In which of the following processes, the bond order has increased and paramagnetic character has cha

Q: Why does $O_2$ have unpaired electrons even though it has an even number of electrons (16)?

Great question! According to Molecular Orbital Theory, the last two electrons in $O_2$ enter the degenerate $\pi^*_{2p_x}$ and $\pi^*_{2p_y}$ orbitals. Following Hund's rule, they occupy these orbitals singly with parallel spins, resulting in two unpaired electrons.

Q: How can I quickly determine if a diatomic molecule is paramagnetic?

A quick trick: If the total number of electrons is odd, it's definitely paramagnetic. If it's even, it's usually diamagnetic, EXCEPT for 10 and 16 electrons (like $B_2$ and $O_2$), which are paramagnetic due to degenerate orbitals.

Q: Why does removing an electron sometimes increase the bond order?

Bond order depends on the difference between bonding and antibonding electrons. If you remove an electron from an *antibonding* orbital, the number of antibonding electrons ($N_a$) decreases. Since $\text{BO} = \frac{N_b - N_a}{2}$, a smaller $N_a$ leads to a higher bond order!

Q: Is $NO^+$ isoelectronic with any other common molecule?

Yes! $NO^+$ has $7 + 8 - 1 = 14$ electrons. This is exactly the same number of electrons as $N_2$ and $CO$. Because they are isoelectronic, they all have a bond order of 3 and are diamagnetic.

Q: Do I need to draw the full MO diagram every time?

Not necessarily! Once you memorize the filling order and the '14-electron rule' (where 14 electrons give a max bond order of 3, and every electron added or removed drops the BO by 0.5), you can solve these questions in seconds.

Question

In which of the following processes, the bond order has increased and paramagnetic character has changed to diamagnetic?

Options:

(A) $O_2 \rightarrow O_2^+$

(B) $N_2 \rightarrow N_2^+$

(C) $O_2 \rightarrow O_2^{2-}$

(D) $NO \rightarrow NO^+$

Sigma Percentile · Accepted Answer

### The Quest for Stability

Imagine you are an architect of molecules. Your goal is to find a specific transformation where a molecule not only becomes more tightly bound together but also pairs up all its lonely, unpaired electrons.

In the language of chemistry, we are looking for a process where the **Bond Order (BO)** increases, and the magnetic nature shifts from **paramagnetic** to **diamagnetic**.

To solve this, we need our trusty blueprint: **Molecular Orbital Theory (MOT)**.

### The Master Equations

Before we dive into the options, let's recall the fundamental rules of MOT.

The strength of a bond is measured by its Bond Order, calculated as:
$$BO = \frac{N_b - N_a}{2}$$
where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.

**Key Principle:** Adding electrons to bonding orbitals or removing them from antibonding orbitals will *increase* the bond order.

For the magnetic nature, the rule is simple:
If there is even one unpaired electron, the species is **paramagnetic**. If all electrons are perfectly paired, it is **diamagnetic**.

### Decoding the Options

Let's put each option to the test.

**Option (a): $O_2 \rightarrow O_2^+$**
The oxygen molecule, $O_2$, has $16$ electrons. Its MO configuration ends with two unpaired electrons in the $\pi^*$ orbitals, making it paramagnetic with a bond order of $2$.

When it loses an electron to form $O_2^+$, the bond order increases to $2.5$. However, it still has one unpaired electron left. It remains paramagnetic. Close, but not what we need!

**Option (b): $N_2 \rightarrow N_2^+$**
Nitrogen, $N_2$, is a highly stable molecule with $14$ electrons. All are paired, making it diamagnetic with a maximum bond order of $3$.

Removing an electron to form $N_2^+$ drops the bond order to $2.5$ and introduces an unpaired electron. It becomes paramagnetic. This is the exact opposite of what we want!

**Option (c): $O_2 \rightarrow O_2^{2-}$**
Starting again with paramagnetic $O_2$ ($BO = 2$), we add two electrons to form the peroxide ion, $O_2^{2-}$.

These electrons fill the $\pi^*$ orbitals, pairing everything up. It successfully becomes diamagnetic! But wait, because we added electrons to *antibonding* orbitals, the bond order drops to $1$. This option fails the first condition.

### The Winning Transition

**Option (d): $NO \rightarrow NO^+$**
Nitric oxide, $NO$, has $15$ electrons. Because $15$ is an odd number, it is guaranteed to be paramagnetic. Specifically, it has one unpaired electron sitting in the $\pi^*_{2p}$ antibonding orbital. Its bond order is $2.5$.

Now, let's remove an electron to form the nitrosonium ion, $NO^+$.

Where does this electron come from? It is removed from the highest energy orbital, which is that exact $\pi^*_{2p}$ antibonding orbital!

By removing an antibonding electron, $N_a$ decreases. 
$$BO = \frac{10 - 4}{2} = 3.0$$
The bond order increases from $2.5$ to $3.0$!

Furthermore, the only unpaired electron is now gone. $NO^+$ has $14$ electrons, making it isoelectronic with $N_2$. All electrons are paired, so it is perfectly **diamagnetic**.

Both conditions are flawlessly met. The correct process is indeed $NO \rightarrow NO^+$.

Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: In which of the following processes, the bond order has increased and paramagnetic character has changed to diamagnetic?

Select Answer:

Visualized Solution

Problem Analysis

Key Concepts of MOT

Option (a): $O_{2} \to O_{2}^{+}$

Option (b): $N_{2} \to N_{2}^{+}$

Option (c): $O_{2} \to O_{2}^{2 -}$

Option (d): $N O \to N O^{+}$

Visualizing the Transition

Final Conclusion

The Sigma Insight: Molecular Orbital Theory

The Quest for Stability

The Master Equations

Decoding the Options

The Winning Transition

Similar Questions

The bond order and magnetic behaviour of $O_{2}^{-}$ ion are, respectively

Which of the following species exhibits diamagnetic behaviour?

Which one of the following properties is not shown by NO?

In the following the correct bond order sequence is

Which of the following species is not paramagnetic?

Which one of the following molecules is expected to exhibit diamagnetic behaviour?

Among the following species, the diamagnetic molecule is

Two pi and half sigma bonds are present in

Increasing order of bond strength of $O_{2}$ , $O_{2}^{-}$ , $O_{2}^{2 -}$ and $O_{2}^{+}$ is

Which one of the following pairs of species have the same bond order?