Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT statement(s) is(are)

The fascinating world of Molecular Orbital (MO) Theory allows us to understand the deep quantum mechanical nature of chemical bonds. Unlike simple Lewis structures, MO theory explains why oxygen is paramagnetic and why certain molecules like ${\text{Ne}}_2$ simply refuse to exist. In this problem, we are tasked with evaluating four statements about homonuclear diatomic molecules. Let's embark on this journey and dissect each option one by one!

Why ${\text{Ne}}_2$ Doesn't Exist

Let's start by looking at the first statement regarding the bond order of ${\text{Ne}}_2$. A single Neon atom is a noble gas with a completely filled valence shell, possessing 10 electrons. When two Neon atoms approach each other to form a hypothetical ${\text{Ne}}_2$ molecule, we have a total of 20 electrons to accommodate in our molecular orbitals.

According to the Aufbau principle, we fill the lowest energy orbitals first. The electronic configuration for ${\text{Ne}}_2$ becomes:

Notice something striking? Every single bonding orbital is perfectly matched by a completely filled anti-bonding orbital.

The formula for Bond Order is beautifully simple: where $N_b$ is the number of bonding electrons and $N_a$ is the number of anti-bonding electrons.

For ${\text{Ne}}_2$, we have 10 bonding electrons and 10 anti-bonding electrons.

A bond order of zero physically means there is no net stabilizing force holding the two atoms together. The molecule simply does not exist under normal conditions. Therefore, Statement A is absolutely correct.

Moving on to the second statement, we need to determine the nature of the Highest Occupied Molecular Orbital (HOMO) for the fluorine molecule, ${\text{F}}_2$.

A single fluorine atom has 9 electrons, so ${\text{F}}_2$ has a total of 18 electrons. Let's fill the molecular orbitals for ${\text{F}}_2$:

The last electrons we placed went into the ${\pi }_{2p_x}^{\ast }$ and ${\pi }_{2p_y}^{\ast }$ orbitals. These are the highest energy orbitals that contain electrons, making them the HOMO of the molecule.

Crucially, these orbitals are formed by the side-by-side overlap of $p$-orbitals, which makes them $\pi$-type orbitals. Statement B claims that the HOMO of ${\text{F}}_2$ is $\sigma$-type. This is a direct contradiction of our derived configuration! Thus, Statement B is incorrect.

The third statement compares the bond energy of the oxygen molecule (${\text{O}}_2$) with its cation (${\text{O}}_2^{+}$). This is where MO theory truly shines.

Let's look at neutral ${\text{O}}_2$, which has 16 electrons. Its configuration ends with two unpaired electrons in the anti-bonding ${\pi }^{\ast }$ orbitals: Calculating its bond order:

Now, what happens when we ionize ${\text{O}}_2$ to form ${\text{O}}_2^{+}$? We must remove one electron. But from where? The electron is removed from the highest energy level, which is the anti-bonding ${\pi }^{\ast }$ orbital.

Removing an electron from an anti-bonding orbital actually increases the stability of the molecule! Let's calculate the new bond order for ${\text{O}}_2^{+}$ (15 electrons):

A higher bond order directly correlates to a stronger, more tightly held bond. Therefore, the bond energy of ${\text{O}}_2^{+}$ is strictly greater than the bond energy of ${\text{O}}_2$.

Statement C claims that the bond energy of ${\text{O}}_2^{+}$ is smaller. This is a classic trap, and it is incorrect.

Finally, let's evaluate the fourth statement regarding the bond lengths of ${\text{Li}}_2$ and ${\text{B}}_2$. This requires us to step back from MO theory for a moment and recall our periodic trends.

Lithium (Li) and Boron (B) are both in the second period of the periodic table. As we move from left to right across a period, the effective nuclear charge increases, pulling the electron cloud closer to the nucleus. Consequently, atomic radius decreases.

This means that a Lithium atom is significantly larger than a Boron atom. When two large Lithium atoms bond to form ${\text{Li}}_2$, their nuclei are kept further apart compared to the smaller Boron atoms in ${\text{B}}_2$.

Therefore, the bond length of ${\text{Li}}_2$ is naturally larger than the bond length of ${\text{B}}_2$. Statement D is correct.

By systematically applying Molecular Orbital Theory and periodic trends, we have successfully navigated through the options. We discovered that the HOMO of ${\text{F}}_2$ is $\pi$-type (making B incorrect) and that removing an anti-bonding electron from ${\text{O}}_2$ increases its bond energy (making C incorrect).

Since the question asks for the INCORRECT statements, our final answers are (B) and (C).