B2 has a bond order of 1, calculated from its molecular orbital configuration.
Questions & Step-by-step Solutions
1 item
Q
Q: What is the bond order of the molecule B2?
Solution: B2 has a bond order of 1, calculated from its molecular orbital configuration.
Steps: 10
Step 1: Understand what bond order means. Bond order is a number that tells us how many bonds are between two atoms in a molecule.
Step 2: Know that B2 is a molecule made of two boron atoms.
Step 3: Find the total number of valence electrons for B2. Each boron atom has 3 valence electrons, so B2 has 3 + 3 = 6 valence electrons.
Step 4: Use the molecular orbital theory to fill the molecular orbitals with these 6 electrons. The order of filling is: sigma(1s), sigma*(1s), sigma(2s), sigma*(2s), sigma(2p), and pi(2p).
Step 5: Fill the orbitals: 2 electrons in sigma(1s), 2 electrons in sigma*(1s), 2 electrons in sigma(2s), and 0 in sigma*(2s) and pi(2p).
Step 6: Count the number of bonding and antibonding electrons. Bonding electrons are in sigma(1s), sigma(2s), and pi(2p) while antibonding electrons are in sigma*(1s) and sigma*(2s).
Step 7: For B2, there are 4 bonding electrons (2 in sigma(1s) and 2 in sigma(2s)) and 2 antibonding electrons (2 in sigma*(1s)).
Step 8: Use the bond order formula: Bond Order = (Number of bonding electrons - Number of antibonding electrons) / 2.
Step 9: Plug in the numbers: Bond Order = (4 - 2) / 2 = 2 / 2 = 1.
Step 10: Conclude that the bond order of the molecule B2 is 1.
