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 o₂⁺ yes no diamagnetic paramagnetic o₂²⁻ yes no diamagnetic paramagnetic he₂⁺ yes no diamagnetic paramagnetic

Explanation:

Step1: Recall molecular - orbital theory for $O_2^+$

The atomic number of oxygen is 8. For $O_2^+$, the number of electrons is $2\times8 - 1=15$. The molecular - orbital configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p})^2(\pi_{2p})^4(\pi_{2p}^*)^1$.

Step2: Determine stability

Since the bond order is non - zero, $O_2^+$ is stable. 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. Here, $N_b = 10$ and $N_a = 5$, so $BO=\frac{1}{2}(10 - 5)=2.5$.

Step3: Determine magnetism

There is one unpaired electron in the $\pi_{2p}^*$ orbital, so $O_2^+$ is paramagnetic.

Step4: Recall molecular - orbital theory for $O_2^{2 - }$

The number of electrons in $O_2^{2 - }$ is $2\times8+2 = 18$. The molecular - orbital configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p})^2(\pi_{2p})^4(\pi_{2p}^*)^4$.

Step5: Determine stability

Using the bond - order formula $BO=\frac{1}{2}(N_b - N_a)$, with $N_b = 10$ and $N_a = 8$, we get $BO=\frac{1}{2}(10 - 8)=1$. Since the bond order is non - zero, $O_2^{2 - }$ is stable.

Step6: Determine magnetism

All electrons are paired, so $O_2^{2 - }$ is diamagnetic.

Step7: Recall molecular - orbital theory for $He_2^+$

The number of electrons in $He_2^+$ is $2\times2 - 1=3$. The molecular - orbital configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^1$.

Step8: Determine stability

Using the bond - order formula $BO=\frac{1}{2}(N_b - N_a)$, with $N_b = 2$ and $N_a = 1$, we get $BO=\frac{1}{2}(2 - 1)=0.5$. Since the bond order is non - zero, $He_2^+$ is stable.

Step9: Determine magnetism

There is one unpaired electron in the $\sigma_{1s}^*$ orbital, so $He_2^+$ is paramagnetic.

Answer:

molecule	stable?	diamagnetic or paramagnetic?	bond order
$O_2^{2 - }$	yes	diamagnetic	1
$He_2^+$	yes	paramagnetic	0.5