Which of the following species is not paramagnetic?
(a) NO
(b) CO
(c) O2
(d) B2

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

This is a classic exception! In $O_2$, the last two electrons enter the degenerate $\pi^* 2p_x$ and $\pi^* 2p_y$ orbitals. According to Hund's rule, they must occupy separate orbitals with parallel spins before pairing up. This leaves two unpaired electrons, making $O_2$ paramagnetic.

Q: Is there a quick trick to identify paramagnetic species without writing the full MO configuration?

Yes! If the total number of electrons is odd, the species is always paramagnetic. If the total is even, it is usually diamagnetic, with two major exceptions: 10 electrons (like $B_2$) and 16 electrons (like $O_2$), which are paramagnetic.

Q: Why does the order of molecular orbitals change for molecules with 14 or fewer electrons?

For molecules with $\le 14$ electrons (like $N_2$, $CO$, $B_2$), the energy difference between the 2s and 2p orbitals is small. This leads to 's-p mixing,' which pushes the $\sigma 2p_z$ orbital higher in energy than the $\pi 2p_x$ and $\pi 2p_y$ orbitals.

Q: What does it physically mean for a substance to be diamagnetic?

Physically, a diamagnetic substance is weakly repelled by an external magnetic field. This happens because all its electrons are paired, and their opposite spins cancel out each other's magnetic moments, leaving no net magnetic moment.

Q: Can a molecule with a fractional bond order be diamagnetic?

No. A fractional bond order (like 1.5 or 2.5) implies that there is an odd number of electrons in the bonding or antibonding molecular orbitals. An odd number of electrons guarantees that at least one electron is unpaired, making the species paramagnetic.

Question

Which of the following species is not paramagnetic?

Options:

(A) NO

(B) CO

(C) $O_2$

(D) $B_2$

Sigma Percentile · Accepted Answer

### The Magnetic Mystery: Paramagnetic vs. Diamagnetic

When we dive into the quantum world of molecules, their magnetic behavior is one of the most fascinating properties to uncover. A molecule's magnetic nature is entirely dictated by the arrangement of its electrons.

If a molecule has even a single **unpaired electron** wandering alone in an orbital, it acts like a tiny magnet and is drawn towards external magnetic fields. We call this **paramagnetic**. On the flip side, if every single electron has a partner and they are all perfectly paired up, their magnetic spins cancel each other out. This makes the molecule weakly repelled by magnetic fields, a state we call **diamagnetic**.

Our mission in this problem is to identify the imposter—the species that is *not* paramagnetic, meaning we are hunting for the diamagnetic molecule.

### The Master Tool: Molecular Orbital Theory

To solve this, we rely on Molecular Orbital Theory (MOT). MOT tells us that atomic orbitals combine to form molecular orbitals, which are then filled with electrons according to the Aufbau principle, Pauli exclusion principle, and Hund's rule.

There is a brilliant, time-saving trick for competitive exams like JEE: **The Electron Counting Rule**.
- If a species has an **odd** total number of electrons, it is **always paramagnetic**.
- If a species has an **even** total number of electrons, it is **usually diamagnetic**.

However, beware of the traps! Molecules with exactly **10 or 16 electrons** are the famous exceptions. Due to degenerate (equal energy) orbitals and Hund's rule, they end up with two unpaired electrons, making them strongly paramagnetic.

### Analyzing the Suspects

Let's put our suspects under the MOT microscope:

**1. Nitric Oxide ($NO$)**
Nitrogen brings 7 electrons, and Oxygen brings 8. 
Total electrons = $7 + 8 = 15$.
Since 15 is an odd number, we instantly know $NO$ must have an unpaired electron. If we write the configuration, the 15th electron sits alone in the $\pi^* 2p_x$ orbital. 
**Verdict:** Paramagnetic.

**2. Oxygen Gas ($O_2$)**
Oxygen brings 8 electrons, so $O_2$ has $8 + 8 = 16$ electrons.
Wait, 16 is an even number! But remember the trap? $O_2$ is the classic exception. The last two electrons enter the degenerate $\pi^* 2p_x$ and $\pi^* 2p_y$ orbitals. According to Hund's rule, they must occupy separate orbitals before pairing. 
**Verdict:** Paramagnetic (with 2 unpaired electrons).

**3. Boron Gas ($B_2$)**
Boron brings 5 electrons, so $B_2$ has $5 + 5 = 10$ electrons.
This is the other famous exception! The 9th and 10th electrons enter the degenerate $\pi 2p_x$ and $\pi 2p_y$ orbitals. Just like in oxygen, they stay unpaired in separate orbitals.
**Verdict:** Paramagnetic (with 2 unpaired electrons).

**4. Carbon Monoxide ($CO$)**
Carbon brings 6 electrons, and Oxygen brings 8. 
Total electrons = $6 + 8 = 14$.
Fourteen is an even number, and it's not an exception. Let's verify with its configuration:
$$\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$$
Every single orbital is completely filled with a pair of electrons. There are absolutely no unpaired electrons left.
**Verdict:** Diamagnetic.

### The Final Conclusion

Out of all the given options, Carbon Monoxide ($CO$) is the only species where all electrons are perfectly paired. Therefore, it is diamagnetic, which means it is the only species that is **not paramagnetic**.

The correct answer is **(b) CO**.

Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Which of the following species is not paramagnetic?

Select Answer:

Visualized Solution

Magnetic Nature in MOT

Analyzing $N O$

Analyzing $C O$

Analyzing $O_{2}$

Analyzing $B_{2}$

Conclusion

The Sigma Insight: Molecular Orbital Theory

The Magnetic Mystery

The Master Tool

Analyzing the Suspects

The Final Conclusion

Similar Questions

Assuming 2s-2p mixing is NOT operative, the paramagnetic species among the following is :

Which of the following species exhibits diamagnetic behaviour?

Among the following species, the diamagnetic molecule is

Number of paramagnetic oxides among the following given oxides is $Li_{2} O, Na_{2} O_{2}, KO_{2}, HgO$ and $K_{2} O$

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

Which of the following molecules/ions does not contain unpaired electrons?

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

Among $H_{2}$ , $He_{2}^{+}$ , $Li_{2}$ , $Be_{2}$ , $B_{2}$ , $C_{2}$ , $N_{2}$ , $O_{2}^{-}$ , and $F_{2}$ , the number of diamagnetic species is - (Atomic number : H = 1, He = 2, Li = 3, Be = 4, B = 5, C = 6, N = 7, O = 8, f = 9)

Among the following molecules/ions, $C_{2}^{2 -}, N_{2}^{2 -}, O_{2}^{2 -}, O_{2}$ Which one is diamagnetic and has the shortest bond length?

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