Two pi and half sigma bonds are present in
(a) O2+
(b) N2
(c) N2+
(d) O2

Q: Why does $N_2$ have a different MO energy order than $O_2$?

In molecules up to $N_2$, the $2s$ and $2p_z$ orbitals are close in energy and interact (s-p mixing), pushing the $\sigma_{2p_z}$ orbital higher in energy than the $\pi_{2p_{x,y}}$ orbitals. In $O_2$ and $F_2$, the energy gap is larger, so s-p mixing is negligible, and $\sigma_{2p_z}$ is lower than $\pi_{2p_{x,y}}$.

Q: Can a bond order be a fraction?

Yes! A fractional bond order, like 0.5 or 1.5, simply means there is an odd number of electrons in the bonding or antibonding orbitals. It indicates a weaker or less stable bond compared to a full integer bond, but it is completely physically valid.

Q: How do we know which electrons form $\pi$ bonds and which form $\sigma$ bonds?

Electrons in $\sigma$ and $\sigma^*$ orbitals contribute to the $\sigma$ bond order, while electrons in $\pi$ and $\pi^*$ orbitals contribute to the $\pi$ bond order.

Q: Why do we ignore the $1s$ and $2s$ electrons when calculating these specific bond orders?

The $1s$ and $2s$ subshells are fully filled in these molecules, meaning their bonding and antibonding orbitals have an equal number of electrons. They cancel each other out, resulting in a net bond order of zero for those inner shells. We only need to focus on the valence $2p$ electrons.

Q: Is $N_2^+$ paramagnetic or diamagnetic?

Because $N_2^+$ has one unpaired electron in the $\sigma_{2p_z}$ orbital, it is paramagnetic.

Question

Two pi and half sigma bonds are present in

Options:

(A) $O_2^+$

(B) $N_2$

(C) $N_2^+$

(D) $O_2$

Sigma Percentile · Accepted Answer

### The Quantum Puzzle

Imagine you are a molecular architect, and your task is to build a molecule with a very specific blueprint: it must contain exactly **two $\pi$ bonds** and **half a $\sigma$ bond**.

This might sound like a strange request. How can you have *half* a bond?

In the classical Lewis dot structure world, bonds are whole numbers. But in the quantum realm of **Molecular Orbital (MO) Theory**, electrons are delocalized over the entire molecule. A fractional bond order simply means there is an imbalance between bonding and antibonding electrons that doesn't perfectly pair up.

### Decoding the Bond Order

To solve this puzzle, we need to look at the $\sigma$ and $\pi$ electrons separately.

The total bond order is the sum of the $\sigma$ bond order and the $\pi$ bond order.

We calculate these using the formulas:
$$ 	ext{B.O.}_{\sigma} = \frac{N_{b,\sigma} - N_{a,\sigma}}{2} $$
$$ 	ext{B.O.}_{\pi} = \frac{N_{b,\pi} - N_{a,\pi}}{2} $$

Our target is $	ext{B.O.}_{\pi} = 2$ and $	ext{B.O.}_{\sigma} = 0.5$. Let's test our candidates!

### The Oxygen Candidates

Let's start with the oxygen molecule, $O_2$. It has 16 electrons.

Its valence MO configuration is:
$$ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ \sigma_{2p_z}^2 \ \pi_{2p_{x,y}}^4 \ \pi_{2p_{x,y}}^{*2} $$

If we calculate the bond orders for the $2p$ subshell, we find that the $\sigma$ bond order is $\frac{2 - 0}{2} = 1$.

The $\pi$ bond order is $\frac{4 - 2}{2} = 1$.

This gives us one $\sigma$ bond and one $\pi$ bond (a double bond). This doesn't match our blueprint.

What if we remove an electron to form $O_2^+$? The electron leaves the highest energy orbital, which is a $\pi^*$ orbital.

The $\pi$ bond order becomes $\frac{4 - 1}{2} = 1.5$. Still not what we are looking for.

### The Nitrogen Candidates

Now, let's turn our attention to nitrogen, $N_2$, which has 14 electrons.

Because of s-p mixing, the energy levels are slightly different. The $\pi_{2p}$ orbitals are lower in energy than the $\sigma_{2p_z}$ orbital.

The configuration is:
$$ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ \pi_{2p_{x,y}}^4 \ \sigma_{2p_z}^2 $$

Here, the $\pi$ bond order is $\frac{4 - 0}{2} = 2$. We have our two $\pi$ bonds!

However, the $\sigma$ bond order is $\frac{2 - 0}{2} = 1$. We have a full $\sigma$ bond, but we only want half.

### The Final Verdict

Let's see what happens when we ionize $N_2$ to form $N_2^+$. We remove one electron.

Where does it come from? It comes from the highest occupied molecular orbital (HOMO), which is the $\sigma_{2p_z}$ orbital.

The new configuration is:
$$ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ \pi_{2p_{x,y}}^4 \ \sigma_{2p_z}^1 $$

Let's recalculate. The $\pi$ bond order remains $\frac{4 - 0}{2} = 2$.

But look at the $\sigma$ bond order! It is now $\frac{1 - 0}{2} = 0.5$.

We have successfully found our molecule. **$N_2^+$ has exactly two $\pi$ bonds and half a $\sigma$ bond.**

This beautiful result showcases the power of Molecular Orbital Theory in predicting the intricate electronic structure of molecules.

Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Two pi and half sigma bonds are present in

Select Answer:

Visualized Solution

Decoding the Question

MO Configuration of $O_{2}$ and $O_{2}^{+}$

MO Configuration of $N_{2}$

MO Configuration of $N_{2}^{+}$

Conclusion

The Sigma Insight: Molecular Orbital Theory

The Quantum Puzzle

Decoding the Bond Order

The Oxygen Candidates

The Nitrogen Candidates

The Final Verdict

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