krit.club logo

Chemical Bonding and Molecular Structure - Molecular Orbital Theory

Grade 11CBSEChemistry

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

🔑Concepts

•

Molecular Orbital Theory (MOT) explains that atomic orbitals combine to form molecular orbitals where electrons are influenced by more than one nucleus. This is based on the Linear Combination of Atomic Orbitals (LCAO) principle.

•

Bonding Molecular Orbitals (BMO) are formed by the additive effect of atomic orbitals, leading to lower energy and higher stability. Antibonding Molecular Orbitals (ABMO) are formed by the subtractive effect, leading to higher energy and lower stability.

•

The number of molecular orbitals formed is always equal to the number of atomic orbitals combined. For example, two 1s1s orbitals combine to form one σ1s\sigma 1s (bonding) and one σ∗1s\sigma^* 1s (antibonding) orbital.

•

For homonuclear diatomic molecules of elements like Li2,Be2,B2,C2,N2Li_2, Be_2, B_2, C_2, N_2 (where Z≤7Z \le 7), the energy level sequence is: σ1s<σ∗1s<σ2s<σ∗2s<(π2px=π2py)<σ2pz<(π∗2px=π∗2py)<σ∗2pz\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.

•

For O2O_2 and F2F_2 (where Z>7Z > 7), the energy level sequence changes due to the lack of s−ps-p mixing: σ1s<σ∗1s<σ2s<σ∗2s<σ2pz<(π2px=π2py)<(π∗2px=π∗2py)<σ∗2pz\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 Order is a measure of the strength of a bond. A molecule is stable if Nb>NaN_b > N_a (Bond Order >0> 0). If Nb=NaN_b = N_a, the molecule is unstable and does not exist (e.g., He2He_2).

•

Magnetic Nature: If all molecular orbitals in a molecule are doubly occupied, the substance is diamagnetic. If one or more molecular orbitals are singly occupied, the substance is paramagnetic.

📐Formulae

Bond Order (B.O.)=12(Nb−Na)\text{Bond Order (B.O.)} = \frac{1}{2} (N_b - N_a) where NbN_b is the number of bonding electrons and NaN_a is the number of antibonding electrons.

ψMO=ψA±ψB\psi_{MO} = \psi_A \pm \psi_B representing the wave function of the molecular orbital.

Bond Strength∝Bond Order\text{Bond Strength} \propto \text{Bond Order}

Bond Length∝1Bond Order\text{Bond Length} \propto \frac{1}{\text{Bond Order}}

💡Examples

Problem 1:

Calculate the bond order and predict the magnetic behavior of the Oxygen molecule (O2O_2).

Solution:

Oxygen (Z=8Z=8) has 1616 electrons. The electronic configuration according to MOT is: (σ1s)2,(σ∗1s)2,(σ2s)2,(σ∗2s)2,(σ2pz)2,(π2px2=π2py2),(π∗2px1=π∗2py1)(\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). Here, Nb=10N_b = 10 and Na=6N_a = 6. Bond Order =12(10−6)=2= \frac{1}{2}(10 - 6) = 2.

Explanation:

Since the bond order is 22, a double bond exists between the two oxygen atoms. Because there are two unpaired electrons in the π∗2px\pi^* 2p_x and π∗2py\pi^* 2p_y orbitals, O2O_2 is paramagnetic.

Problem 2:

Why does He2He_2 molecule not exist?

Solution:

Helium (Z=2Z=2) has 22 electrons per atom, so He2He_2 would have 44 electrons. The MOT configuration is: (σ1s)2,(σ∗1s)2(\sigma 1s)^2, (\sigma^* 1s)^2. Here, Nb=2N_b = 2 and Na=2N_a = 2. Bond Order =12(2−2)=0= \frac{1}{2}(2 - 2) = 0.

Explanation:

A bond order of zero implies that no net force of attraction exists between the two atoms, meaning the molecule is unstable and cannot exist.

Problem 3:

Compare the stability of N2N_2 and N2+N_2^+.

Solution:

For N2N_2 (14e−14 e^-): B.O.=12(10−4)=3B.O. = \frac{1}{2}(10 - 4) = 3. For N2+N_2^+ (13e−13 e^-): One electron is removed from a bonding orbital (σ2pz\sigma 2p_z), so B.O.=12(9−4)=2.5B.O. = \frac{1}{2}(9 - 4) = 2.5.

Explanation:

Since the bond order of N2N_2 (3.03.0) is greater than that of N2+N_2^+ (2.52.5), N2N_2 is more stable than N2+N_2^+.

Molecular Orbital Theory - Revision Notes & Key Formulas | CBSE Class 11 Chemistry