Question

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part a

use molecular orbital theory to determine the bond order in b₂.

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Explanation:

Step1: Determine the number of valence electrons in B₂

Boron (B) has an atomic number of 5. The electron configuration of B is \(1s^{2}2s^{2}2p^{1}\). Each B atom has 3 valence electrons (\(2s^{2}2p^{1}\)). For \(B_{2}\), the total number of valence electrons is \(3 + 3=6\).

Step2: Write the molecular orbital configuration for \(B_{2}\)

The molecular orbital configuration for \(B_{2}\) (using the order \(\sigma_{2s}<\sigma_{2s}^{*}<\pi_{2p_{x}}=\pi_{2p_{y}}<\sigma_{2p_{z}}<\pi_{2p_{x}}^{*}=\pi_{2p_{y}}^{*}<\sigma_{2p_{z}}^{*}\)) is \((\sigma_{2s})^{2}(\sigma_{2s}^{*})^{2}(\pi_{2p_{x}})^{1}(\pi_{2p_{y}})^{1}\)

Step3: Calculate the bond order

The bond - order formula is \(BO=\frac{1}{2}(N_{b}-N_{a})\), where \(N_{b}\) is the number of electrons in bonding orbitals and \(N_{a}\) is the number of electrons in antibonding orbitals.
Here, \(N_{b} = 4\) (2 from \(\sigma_{2s}\) and 2 from \(\pi_{2p}\) orbitals, but wait, no: actually, for \(B_{2}\), the correct count: bonding electrons \(N_{b}=4\) (\(2\) in \(\sigma_{2s}\) and \(2\) in \(\pi_{2p}\) orbitals considering the correct filling. Antibonding electrons \(N_{a} = 2\) (\(2\) in \(\sigma_{2s}^{*}\)).

$$BO=\frac{1}{2}(4 - 2)$$

Answer:

\(1\)