Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Which one of the following pairs of species have the same bond order?

Have you ever looked at a Molecular Orbital Theory question and felt a wave of dread? Drawing those massive energy level diagrams for every single molecule in the options can eat up precious minutes in an exam. But what if I told you there's a cheat code? A beautiful, elegant shortcut that can solve this in seconds?

In the quantum world of molecules, electrons don't really care which atoms they came from once they are shared. They just fill up the available molecular orbitals from the bottom up. This leads to a profound realization: if two diatomic species have the exact same total number of electrons, they will fill the exact same molecular orbitals.

We call these isoelectronic species. And because their electron configurations are identical, their bond orders must be identical too!

Let's put this shortcut to the test with our first pair: ${\text{CN}}^{-}$ and ${\text{NO}}^{+}$.

First, the cyanide ion, ${\text{CN}}^{-}$. Carbon brings $6$ electrons to the table. Nitrogen brings $7$. And that little negative charge? That's one extra electron joining the party. So, $6+7+1=14$ electrons.

Next, the nitrosonium ion, ${\text{NO}}^{+}$. Nitrogen has $7$ electrons, and oxygen has $8$. But wait, there's a positive charge! That means one electron has been kicked out. So, $7+8-1=14$ electrons.

Look at that! Both species have exactly $14$ electrons. They are perfectly isoelectronic.

Now, think of the most famous $14$-electron molecule: Nitrogen gas, ${\text{N}}_2$. We know ${\text{N}}_2$ has a triple bond. Therefore, any diatomic species with $14$ electrons will also have a Bond Order of 3.

Without drawing a single orbital, we've proven that both ${\text{CN}}^{-}$ and ${\text{NO}}^{+}$ have a bond order of $3$. They are a perfect match, making option (a) our undisputed correct answer. The next time you see a bond order question, don't panic—just count the electrons!