Molecular Orbital Theory (MOT) & MO Diagrams

Why MOT? The Theory That Explains O₂

Valence Bond Theory gave us bonds and shapes, but it stumbled on one stubborn fact: liquid oxygen is attracted to a magnet — it is paramagnetic, meaning it has unpaired electrons. The Lewis structure $O=O$ shows everything paired. Molecular Orbital Theory (MOT), developed by Hund and Mulliken, finally explains this.

Core idea: When atoms combine, their atomic orbitals merge to form new molecular orbitals (MOs) that belong to the whole molecule, not to individual atoms. Electrons then fill these MOs following the same rules as atomic orbitals (Aufbau, Pauli, Hund).

Formation of MOs (LCAO)

Molecular orbitals are formed by the Linear Combination of Atomic Orbitals (LCAO). Two atomic orbitals combine in two ways:

• Constructive (additive) combination → bonding MO (lower energy, electron density between nuclei). Denoted $\sigma$, $\pi$.
• Destructive (subtractive) combination → antibonding MO (higher energy, a node between nuclei). Denoted $\sigma^*$, $\pi^*$.

Key Point: $N$ atomic orbitals always produce $N$ molecular orbitals — half bonding, half antibonding. A bonding MO is more stable (lower energy) than the atomic orbitals; an antibonding MO is less stable (higher energy).

[JEE Tip] Bonding MO = electrons help hold the molecule together; antibonding MO (the starred ones) = electrons work against bonding.

Energy Ordering of Molecular Orbitals

Electrons fill MOs from lowest to highest energy. There are two orderings, depending on the molecule.

For $O_2$, $F_2$, $Ne_2$ (and heavier):

$$\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$$

For $Li_2$, $Be_2$, $B_2$, $C_2$, $N_2$ (lighter than $O_2$):

Due to s-p mixing, the $\sigma 2p_z$ rises above the two $\pi 2p$ orbitals: $$\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$$

Key Point: The ONLY difference is the relative order of $\sigma 2p_z$ and the $\pi 2p$ pair. Up to and including $N_2$, the π orbitals fill first; from $O_2$ onward, $\sigma 2p_z$ fills first.

[NEET Important] Remember the cutoff: 'till $N_2$, π is below σ2pz; from $O_2$, σ2pz is below π'. This is the single most error-prone point in MOT.

Bond Order, Stability & Magnetism

The most useful number MOT gives us is the bond order.

Definition: $$\text{Bond order} = \frac{1}{2}\,(N_b - N_a)$$ where $N_b$ = number of electrons in bonding MOs and $N_a$ = number in antibonding MOs.

What bond order tells us:

• Bond order > 0 → the molecule/ion exists and is stable. Bond order = 0 → it does not exist (e.g. $He_2$).
• Higher bond order → shorter bond, greater bond enthalpy, more stable molecule.
• It can be fractional (e.g. $O_2^+$ has bond order 2.5).

Magnetic behaviour

• All electrons paired → diamagnetic (weakly repelled by a magnet).
• One or more unpaired electrons → paramagnetic (attracted by a magnet); more unpaired electrons → stronger paramagnetism.

Key Point: MOT's triumph — $O_2$ has two unpaired electrons in its $\pi^* 2p$ orbitals, so it is paramagnetic. This is the fact VBT could never explain.

[JEE Tip] Always write the full MO configuration, then read off both bond order and the number of unpaired electrons (for magnetism).

MO Configurations of Key Diatomics

Let's build the important ones. (Core $\sigma 1s$, $\sigma^*1s$ are filled for second-period; we show valence MOs.)

Hydrogen $H_2$ (2 e⁻)

$\sigma 1s^2$. Bond order $= \tfrac{1}{2}(2-0) = 1$. Diamagnetic. Stable.

Helium $He_2$ (4 e⁻)

$\sigma 1s^2\,\sigma^*1s^2$. Bond order $= \tfrac{1}{2}(2-2) = 0$. Does not exist.

