What is the bond order of the molecule B2 according to molecular orbital theory?
Practice Questions
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What is the bond order of the molecule B2 according to molecular orbital theory?
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B2 has a bond order of 1, calculated as (2 bonding - 0 antibonding)/2.
Questions & Step-by-step Solutions
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Q: What is the bond order of the molecule B2 according to molecular orbital theory?
Solution: B2 has a bond order of 1, calculated as (2 bonding - 0 antibonding)/2.
Steps: 7
Step 1: Identify the total number of electrons in the B2 molecule. Each boron atom has 5 electrons, so B2 has a total of 10 electrons.
Step 2: Fill the molecular orbitals according to the molecular orbital theory. The order of filling is: sigma(1s), sigma*(1s), sigma(2s), sigma*(2s), sigma(2p), pi(2p), pi(2p)*, sigma(2p)*.
Step 3: For B2, the first 10 electrons fill the orbitals as follows: 2 in sigma(1s), 2 in sigma*(1s), 2 in sigma(2s), 2 in sigma*(2s), and 2 in sigma(2p).
Step 4: Count the number of bonding and antibonding electrons. Bonding electrons are in sigma(1s), sigma(2s), and sigma(2p) orbitals. Antibonding electrons are in sigma*(1s) and sigma*(2s) orbitals.
Step 5: For B2, there are 2 bonding electrons (from sigma(2p)) and 0 antibonding electrons (since the pi(2p) orbitals are not filled).
Step 6: Calculate the bond order using the formula: Bond Order = (Number of bonding electrons - Number of antibonding electrons) / 2.
Step 7: Substitute the values: Bond Order = (2 - 0) / 2 = 1.
