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L.C.A.O. Principle is involved in the fo...

L.C.A.O. Principle is involved in the formation of the molecular orbitals according to molecular orbital theory. The energy of the bonding molecular orbital is less than that of the combining atomic orbitals while that of the antibonding molecular orbitals while that of the order `(B.O.)=1/2(N_(b)-N_(a))` helps in predicting (i) formation of molecules/molecular ions, bond dissociation energy, stability and bond length. Only the molecules or ions with positive B.O. can be formed. These will be diamagnetic if all molecular orbitals are dilled and paramagnetic if one of more are half filled. The atomic orbitals at the time of overlap must have the same symmetry as well.
The bond order (B.O.) in `B_(2)` molecule is:

A

0

B

1

C

2

D

3

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To calculate the bond order (B.O.) of the B2 molecule using the molecular orbital theory, we can follow these steps: ### Step-by-Step Solution: 1. **Determine the Total Number of Electrons**: - Each boron atom has 5 electrons. Since B2 has two boron atoms, the total number of electrons is: \[ 2 \times 5 = 10 \text{ electrons} \] 2. **Construct the Molecular Orbital Diagram**: - The molecular orbitals for B2 are filled according to the energy levels. The order of filling for B2 is: - σ(1s) - σ*(1s) - σ(2s) - σ*(2s) - π(2p_x) - π(2p_y) - σ(2p_z) - σ*(2p_z) - For B2, the molecular orbital filling will look like this: - σ(1s)² - σ*(1s)² - σ(2s)² - σ*(2s)² - π(2p_x)¹ - π(2p_y)¹ 3. **Fill the Molecular Orbitals**: - Fill the molecular orbitals with the 10 electrons: - σ(1s)² → 2 electrons - σ*(1s)² → 2 electrons - σ(2s)² → 2 electrons - σ*(2s)² → 2 electrons - π(2p_x)¹ → 1 electron - π(2p_y)¹ → 1 electron - Total: 2 + 2 + 2 + 2 + 1 + 1 = 10 electrons 4. **Identify Bonding and Antibonding Electrons**: - Bonding electrons are those in the bonding molecular orbitals: - σ(1s), σ(2s), π(2p_x), π(2p_y) - Antibonding electrons are those in the antibonding molecular orbitals: - σ*(1s), σ*(2s) 5. **Count the Electrons**: - Bonding electrons = 2 (σ(1s)) + 2 (σ(2s)) + 1 (π(2p_x)) + 1 (π(2p_y)) = 6 - Antibonding electrons = 2 (σ*(1s)) + 2 (σ*(2s)) = 4 6. **Calculate Bond Order (B.O.)**: - Use the formula for bond order: \[ \text{Bond Order (B.O.)} = \frac{1}{2} (N_b - N_a) \] - Where \(N_b\) is the number of bonding electrons and \(N_a\) is the number of antibonding electrons: \[ \text{B.O.} = \frac{1}{2} (6 - 4) = \frac{1}{2} \times 2 = 1 \] ### Final Answer: The bond order (B.O.) in the B2 molecule is **1**. ---
To calculate the bond order (B.O.) of the B2 molecule using the molecular orbital theory, we can follow these steps: ### Step-by-Step Solution: 1. **Determine the Total Number of Electrons**: - Each boron atom has 5 electrons. Since B2 has two boron atoms, the total number of electrons is: \[ 2 \times 5 = 10 \text{ electrons} ...
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