Nitrogen $N_2$ (14 e⁻)

$KK\,\sigma 2s^2\,\sigma^*2s^2\,\pi 2p_x^2\,\pi 2p_y^2\,\sigma 2p_z^2$. $N_b=8, N_a=2$. Bond order $= \tfrac{1}{2}(8-2) = 3$. Diamagnetic.

Oxygen $O_2$ (16 e⁻)

$KK\,\sigma 2s^2\,\sigma^*2s^2\,\sigma 2p_z^2\,\pi 2p_x^2\,\pi 2p_y^2\,\pi^*2p_x^1\,\pi^*2p_y^1$. $N_b=8, N_a=4$. Bond order $= \tfrac{1}{2}(8-4) = 2$. Paramagnetic (2 unpaired in π*).

Fluorine $F_2$ (18 e⁻)

Fills both $\pi^*$ completely. Bond order $= \tfrac{1}{2}(8-6) = 1$. Diamagnetic.

Oxygen-family ions

• $O_2^+$ (15 e⁻): bond order $2.5$, paramagnetic.
• $O_2^-$ (17 e⁻): bond order $1.5$, paramagnetic.
• $O_2^{2-}$ (18 e⁻): bond order $1$, diamagnetic. Stability/bond-length order: $O_2^+ > O_2 > O_2^- > O_2^{2-}$.

[NEET Important] $B_2$ and $O_2$ are both paramagnetic; $C_2$ and $N_2$ are diamagnetic. These four are exam favourites.

Solved Examples

Example 1: Bond order of N₂

Calculate the bond order of $N_2$ (14 electrons).

Solution:

1. Configuration: bonding electrons $N_b = 8$ (σ2s, π2p×2, σ2pz), antibonding $N_a = 2$ (σ*2s).
2. Formula: bond order $= \tfrac{1}{2}(N_b - N_a)$.
3. Compute: $\tfrac{1}{2}(8 - 2) = 3$.

Takeaway: $N_2$ has bond order 3, consistent with its triple bond and high stability.

Example 2: Why O₂ is paramagnetic

Use MOT to explain the paramagnetism of $O_2$.

Solution:

1. 16 electrons fill up to the $\pi^*2p$ orbitals.
2. The last two electrons go singly into the two degenerate $\pi^*2p_x$ and $\pi^*2p_y$ orbitals (Hund's rule).
3. Two unpaired electrons → $O_2$ is paramagnetic.

Takeaway: The two unpaired electrons in the antibonding π* orbitals make $O_2$ paramagnetic — MOT's signature success.

Example 3: Existence of He₂

Does the $He_2$ molecule exist? Justify with bond order.

Solution:

1. 4 electrons: $\sigma 1s^2\,\sigma^*1s^2$.
2. $N_b = 2, N_a = 2$.
3. Bond order $= \tfrac{1}{2}(2-2) = 0$.
4. Conclusion: zero bond order → no net bonding → $He_2$ does not exist.

Takeaway: Bond order 0 means the molecule is not formed.

Example 4: Bond order of O₂⁺

Find the bond order of the dioxygenyl cation $O_2^+$ (15 electrons).

Solution:

1. Remove 1 electron from $O_2$'s $\pi^*$ orbital → 15 electrons.
2. $N_b = 8, N_a = 3$.
3. Bond order $= \tfrac{1}{2}(8-3) = 2.5$.

Takeaway: Removing an antibonding electron increases bond order; $O_2^+$ (2.5) is stronger than $O_2$ (2).

Example 5: Comparing O₂, O₂⁺, O₂⁻, O₂²⁻

Arrange these in increasing order of bond length.

Solution:

1. Bond orders: $O_2^+$ (2.5), $O_2$ (2.0), $O_2^-$ (1.5), $O_2^{2-}$ (1.0).
2. Higher bond order → shorter bond.
3. Increasing bond length: $O_2^+ < O_2 < O_2^- < O_2^{2-}$.

Takeaway: Bond length is inversely related to bond order across an isoelectronic-like series.

