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Chemical Bonding and Molecular Structure - Molecular Orbital Theory

Grade 11ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Molecular Orbital Theory (MOT) states that atomic orbitals (AOAOs) lose their individual identity and combine to form Molecular Orbitals (MOMOs) which belong to the molecule as a whole.

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Linear Combination of Atomic Orbitals (LCAO) is the primary method for forming MOMOs. When two AOAOs combine, they form two MOMOs: a Bonding Molecular Orbital (BMO) and an Antibonding Molecular Orbital (ABMO).

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Bonding Molecular Orbitals (σσ, ππ) have lower energy and higher stability than the parent AOAOs, while Antibonding Molecular Orbitals (σ∗σ^*, π∗π^*) have higher energy and lower stability.

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The electronic configuration of MOMOs follows the Aufbau Principle, Pauli's Exclusion Principle, and Hund's Rule of Maximum Multiplicity.

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Bond Order (BOBO) is defined as half the difference between the number of electrons in bonding orbitals (NbN_b) and antibonding orbitals (NaN_a). A positive BOBO implies a stable molecule.

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The magnetic nature of a molecule depends on the presence of unpaired electrons: paramagnetic (at least one unpaired electron) or diamagnetic (all electrons paired).

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For molecules like Li2,Be2,B2,C2,N2Li_2, Be_2, B_2, C_2, N_2 (up to Z=7Z=7), the energy of π2px\pi 2p_x and π2py\pi 2p_y is lower than σ2pz\sigma 2p_z due to s−ps-p mixing.

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For molecules like O2,F2,Ne2O_2, F_2, Ne_2 (where Z>7Z > 7), the σ2pz\sigma 2p_z orbital is lower in energy than π2px\pi 2p_x and π2py\pi 2p_y.

📐Formulae

Bond Order (BO)=12(Nb−Na)Bond\,Order\,(BO) = \frac{1}{2}(N_b - N_a)

Stability:Nb>Na  ⟹  Stable;Nb≤Na  ⟹  UnstableStability: N_b > N_a \implies Stable; N_b \leq N_a \implies Unstable

Energy Order (Z≤7):σ1s<σ∗1s<σ2s<σ∗2s<(π2px=π2py)<σ2pz<(π∗2px=π∗2py)<σ∗2pzEnergy\,Order\,(Z \leq 7): \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

Energy Order (Z>7):σ1s<σ∗1s<σ2s<σ∗2s<σ2pz<(π2px=π2py)<(π∗2px=π∗2py)<σ∗2pzEnergy\,Order\,(Z > 7): \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

Bond Length∝1Bond OrderBond\,Length \propto \frac{1}{Bond\,Order}

💡Examples

Problem 1:

Calculate the bond order and determine the magnetic behavior of the oxygen molecule (O2O_2).

Solution:

  1. Total electrons in O2=8+8=16O_2 = 8 + 8 = 16.
  2. Filling according to energy order (Z>7Z > 7): (σ1s)2,(σ∗1s)2,(σ2s)2,(σ∗2s)2,(σ2pz)2,(π2px)2,(π2py)2,(π∗2px)1,(π∗2py)1(\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)^1, (\pi^* 2p_y)^1.
  3. Nb=10N_b = 10 (2 in 1s1s, 2 in 2s2s, 6 in 2p2p); Na=6N_a = 6 (2 in 1s∗1s^*, 2 in 2s∗2s^*, 2 in 2p∗2p^*).
  4. BO=12(10−6)=2BO = \frac{1}{2}(10 - 6) = 2.
  5. Since there are two unpaired electrons in the π∗2p\pi^* 2p orbitals, the molecule is paramagnetic.

Explanation:

Even though O2O_2 has an even number of electrons, MOT correctly predicts its paramagnetic nature due to the presence of two unpaired electrons in the degenerate antibonding pi orbitals, which Valence Bond Theory fails to explain.

Problem 2:

Determine if He2He_2 can exist as a stable molecule using MOT.

Solution:

  1. Total electrons in He2=2+2=4He_2 = 2 + 2 = 4.
  2. MOMO configuration: (σ1s)2,(σ∗1s)2(\sigma 1s)^2, (\sigma^* 1s)^2.
  3. Nb=2N_b = 2, Na=2N_a = 2.
  4. BO=12(2−2)=0BO = \frac{1}{2}(2 - 2) = 0.

Explanation:

A bond order of 00 indicates that no net force of attraction exists between the two atoms to hold them together. Therefore, the He2He_2 molecule does not exist under normal conditions.