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Question

Which of these species has a longer bond, B2\mathrm{B}_2 or B2+\mathrm{B}_2^{+}? Explain in terms of molecular orbital theory.

Solution

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Answered 1 year ago
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To determine which species has the longer bond, we have to draw a molecular orbital (MO) energy level diagram for BX2\ce{B2} and BX2X+\ce{B2+} and calculate the bond order. The species with a higher value of the bond order will have a smaller bond because the attraction between electrons is stronger which leads to a smaller bond. The formula for a bond order is:

bond order=# of e in the bonding MO# of e in antibonding MO2\small \begin{align*} \text{bond order}=\dfrac{\text{\# of e$^-$ in the bonding MO}-\text{\# of $e^-$ in antibonding MO}}{2} \end{align*}

Drawing the diagram, we have to follow the rules: Hund's rule, each orbital can't have more than 2 electrons with opposite spins, and we have to fill the lower energy orbitals first.

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