Example 6: Magnetic nature of B₂

Is $B_2$ (10 electrons) paramagnetic or diamagnetic?

Solution:

1. For B₂ (lighter than O₂), π2p is below σ2pz.
2. Configuration: …$\sigma 2s^2\,\sigma^*2s^2\,\pi 2p_x^1\,\pi 2p_y^1$.
3. The last two electrons go singly into the two degenerate π2p orbitals → 2 unpaired.
4. Conclusion: $B_2$ is paramagnetic (bond order $= \tfrac{1}{2}(6-4) = 1$).

Takeaway: The s-p mixing ordering makes $B_2$ paramagnetic — a classic JEE trap.

Example 7: Bond order of C₂

Calculate the bond order and magnetism of $C_2$ (12 electrons).

Solution:

1. Ordering (lighter than O₂): …$\sigma 2s^2\,\sigma^*2s^2\,\pi 2p_x^2\,\pi 2p_y^2$.
2. $N_b = 6, N_a = 2$. Bond order $= \tfrac{1}{2}(6-2) = 2$.
3. All electrons paired → diamagnetic.

Takeaway: $C_2$ has bond order 2 and is diamagnetic (both π bonds, no σ2pz electrons).

Example 8: Bond order of F₂

Find the bond order of $F_2$ (18 electrons).

Solution:

1. Configuration fills both $\pi^*$ completely: $N_b = 8, N_a = 6$.
2. Bond order $= \tfrac{1}{2}(8-6) = 1$.
3. All paired → diamagnetic.

Takeaway: $F_2$ has a single bond (bond order 1), the weakest among $N_2$, $O_2$, $F_2$.

Example 9: Difference between bonding and antibonding MOs

State two differences between a bonding and an antibonding molecular orbital.

Solution:

1. Energy: bonding MO is lower in energy than the parent atomic orbitals; antibonding MO is higher.
2. Electron density: bonding MO has high density between the nuclei; antibonding MO has a node (zero density) between them.
3. Effect: electrons in bonding MOs stabilise the molecule; in antibonding MOs they destabilise it.

Takeaway: Bonding MOs glue the molecule; antibonding (starred) MOs pull it apart.

Example 10: Why N₂ is more stable than N₂⁺

Compare the bond orders of $N_2$ and $N_2^+$.

Solution:

1. $N_2$ (14 e⁻): bond order 3.
2. $N_2^+$ (13 e⁻): remove one bonding electron (from σ2pz) → $N_b = 7, N_a = 2$ → bond order $= \tfrac{1}{2}(7-2) = 2.5$.
3. Conclusion: $N_2$ (3) > $N_2^+$ (2.5), so $N_2$ is more stable with a shorter bond.

Takeaway: Removing a bonding electron lowers bond order and stability (contrast with $O_2^+$).

Example 11: Number of unpaired electrons in O₂⁻

How many unpaired electrons are in the superoxide ion $O_2^-$ (17 electrons)?

Solution:

1. Add one electron to $O_2$'s $\pi^*$ orbitals → 17 electrons.
2. The π* orbitals now hold 3 electrons: one is paired, one remains unpaired.
3. Conclusion: 1 unpaired electron → paramagnetic; bond order $= \tfrac{1}{2}(8-5) = 1.5$.

Takeaway: $O_2^-$ has bond order 1.5 and one unpaired electron (paramagnetic).

Example 12: Predicting whether Be₂ exists

Use MOT to decide if $Be_2$ (8 electrons) exists.

Solution:

1. Configuration: $\sigma 1s^2\,\sigma^*1s^2\,\sigma 2s^2\,\sigma^*2s^2$.
2. $N_b = 4, N_a = 4$.
3. Bond order $= \tfrac{1}{2}(4-4) = 0$.
4. Conclusion: like $He_2$, $Be_2$ has zero bond order and does not exist as a stable molecule.

Takeaway: Filled bonding and antibonding pairs cancel → bond order 0 → no stable molecule.