Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The difference between bond orders of $\text{CO}$ and ${\text{NO}}^{\oplus }$ is $x/2$. where, $x=\dots \dots$ . (Round off to the nearest integer).

Decoding Bond Orders

Chemical bonding is filled with beautiful symmetries, and one of the most elegant concepts is the Molecular Orbital Theory (MOT). In this problem, we are tasked with finding the difference between the bond orders of carbon monoxide ($\text{CO}$) and the nitrosonium ion (${\text{NO}}^{\oplus }$), and then equating that difference to $x/2$ to find the integer $x$. Let's break this down step by step.

To find the bond order using Molecular Orbital Theory, our first step is always to determine the total electron count of the molecule. For carbon monoxide, we look at the constituent atoms: - Carbon ($\text{C}$) contributes $6$ electrons. - Oxygen ($\text{O}$) contributes $8$ electrons.

Adding these together, we get a total of $14$ electrons.

Now, we write down the molecular orbital configuration for a $14$-electron system. Following the standard energy level filling order, the configuration is:

To calculate the bond order, we use the formula:

Where $N_b$ is the number of electrons in bonding orbitals and $N_a$ is the number of electrons in anti-bonding orbitals (the ones with the asterisk $\ast$ mark). Counting them up, we have $10$ bonding electrons and $4$ anti-bonding electrons.

So, carbon monoxide features a strong triple bond.

Next, let's evaluate the nitrosonium ion. We calculate its total electrons similarly: - Nitrogen ($\text{N}$) contributes $7$ electrons. - Oxygen ($\text{O}$) contributes $8$ electrons. - The positive charge ($\oplus$) indicates the loss of $1$ electron.

Total electrons = $7+8-1=14$ electrons.

Notice something fascinating? Both $\text{CO}$ and ${\text{NO}}^{\oplus }$ have exactly $14$ electrons. In chemistry, we call such species isoelectronic. Because they have the same number of electrons, they will fill their molecular orbitals in the exact same pattern.

Consequently, the number of bonding and anti-bonding electrons remains identical ($N_b=10$, $N_a=4$).

Just like $\text{CO}$, the ${\text{NO}}^{\oplus }$ ion also possesses a triple bond.

The problem asks for the difference between these two bond orders, which is given as $x/2$. Let's calculate the difference:

Now, we equate this difference to the expression provided in the question:

Solving for $x$, we simply multiply both sides by $2$:

This problem serves as a brilliant reminder of the power of the isoelectronic principle. Whenever you spot species with $14$ electrons—like ${\text{N}}_2$, $\text{CO}$, ${\text{NO}}^{\oplus }$, or ${\text{CN}}^{-}$—you can confidently predict that they will have a bond order of $3$. Recognizing these patterns is a massive time-saver in competitive exams!