Which of the following molecules/ions does not contain unpaired electrons?
(a) O2^2-
(b) B2
(c) N2^+

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

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 2 unpaired electrons.

Q: Why is the MO energy order different for molecules with 14 or fewer electrons compared to those with more than 14?

For molecules with $\le 14$ electrons (like $B_2, C_2, N_2$), there is significant s-p mixing which pushes the $\sigma 2p_z$ orbital higher in energy than the $\pi 2p_x$ and $\pi 2p_y$ orbitals. For $> 14$ electrons (like $O_2, F_2$), s-p mixing is negligible, so $\sigma 2p_z$ is lower in energy.

Q: How do we calculate the total electrons for an ion like $N_2^+$?

First, find the total electrons for the neutral molecule. Nitrogen has atomic number 7, so $N_2$ has $7 + 7 = 14$ electrons. The positive charge indicates the loss of one electron, so $N_2^+$ has $14 - 1 = 13$ electrons.

Q: What does it mean if a species is diamagnetic?

A species is diamagnetic if all of its electrons are paired in its molecular orbitals. Such substances are weakly repelled by a magnetic field. If there is even one unpaired electron, the species is paramagnetic.

Q: Can we use Lewis dot structures to predict unpaired electrons?

Lewis structures often fail to predict the correct magnetic behavior. For example, the Lewis structure of $O_2$ shows all electrons paired (a double bond and lone pairs), but experimentally it is paramagnetic. Molecular Orbital Theory correctly predicts its 2 unpaired electrons.

Question

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

Options:

(A) $O_2^{2-}$

(B) $B_2$

(C) $N_2^+$

(D) $O_2$

Sigma Percentile · Accepted Answer

### Unmasking the Magnetic Nature of Molecules: A Journey through Molecular Orbital Theory

Have you ever wondered why liquid oxygen sticks to a magnet? The answer lies hidden in the quantum world of Molecular Orbital (MO) Theory. In this problem, we are on a mission to find the odd one out: a species that has absolutely zero unpaired electrons, making it diamagnetic.

### The Master Tool: Molecular Orbital Theory

To solve this, we cannot rely on simple Lewis dot structures. We must calculate the total electron count for each species and fill them into molecular orbitals according to the Aufbau principle, Pauli exclusion principle, and Hund's rule of maximum multiplicity.

Let's analyze our suspects one by one.

### Analyzing the Suspects

**1. The Boron Molecule ($B_2$)**
Boron has an atomic number of 5, so $B_2$ has a total of 10 electrons. When we write its MO configuration:

$\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1 \approx \pi 2p_y^1$

The last two electrons enter the degenerate $\pi 2p_x$ and $\pi 2p_y$ orbitals. According to Hund's rule, they occupy these orbitals singly. Thus, $B_2$ has **2 unpaired electrons**.

**2. The Nitrogen Cation ($N_2^+$)**
Nitrogen has an atomic number of 7, so $N_2$ has 14 electrons. The positive charge means we remove one electron, leaving us with 13 electrons. Its configuration is:

$\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 \approx \pi 2p_y^2, \sigma 2p_z^1$

The final electron sits alone in the $\sigma 2p_z$ orbital. So, $N_2^+$ has **1 unpaired electron**.

**3. The Oxygen Molecule ($O_2$)**
Oxygen has an atomic number of 8, giving $O_2$ a total of 16 electrons. Its configuration is:

$\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 \approx \pi 2p_y^2, \pi^* 2p_x^1 \approx \pi^* 2p_y^1$

Just like $B_2$, the last two electrons are forced to occupy the degenerate $\pi^* 2p_x$ and $\pi^* 2p_y$ orbitals singly. This gives $O_2$ **2 unpaired electrons**, explaining its paramagnetic nature!

### The Culprit: The Peroxide Ion ($O_2^{2-}$)

Finally, we arrive at the peroxide ion, $O_2^{2-}$. It starts with 16 electrons from the two oxygen atoms, plus 2 extra electrons from the negative charge, totaling 18 electrons.

Let's look at its configuration:

$\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 \approx \pi 2p_y^2, \pi^* 2p_x^2 \approx \pi^* 2p_y^2$

Those two extra electrons perfectly pair up with the single electrons that were present in the $\pi^*$ orbitals of neutral $O_2$. Now, every single orbital is fully occupied. There are **zero unpaired electrons**!

### Conclusion

Because all of its electrons are paired, $O_2^{2-}$ is diamagnetic. It is the only species in our list that does not contain any unpaired electrons. The correct option is indeed **(a)**.

Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Which of the following molecules/ions does not contain unpaired electrons?

Select Answer:

Visualized Solution

$O bj ec t i v e$

$M o l ec u l a r O r bi t a l T h eor y$

$A na l y z in g B_{2}$

$A na l y z in g N_{2}^{+}$

$A na l y z in g O_{2}$

$A na l y z in g O_{2}^{2 -}$

$E l ec t r o n P ai r in g in O_{2}^{2 -}$

$F ina l C o n c l u s i o n$

The Sigma Insight: Molecular Orbital Theory

Unmasking the Magnetic Nature of Molecules

The Master Tool

Analyzing the Suspects

The Culprit

Conclusion

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