According to molecular orbital theory, the number of unpaired electron(s) in O2^2- is ....... .

Q: Why does $\sigma 2p_z$ come before $\pi 2p_x$ in $O_2$ but not in $N_2$?

In atoms like O and F, the effective nuclear charge is very high, which pulls the 2s orbital much lower in energy than the 2p orbitals. Because the energy gap between 2s and 2p is so large, they do not mix (no s-p mixing). Without this mixing, the $\sigma 2p_z$ orbital naturally falls below the $\pi 2p$ orbitals. In $N_2$ and lighter elements, s-p mixing pushes the $\sigma 2p_z$ orbital higher.

Q: What is the bond order of the $O_2^{2-}$ ion?

Bond order is calculated as $\frac{N_b - N_a}{2}$. For $O_2^{2-}$, there are 10 bonding electrons and 8 anti-bonding electrons. So, Bond Order = $\frac{10 - 8}{2} = 1$. This means there is a single bond between the oxygen atoms, just like in hydrogen peroxide ($H_2O_2$).

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

This is the greatest triumph of Molecular Orbital Theory! When you fill 16 electrons into the MOs of $O_2$, the last two electrons enter the degenerate $\pi^* 2p_x$ and $\pi^* 2p_y$ orbitals. According to Hund's rule, they must occupy these orbitals singly with parallel spins before pairing up. These 2 unpaired electrons make $O_2$ paramagnetic.

Q: Does the $O_2^{2-}$ ion exist in real life?

Absolutely! It is known as the peroxide ion. You encounter it frequently in chemistry in compounds like Sodium peroxide ($Na_2O_2$), Barium peroxide ($BaO_2$), and Hydrogen peroxide ($H_2O_2$).

Q: How can I quickly remember the MO sequence for exams?

For $O_2$ and $F_2$, remember the pattern '1-2-2-1' for the 2p orbitals: 1 $\sigma$, 2 $\pi$, 2 $\pi^*$, 1 $\sigma^*$. For $N_2$ and lighter diatomic molecules, the pattern is '2-1-2-1': 2 $\pi$, 1 $\sigma$, 2 $\pi^*$, 1 $\sigma^*$.

Question

According to molecular orbital theory, the number of unpaired electron(s) in $O_2^{2-}$ is ....... .

Sigma Percentile · Accepted Answer

### The Magic of Molecular Orbitals

When we dive into the microscopic world of molecules, the simple Lewis dot structures often fail to tell the whole story. This is where **Molecular Orbital (MO) Theory** steps in like a superhero, explaining the magnetic properties and bond strengths that classical theories cannot. Today, we are going to decode the peroxide ion, $O_2^{2-}$, and find out exactly how its electrons are arranged.

### Step 1: Counting the Electrons

Before we can place electrons into orbitals, we need to know how many we have.

A neutral oxygen atom ($O$) has an atomic number of 8, meaning it has 8 electrons. In an $O_2$ molecule, we have $8 + 8 = 16$ electrons. However, we are dealing with the peroxide ion, $O_2^{2-}$. That $2-$ charge indicates that the molecule has gained two extra electrons from an external source (like two sodium atoms in $Na_2O_2$).

Total electrons $= 16 + 2 = \mathbf{18 	ext{ electrons}}$.

### Step 2: The MO Energy Sequence

To fill these 18 electrons, we must follow the specific energy sequence for oxygen. Because oxygen is highly electronegative, the energy gap between its $2s$ and $2p$ orbitals is large. This prevents $s-p$ mixing, causing the $\sigma 2p_z$ orbital to drop lower in energy than the $\pi 2p_x$ and $\pi 2p_y$ orbitals.

The sequence is:
$\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < \pi 2p_x = \pi 2p_y < \pi^* 2p_x = \pi^* 2p_y < \sigma^* 2p_z$

### Step 3: Filling the Orbitals

Now, we play the role of an architect, placing electrons into these orbital 'rooms' following the Aufbau principle, Pauli's exclusion principle, and Hund's rule.

The first 14 electrons comfortably fill the lower energy levels:
$\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$

We have $18 - 14 = 4$ electrons left. These remaining 4 electrons must enter the degenerate anti-bonding orbitals: $\pi^* 2p_x$ and $\pi^* 2p_y$.

According to **Hund's Rule of Maximum Multiplicity**, electrons will first occupy degenerate orbitals singly before pairing up. So, the first two electrons go in singly: $\pi^* 2p_x^1 = \pi^* 2p_y^1$. The last two electrons will then pair up with them, resulting in:
$\pi^* 2p_x^2 = \pi^* 2p_y^2$

### The Final Verdict

Looking at the complete electronic configuration of $O_2^{2-}$:
$\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$

Every single orbital is completely filled with a pair of electrons. There are absolutely **zero** unpaired electrons.

Because all electrons are paired, the $O_2^{2-}$ ion is **diamagnetic** (it is weakly repelled by a magnetic field). This is a beautiful contrast to neutral $O_2$ gas, which only has 16 electrons, leaving two unpaired electrons in the $\pi^*$ orbitals, making it paramagnetic!

Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: According to molecular orbital theory, the number of unpaired electron(s) in $O_{2}^{2 -}$ is ....... .

Enter Numerical Value:

Visualized Solution

$O_{2}^{2 -}$ $Electron Count$

$MO Energy Sequence$

$Filling Lower Orbitals$

$Filling Highest Occupied MOs$

$Final Answer$

$Magnetic Property$

The Sigma Insight: Molecular Orbital Theory

The Magic of Molecular Orbitals

Step 1

Step 2

Step 3

The Final Verdict

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