Among the following species, the diamagnetic molecule is
(a) CO
(b) B2
(c) NO
(d) O2

Q: Why is $O_2$ paramagnetic even though it has an even number of electrons (16)?

Because the last two electrons enter two degenerate $\pi^*$ antibonding orbitals singly, following Hund's rule of maximum multiplicity. This leaves two unpaired electrons, making it paramagnetic.

Q: Does the MO energy level order change for different molecules?

Yes! For molecules with 14 or fewer electrons (like $N_2$, $CO$, $B_2$), the $\sigma 2p_z$ orbital is higher in energy than the $\pi 2p_{x,y}$ orbitals due to s-p mixing. For molecules with more than 14 electrons (like $O_2$, $F_2$), $\sigma 2p_z$ is lower in energy.

Q: How do I quickly find if a molecule is paramagnetic without drawing the whole diagram?

Odd electron molecules are always paramagnetic. For even electron molecules, they are usually diamagnetic, EXCEPT for those with exactly 10 or 16 electrons, which are paramagnetic.

Q: What is the bond order of CO?

CO has 14 electrons. It has 10 bonding electrons and 4 antibonding electrons. The bond order is calculated as $(10 - 4) / 2 = 3$, indicating a strong triple bond.

Q: Why do we use MOT instead of Valence Bond Theory (VBT) here?

VBT fails to explain the paramagnetic nature of $O_2$. According to simple Lewis structures and VBT, all electrons in $O_2$ appear paired. MOT correctly predicts the two unpaired electrons in the antibonding orbitals.

Question

Among the following species, the diamagnetic molecule is

Options:

(A) $	ext{CO}$

(B) $	ext{B}_2$

(C) $	ext{NO}$

(D) $	ext{O}_2$

Sigma Percentile · Accepted Answer

### Unveiling the Magnetic Mysteries of Molecules

When we dive into the microscopic world of molecules, their magnetic behavior is one of the most fascinating properties to explore. The core principle is simple: if a molecule has all its electrons neatly paired up, it is **diamagnetic** (repelled by magnetic fields). However, if there is even a single unpaired electron wandering in an orbital, the molecule becomes **paramagnetic** (attracted to magnetic fields).

To determine the magnetic nature of diatomic molecules, we rely on the elegant **Molecular Orbital Theory (MOT)**. MOT provides us with an energy level sequence to fill electrons, much like the Aufbau principle for atoms.

### The Power of Molecular Orbital Theory

The filling order of molecular orbitals depends on the total number of electrons in the molecule:

For molecules with $\le 14$ electrons, the $\sigma 2p_z$ orbital is higher in energy than the degenerate $\pi 2p_x$ and $\pi 2p_y$ orbitals:
$$ \sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \pi 2p_x = \pi 2p_y < \sigma 2p_z < \pi^* 2p_x = \pi^* 2p_y < \sigma^* 2p_z $$

For molecules with $> 14$ electrons, the $\sigma 2p_z$ orbital drops below the $\pi 2p$ orbitals:
$$ \sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < \pi 2p_x = \pi 2p_y < \pi^* 2p_x = \pi^* 2p_y < \sigma^* 2p_z $$

### Analyzing the Suspects

Let's put our options to the test by counting their total electrons and filling the orbitals.

**1. The $B_2$ Molecule:**
Boron has an atomic number of 5, so $B_2$ has a total of $10$ electrons. Following the $\le 14$ electron rule, the valence configuration is:
$$ \dots \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1 = \pi 2p_y^1 $$
According to Hund's rule, the last two electrons enter the degenerate $\pi$ orbitals singly. With **two unpaired electrons**, $B_2$ is paramagnetic.

**2. The $NO$ Molecule:**
Nitrogen (7) and Oxygen (8) give $NO$ a total of $15$ electrons. Since it's an odd number, it must have an unpaired electron! The valence configuration is:
$$ \dots \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 $$
With **one unpaired electron** in the antibonding $\pi^*$ orbital, $NO$ is paramagnetic.

**3. The $O_2$ Molecule:**
Oxygen has $16$ electrons. This is a classic trap! Even though 16 is an even number, let's look at the filling:
$$ \dots \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1 $$
The last two electrons occupy the degenerate $\pi^*$ orbitals singly. Thus, $O_2$ has **two unpaired electrons** and is highly paramagnetic.

### The Diamagnetic Champion

Finally, let's examine **Carbon Monoxide ($CO$)**. Carbon (6) and Oxygen (8) yield a total of $14$ electrons. Filling the orbitals according to the $\le 14$ rule:
$$ \dots \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2 $$
Every single electron is perfectly paired. There are **zero unpaired electrons**. Therefore, $CO$ is our diamagnetic molecule!

### The Golden Trick for Exams

To save precious time in competitive exams like JEE and NEET, remember this golden rule:
- **Odd electron species** are ALWAYS paramagnetic.
- **Even electron species** are GENERALLY diamagnetic.
- **The Exceptions:** Molecules with exactly **10 or 16 electrons** (like $B_2$ and $O_2$) are paramagnetic due to the presence of degenerate orbitals that fill singly first.

Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Among the following species, the diamagnetic molecule is

Select Answer:

Visualized Solution

Magnetic Nature

MOT Filling Rules

Analyzing $B_{2}$ Molecule

Analyzing $N O$ Molecule

Analyzing $O_{2}$ Molecule

Analyzing $C O$ Molecule

Pro Tip for Exams

The Sigma Insight: Molecular Orbital Theory

Unveiling the Magnetic Mysteries of Molecules

The Power of Molecular Orbital Theory

Analyzing the Suspects

The Diamagnetic Champion

The Golden Trick for Exams

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