QUESTION IMAGE
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 |
|---|---|---|---|
| he₂ | yes<br>no | diamagnetic<br>paramagnetic | |
| f₂⁻ | yes<br>no | diamagnetic<br>paramagnetic |
Step1: Determine stability, magnetism and bond - order for $O_2$
Oxygen has 8 electrons per atom. In $O_2$, total electrons = 16. Using molecular - orbital theory, the molecular - orbital configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^1(\pi_{2p_y}^*)^1$. Since bond order > 0, it is stable. Unpaired electrons make it paramagnetic. Bond - order formula is $\text{Bond Order}=\frac{1}{2}(N_b - N_a)$, where $N_b$ is the number of bonding electrons and $N_a$ is the number of anti - bonding electrons. Here, $N_b = 10$ and $N_a = 6$, so $\text{Bond Order}=\frac{1}{2}(10 - 6)=2$.
Step2: Determine stability, magnetism and bond - order for $He_2$
Helium has 2 electrons per atom. In $He_2$, total electrons = 4. The molecular - orbital configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^2$. Using the bond - order formula $\text{Bond Order}=\frac{1}{2}(N_b - N_a)$, with $N_b = 2$ and $N_a = 2$, we get $\text{Bond Order}=\frac{1}{2}(2 - 2)=0$. So it is not stable. All electrons are paired, so it would be diamagnetic if it were stable.
Step3: Determine stability, magnetism and bond - order for $F_2$
Fluorine has 9 electrons per atom. In $F_2$, total electrons = 18. The molecular - orbital configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^2(\pi_{2p_y}^*)^2$. Using the bond - order formula $\text{Bond Order}=\frac{1}{2}(N_b - N_a)$, with $N_b = 10$ and $N_a = 8$, we get $\text{Bond Order}=\frac{1}{2}(10 - 8)=1$. All electrons are paired, so it is diamagnetic.
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| molecule | stable? | diamagnetic or paramagnetic? | bond order |
|---|---|---|---|
| $He_2$ | no | diamagnetic | 0 |
| $F_2$ | yes | diamagnetic | 1 |