Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Among the following molecules/ions, $C_2^{2-},N_2^{2-},O_2^{2-},O_2$ Which one is diamagnetic and has the shortest bond length?

Imagine you are an architect of molecules. Your job is to build the strongest, tightest bridge between two atoms. How do you do it? You look at the blueprints provided by Molecular Orbital Theory (MOT).

In this thrilling problem, we are tasked with finding a molecule or ion that satisfies two strict conditions: it must be diamagnetic (meaning all its electrons are perfectly paired up, leaving no magnetic loose ends), and it must have the shortest bond length.

Bond Order

Before we dive into the electron configurations, we need to establish a fundamental rule of chemical architecture:

This simple relationship tells us everything. The Bond Order is essentially the net number of chemical bonds between two atoms. A higher bond order means more electrons are actively pulling the nuclei together, resulting in a stronger, tighter, and therefore shorter bond.

To find the bond order, we use the formula:

Where $N_b$ is the number of electrons in bonding orbitals (the glue) and $N_a$ is the number of electrons in anti-bonding orbitals (the anti-glue).

Drawing full MO diagrams for every species in an exam is a recipe for running out of time. Instead, use the 14-electron rule.

Any diatomic species with exactly 14 electrons (like $N_2$, $CO$, or $C{N}^{-}$) has a perfectly filled set of bonding orbitals up to ${\sigma }_{2p_z}$, giving it a maximum bond order of 3. For every electron you add or remove from 14, the bond order drops by exactly 0.5.

Let's put our contenders in the ring and see how they stack up.

1. The Champion: $C_2^{2-}$ Let's count the electrons. A neutral carbon atom has 6 electrons. Two carbons give us 12. Add 2 more for the negative charge, and we have exactly 14 electrons.

Because it has 14 electrons, it is isoelectronic with $N_2$. Its configuration fills up perfectly without any unpaired electrons:

Calculating the bond order: $(10-4)/2=3$. Since all electrons are paired, it is diamagnetic. With a massive bond order of 3, it boasts the shortest bond length. We have a very strong candidate here!

2. The Classic Case: $O_2$ Oxygen is the classic textbook example of MOT. It has $8\times 2=16$ electrons. When we fill the orbitals, the last two electrons must go into the degenerate ${\pi }^{\ast }$ anti-bonding orbitals. According to Hund's rule, they take separate orbitals and remain unpaired:

Because of these two unpaired electrons, $O_2$ is paramagnetic. Its bond order is $(10-6)/2=2$. It fails our diamagnetic test immediately.

3. The Peroxide Ion: $O_2^{2-}$ Take an $O_2$ molecule and force two more electrons into it. Now we have 18 electrons. Those two extra electrons pair up with the lonely electrons in the ${\pi }^{\ast }$ orbitals:

Now, all electrons are paired, making it diamagnetic. However, because we added electrons to anti-bonding orbitals, the bond order plummets: $(10-8)/2=1$. A bond order of 1 means a very long, weak bond. It fails the 'shortest bond' test.

4. The Imposter: $N_2^{2-}$ Nitrogen usually has 14 electrons, but this ion has an extra 2, bringing the total to 16 electrons. Wait a minute... 16 electrons? That's exactly the same number as $O_2$!

Because it is isoelectronic with $O_2$, it will have the exact same valence electron configuration. It will have two unpaired electrons in its ${\pi }^{\ast }$ orbitals, making it paramagnetic with a bond order of 2. It fails the test.

After evaluating all the candidates, $C_2^{2-}$ stands victorious. It is the only species that is both diamagnetic and possesses the highest possible bond order of 3, guaranteeing the shortest bond length.

Mastering these electron counts and the resulting magnetic properties is a superpower in chemistry. Keep practicing, and soon you'll be visualizing these orbitals in your sleep!