the table below lists information about some diatomic molecules or molecular ions. for each molecule in the…

Explanation:

Step1: Recall molecular - orbital theory rules

Stability is determined by bond order (a positive bond order means stable). Diamagnetism occurs when all electrons are paired, and paramagnetism when there are un - paired electrons. The bond - order formula is $BO=\frac{1}{2}(N_{b}-N_{a})$, where $N_{b}$ is the number of bonding electrons and $N_{a}$ is the number of antibonding electrons.

Step2: Analyze $B_{2}^{-}$

The electron configuration of $B$ is $1s^{2}2s^{2}2p^{1}$. $B_{2}^{-}$ has a total of $2\times5 + 1=11$ electrons. The molecular - orbital configuration is $(\sigma_{1s})^{2}(\sigma_{1s}^{})^{2}(\sigma_{2s})^{2}(\sigma_{2s}^{})^{2}(\pi_{2p_{x}})^{2}(\pi_{2p_{y}})^{1}$.

• Bond order: $N_{b}=7$, $N_{a}=4$, $BO=\frac{1}{2}(7 - 4)=\frac{3}{2}$. Since $BO>0$, it is stable.
• There is an un - paired electron in the $\pi_{2p_{y}}$ orbital, so it is paramagnetic.

Step3: Analyze $B_{2}$

$B_{2}$ has $2\times5 = 10$ electrons. The molecular - orbital configuration is $(\sigma_{1s})^{2}(\sigma_{1s}^{})^{2}(\sigma_{2s})^{2}(\sigma_{2s}^{})^{2}(\pi_{2p_{x}})^{1}(\pi_{2p_{y}})^{1}$.

• Bond order: $N_{b}=6$, $N_{a}=4$, $BO=\frac{1}{2}(6 - 4)=1$. Since $BO>0$, it is stable.
• There are un - paired electrons in the $\pi_{2p_{x}}$ and $\pi_{2p_{y}}$ orbitals, so it is paramagnetic.

Step4: Analyze $Be_{2}$

The electron configuration of $Be$ is $1s^{2}2s^{2}$. $Be_{2}$ has $2\times4 = 8$ electrons. The molecular - orbital configuration is $(\sigma_{1s})^{2}(\sigma_{1s}^{})^{2}(\sigma_{2s})^{2}(\sigma_{2s}^{})^{2}$.

• Bond order: $N_{b}=4$, $N_{a}=4$, $BO=\frac{1}{2}(4 - 4)=0$. Since $BO = 0$, it is unstable.

Answer:

molecule	stable?	diamagnetic or paramagnetic?	bond order
$B_{2}^{-}$	yes	paramagnetic	$\frac{3}{2}$
$B_{2}$	yes	paramagnetic	$1$
$Be_{2}$	no	N/A (not applicable as unstable)	$0$