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So, I was wondering, I have these molecules $\ce{O^+2}$, $\ce{O^-2}$, $\ce{O2}$ and $\ce{C^+2}$.

When I try to find their bond order and their magnetic nature I get this:

$\ce{O^+2} = \frac{1}{2} \left( 10 - 5 \right) = 2.5$ - Paramagnetic

$\ce{O^-2} = \frac{1}{2} \left( 10 - 7 \right) = 1.5$ - Paramagnetic

$\ce{O2} = \frac{1}{2} \left( 10 - 6 \right) = 2$ - Paramagnetic

$\ce{C^+_2} = \frac{1}{2} \left( 7 - 4 \right) = 1.5$ - Paramagnetic

You can note that $\ce{O^-_2}$ and $\ce{C^+_2}$ have the same bond order and are also paramagnetic, but how can I know which one is more stable so I can arrange them in stability order if $\ce{O^-_2}$ and $\ce{C^+_2}$ have the same orders?

Thanks!

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    $\begingroup$ For future reference: In addition to MathJax, the bodies of questions, answers and comments in ChemSE may use mhchem which eases the input and improves the rendering of chemical formulae and equations. Because it is somewhat special not all engines understand and render properly, MathJax and mhchem shall not be used in the titles. $\endgroup$ – Buttonwood Dec 19 '20 at 8:39
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To recognize the more stable molecule you may extend the description with bond orders is the one considering how the mathematical concept to mix atomic orbitals by LCAO and visualizing the results in Molecular orbital diagrams and compute the overall energy of such a molecule. Eventually, you compare the total energy of $\ce{O^-_2}$ with $\ce{C^+_2}$ using the same external reference.

As about the hydrogen molecule shown below:

enter image description here

Two important details about symmetric diatomic moleculs to retain are:

  • For $n$ atomic orbitals, LCAO yields $n$ molecular orbitals.
  • The split between the lowered bonding molecular orbital and the higher antibonding orbital is not symmetric. The difference between the energy level of the initial atomic orbitals and the the antibinding molecular orbital is higher, than the difference between the initial atomic orbitals and the binding molecular orbital.

So you draw the diagram including the energetic levels, and populate the levels with electrons from the bottom up. If there multiple molecular orbitals of same energy, you fill the orbitals first with one, and later with the second electron. You then compute the the overall energy of the molecule and later compare the results about the molecules in question against a reference in common, e.g. vacuum.

Note that the notation $\sigma$ and $\sigma{}^*$ is about the rotation symmetry around the bond between the molecule (contrasting to $\pi$ and $\pi{}^*$ as perpendicular to this), and not if the molecular orbital in question is conceptually formed from $s$, or $p$ atomic orbitals. Equally, the energetic consecution of levels of the molecular orbitals may change, e.g. between nitrogen and oxygen:

enter image description here

(using material from here)

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  • $\begingroup$ Thank you! I did some research and now I see that C energy levels are a little higher than O levels due to electronegativity values. $\endgroup$ – Adolf Dec 19 '20 at 23:58
  • $\begingroup$ So I ended up like this O2+>O2>C2+>O2- From higher to lower stability, would you agree? $\endgroup$ – Adolf Dec 20 '20 at 1:36
  • $\begingroup$ @Adolf O2+ more stable than O2? I see why you ended up here, but that energy should be counted for electron number and is differently shifted compared to vacuum. $\endgroup$ – Alchimista Dec 20 '20 at 13:33
  • $\begingroup$ Thanks for commenting @Alchimista, Indeed its also weird to me that O2 its more stable than O2+ but when calculating bond order O2 its a 2 while O2+ its a 2.5, since they have the same number of protons in the nucleus I think this is because O2 has 2 unpaired electrons while O2+ only has one therefore making it more stable. What do you think? $\endgroup$ – Adolf Dec 20 '20 at 20:33
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    $\begingroup$ @Adolf well there is one term missing. No ion is more stable than a neutral species in term of its absolute energy vs some and the same reference. What you are at the moment is more or less that a second ionisation potential is higher than the first one, and it is not a surprise, isn't? $\endgroup$ – Alchimista Dec 21 '20 at 8:59

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