Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: In the following the correct bond order sequence is

Imagine you are looking at the very bonds holding atoms together. In the quantum world, electrons don't just orbit nuclei; they occupy molecular orbitals that span across the entire molecule. Some of these orbitals pull the atoms together (bonding orbitals), while others actively push them apart (anti-bonding orbitals).

In this classic Molecular Orbital Theory problem, we are tasked with finding the correct sequence of bond orders for oxygen ($O_2$) and its various ionic forms: $O_2^{+}$, $O_2^{-}$, and $O_2^{2-}$. To do this, we need to understand how adding or removing electrons affects the delicate balance of power between bonding and anti-bonding forces.

The strength of a bond is quantified by its Bond Order, which is calculated using a beautifully simple formula:

Here, $N_b$ represents the number of electrons in bonding orbitals, and $N_a$ represents the number of electrons in anti-bonding orbitals. For all the oxygen species we are considering, the inner molecular orbitals are completely filled, giving us a constant $N_b=10$. The real action happens in the highest energy levels—the anti-bonding ${\pi }^{\ast }2p$ orbitals. Every electron added here increases $N_a$ and weakens the bond.

Let's break down each species one by one:

1. The $O_2^{+}$ Ion (15 electrons): With one electron removed from neutral oxygen, we have $10$ bonding electrons and only $5$ anti-bonding electrons. This gives $O_2^{+}$ the strongest and shortest bond of the group.

2. The Neutral $O_2$ Molecule (16 electrons): This is the oxygen we breathe. It has $10$ bonding electrons and $6$ anti-bonding electrons. It has a solid double bond.

3. The Superoxide Ion, $O_2^{-}$ (17 electrons): By adding an extra electron, it goes straight into the anti-bonding ${\pi }^{\ast }2p$ orbital, increasing $N_a$ to $7$. The bond is now noticeably weaker.

4. The Peroxide Ion, $O_2^{2-}$ (18 electrons): Adding yet another electron fills the anti-bonding ${\pi }^{\ast }2p$ orbitals further, bringing $N_a$ to $8$. This is the weakest bond, essentially a single bond.

Comparing our results, we can clearly see the trend. As we add electrons to the anti-bonding orbitals, the bond order steadily decreases.

The correct sequence is: $O_2^{+}>O_2>O_2^{-}>O_2^{2-}$

This perfectly matches option (c). Always remember this fundamental principle: electrons in anti-bonding orbitals are the enemies of bond stability!