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Chemical Bonding and Molecular Structure - Bonding in Some Homonuclear Diatomic Molecules

Grade 11CBSEChemistry

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

🔑Concepts

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Molecular Orbital Theory (MOT) describes the distribution of electrons in molecules using molecular orbitals formed by the Linear Combination of Atomic Orbitals (LCAO).

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Bonding Molecular Orbitals (BMO) are formed by the additive effect of atomic wave functions (ψMO=ψA+ψBψ_{MO} = ψ_A + ψ_B), having lower energy and higher stability.

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Antibonding Molecular Orbitals (ABMO) are formed by the subtractive effect of atomic wave functions (ψMO∗=ψA−ψBψ^*_{MO} = ψ_A - ψ_B), having higher energy and lower stability.

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

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For molecules like Li2Li_2, Be2Be_2, B2B_2, C2C_2, and N2N_2 (where Z≤7Z \le 7), the σ2pz\sigma 2p_z orbital is higher in energy than the π2px\pi 2p_x and π2py\pi 2p_y orbitals due to s−ps-p mixing.

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For molecules like O2O_2 and F2F_2 (where Z>7Z > 7), the σ2pz\sigma 2p_z orbital is lower in energy than the π2px\pi 2p_x and π2py\pi 2p_y orbitals.

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A molecule is paramagnetic if it contains one or more unpaired electrons and diamagnetic if all electrons are paired.

📐Formulae

Bond Order (B.O.)=12(Nb−Na)Bond\ Order\ (B.O.) = \frac{1}{2}(N_b - N_a)

Configuration for Z≤7:σ1s<σ∗1s<σ2s<σ∗2s<(π2px=π2py)<σ2pz<(π∗2px=π∗2py)<σ∗2pz\text{Configuration for } Z \le 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

Configuration for Z>7:σ1s<σ∗1s<σ2s<σ∗2s<σ2pz<(π2px=π2py)<(π∗2px=π∗2py)<σ∗2pz\text{Configuration for } 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

💡Examples

Problem 1:

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

Solution:

Total electrons in O2=8+8=16O_2 = 8 + 8 = 16. The MO electronic configuration 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. B.O.=12(10−6)=2B.O. = \frac{1}{2}(10 - 6) = 2 Since there are two unpaired electrons in the π∗\pi^* orbitals, O2O_2 is paramagnetic.

Explanation:

The bond order of 2 corresponds to a double bond. The presence of unpaired electrons in antibonding π∗\pi^* orbitals explains its experimental paramagnetic nature, which Valence Bond Theory failed to explain.

Problem 2:

Explain why the He2He_2 molecule does not exist based on Bond Order.

Solution:

Total electrons in He2=2+2=4He_2 = 2 + 2 = 4. The MO electronic configuration is: (σ1s)2(σ∗1s)2(\sigma 1s)^2 (\sigma^* 1s)^2 Here, Nb=2N_b = 2 and Na=2N_a = 2. B.O.=12(2−2)=0B.O. = \frac{1}{2}(2 - 2) = 0

Explanation:

A Bond Order of 0 means that no net force of attraction exists between the two Helium atoms. Therefore, the He2He_2 molecule is unstable and does not exist.

Problem 3:

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

Solution:

For N2N_2 (14 electrons): B.O.=12(10−4)=3.0B.O. = \frac{1}{2}(10 - 4) = 3.0 For N2+N_2^+ (13 electrons): The configuration loses one electron from a bonding orbital (σ2pz\sigma 2p_z). B.O.=12(9−4)=2.5B.O. = \frac{1}{2}(9 - 4) = 2.5

Explanation:

Since N2N_2 has a higher bond order (3.03.0) than N2+N_2^+ (2.52.5), N2N_2 is more stable and has a shorter bond length than N2+N_2^+.