Question

the table below lists information about some diatomic molecules or molecular ions. for each molecule in the table: first, decide if the molecule is stable or not. then, if your answer to this question is \yes\: • decide whether the molecule would be diamagnetic or paramagnetic. • calculate the molecules bond order.

molecule	stable?	diamagnetic or paramagnetic?	bond order
$be_2^+$	$\begin{array}{l}\text{yes}\\text{no}end{array}$	$\begin{array}{l}\text{diamagnetic}\\text{paramagnetic}end{array}$	$square$
$he_2$	$\begin{array}{l}\text{yes}\\text{no}end{array}$	$\begin{array}{l}\text{diamagnetic}\\text{paramagnetic}end{array}$	$square$

Explanation:

Step1: Determine stability of $O_2^{-}$

Use molecular - orbital theory. $O_2^{-}$ has 13 valence electrons. The molecular - orbital configuration is $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p})^2(\pi_{2p})^4(\pi_{2p}^*)^3$. Bond order $=\frac{1}{2}(8 - 5)=1.5$. Since bond order > 0, it is stable.

Step2: Determine magnetism of $O_2^{-}$

There is one unpaired electron in the $\pi_{2p}^*$ orbital. So, it is paramagnetic.

Step3: Determine stability of $Be_2^{+}$

$Be_2^{+}$ has 7 valence electrons. The molecular - orbital configuration is $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p})^1$. Bond order $=\frac{1}{2}(3 - 2)=0.5$. Since bond order > 0, it is stable.

Step4: Determine magnetism of $Be_2^{+}$

There is one unpaired electron in the $\sigma_{2p}$ orbital. So, it is paramagnetic.

Step5: Determine stability of $He_2$

$He_2$ has 4 valence electrons. The molecular - orbital configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^2$. Bond order $=\frac{1}{2}(2 - 2)=0$. Since bond order = 0, it is not stable.

Answer:

molecule	stable?	diamagnetic or paramagnetic?	bond order
$Be_2^{+}$	yes	paramagnetic	0.5
$He_2$	no	N/A	N/A