Using LCAO-MO theory, what is bond order of C2- (the one minus ion)
The Correct Answer and Explanation is:
The bond order of the C2−\text{C}_2^-C2− ion can be determined using the Molecular Orbital (MO) theory and the Linear Combination of Atomic Orbitals (LCAO) approach.
First, let’s review the molecular orbitals formed by the combination of atomic orbitals of carbon atoms. Each carbon atom has 6 electrons, so in the neutral C2\text{C}_2C2 molecule, the total number of electrons is 12. The molecular orbitals are filled in increasing energy levels based on the order of their formation.
For C2−\text{C}_2^-C2− (the C2\text{C}_2C2 ion with one extra electron):
- The total number of electrons in C2−\text{C}_2^-C2− is 13 (12 from the neutral molecule and 1 extra due to the negative charge).
- The molecular orbitals for C2\text{C}_2C2 in order of increasing energy are:
- σ1s\sigma_{1s}σ1s, σ1s∗\sigma^*_{1s}σ1s∗ (from the 1s atomic orbitals),
- σ2s\sigma_{2s}σ2s, σ2s∗\sigma^*_{2s}σ2s∗ (from the 2s atomic orbitals),
- σ2pz\sigma_{2p_z}σ2pz, π2px\pi_{2p_x}π2px, π2py\pi_{2p_y}π2py, π2px∗\pi^*_{2p_x}π2px∗, π2py∗\pi^*_{2p_y}π2py∗, and σ2pz∗\sigma^*_{2p_z}σ2pz∗ (from the 2p atomic orbitals).
Electron configuration of C2−\text{C}_2^-C2−:
- The electrons are filled into the molecular orbitals as follows:
- σ1s2\sigma_{1s}^2σ1s2, \sigma^*_{1s}^2 (2 electrons each),
- σ2s2\sigma_{2s}^2σ2s2, \sigma^*_{2s}^2 (2 electrons each),
- σ2pz2\sigma_{2p_z}^2σ2pz2,
- π2px2\pi_{2p_x}^2π2px2, π2py2\pi_{2p_y}^2π2py2 (4 electrons total in the degenerate π\piπ orbitals),
- \pi^*_{2p_x}^1 (1 electron in the anti-bonding π∗\pi^*π∗ orbital).
Calculation of bond order:
Bond order is given by the formula:Bond Order=12(Number of bonding electrons−Number of anti-bonding electrons)\text{Bond Order} = \frac{1}{2} \left( \text{Number of bonding electrons} – \text{Number of anti-bonding electrons} \right)Bond Order=21(Number of bonding electrons−Number of anti-bonding electrons)
From the configuration, the bonding electrons are those in σ1s\sigma_{1s}σ1s, σ2s\sigma_{2s}σ2s, σ2pz\sigma_{2p_z}σ2pz, and π2px\pi_{2p_x}π2px, π2py\pi_{2p_y}π2py orbitals, giving us:
- Bonding electrons: 2(σ1s)+2(σ2s)+2(σ2pz)+4(π2px+π2py)=102 (\sigma_{1s}) + 2 (\sigma_{2s}) + 2 (\sigma_{2p_z}) + 4 (\pi_{2p_x} + \pi_{2p_y}) = 102(σ1s)+2(σ2s)+2(σ2pz)+4(π2px+π2py)=10
- Anti-bonding electrons: 2(σ1s∗)+2(σ2s∗)+1(π2px∗)=52 (\sigma^*_{1s}) + 2 (\sigma^*_{2s}) + 1 (\pi^*_{2p_x}) = 52(σ1s∗)+2(σ2s∗)+1(π2px∗)=5
So, the bond order is:Bond Order=12(10−5)=12×5=2.5\text{Bond Order} = \frac{1}{2} \left( 10 – 5 \right) = \frac{1}{2} \times 5 = 2.5Bond Order=21(10−5)=21×5=2.5
Conclusion:
The bond order of the C2−\text{C}_2^-C2− ion is 2.5, which indicates a bond that is stronger than a single bond but not as strong as a double bond, reflecting the presence of the extra electron in an anti-bonding orbital.
