_2$ (2.0), the bond energy of $O_2^{+}$ is larger than the bond energy of $O_2$.
5. The statement claims the opposite.
6. Thus, the statement is INCORRECT.

Analysis of Statement D:

D: Bond length of $Li_2$ is larger than the bond length of $B_2$.

1. Bond length is inversely proportional to bond order.
2. For $Li_2$ (6 electrons): The electronic configuration is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2$. $N_b = 4$, $N_a = 2$. Bond order = $\frac{1}{2}(4-2) = 1$.
3. For $B_2$ (10 electrons): The electronic configuration is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^1 (\pi 2p_y)^1$. $N_b = 6$, $N_a = 4$. Bond order = $\frac{1}{2}(6-4) = 1$.
4. Both $Li_2$ and $B_2$ have a bond order of 1. When bond orders are the same, bond length is determined by the atomic size of the constituent atoms.
5. Lithium (Li) and Boron (B) are in the same period (Period 2). Atomic radius decreases across a period from left to right. Therefore, the atomic radius of Li is larger than that of B.
6. Consequently, the bond length of $Li_2$ will be larger than the bond length of $B_2$.
7. Thus, the statement is CORRECT.

The question asks for the INCORRECT statements, which are B and C.