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Why is the bond length of CO+ less than that of CO?
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Lou Heavner
Why is the bond length of CO+ less than that of CO?
Unfortunately, the arguments presented by buckminst and Uncle Al aren't completely right. The MO schemes are correct but the HOMO-$\sigma$ orbital ($s_{\sigma}^{*}(5\sigma)$ in buckminst's diagramm, $\sigma_{3}$ in Uncle Al's diagram) is not antibonding but slightly bonding in character because there is some mixing with the $\ce{p}$ atomic orbitals of the right symmetry (in this respect the MO scheme of Uncle Al is better than the one of buckminst because it indicates this mixing).
So, the original problem still persists.
One possible explanation for the bond shortening after ionization is that the ionization leads to a shift of the $\ce{CO}$-electron-polarization (on ionization an electron is lost from the mostly $\ce{C}$-centered HOMO-$\sigma$ orbital and this leads to a formation of a positive partial charge on the $\ce{C}$ atom).
This strengthens the covalence of the $\ce{CO}$-bond and thus reduces the bond length (since the HOMO-$\sigma$ orbital is only slightly bonding there is not much bonding lost by taking away one of its electrons - the lost bonding is outweighed by the gain in covalency).
You can think of this strenghtening of the bond covalence in the following way: Two atomic orbitals can interact better (form stronger bonds) if their energies are close. Without the positive partial charge on $\ce{C}$ the AOs of $\ce{O}$ lie energetically a lot below the AOs of $\ce{C}$ (this is most pronounced for the $\ce{s}$ AOs which form the HOMO-$\sigma*$ orbital). But with the positive partial charge on $\ce{C}$ the AOs of $\ce{C}$ are shifted down in energy and thus their energies are closer to the related AOs of $\ce{O}$ which leads to a stronger interaction when bonds are formed.
Unfortunately, the arguments presented by buckminst and Uncle Al aren't completely right. The MO schemes are correct but the HOMO-$\sigma$ orbital ($s_{\sigma}^{*}(5\sigma)$ in buckminst's diagramm, $\sigma_{3}$ in Uncle Al's diagram) is not antibonding but slightly bonding in character because there is some mixing with the $\ce{p}$ atomic orbitals of the right symmetry (in this respect the MO scheme of Uncle Al is better than the one of buckminst because it indicates this mixing).
So, the original problem still persists.One possible explanation for the bond shortening after ionization is that the ionization leads to a shift of the $\ce{CO}$-electron-polarization (on ionization an electron is lost from the mostly $\ce{C}$-centered HOMO-$\sigma$ orbital and this leads to a formation of a positive partial charge on the $\ce{C}$ atom).This strengthens the covalence of the $\ce{CO}$-bond and thus reduces the bond length (since the HOMO-$\sigma$ orbital is only slightly bonding there is not much bonding lost by taking away one of its electrons - the lost bonding is outweighed by the gain in covalency).You can think of this strenghtening of the bond covalence in the following way: Two atomic orbitals can interact better (form stronger bonds) if their energies are close. Without the positive partial charge on $\ce{C}$ the AOs of $\ce{O}$ lie energetically a lot below the AOs of $\ce{C}$ (this is most pronounced for the $\ce{s}$ AOs which form the HOMO-$\sigma*$ orbital). But with the positive partial charge on $\ce{C}$ the AOs of $\ce{C}$ are shifted down in energy and thus their energies are closer to the related AOs of $\ce{O}$ which leads to a stronger interaction when bonds are formed.
In LCAO terms, perhaps the sp-hybrid leaves two non-bonding lobes. The carbon lobe has more than half the electron density. and preferentially ionizes.More
There was definitely more to this question than I initially thought. The matter has some aspects that are not very intuitive - the MO diagram of CO makes the answer seem obvious but actually leads to the wrong reasoning (see my answer).More
@Martin Im not sure whether I understand your comment correctly. Are you saying that my answer is incorrect or that it might only be missing the electrostatic argument youre making? Also, you state "Upon ionisation, the former HOMO of CO also drops below the $\pi$ orbitals in CO(+1)". That is interesting: Have you calculated this yourself or how do you know that? But I fear Im not able to draw the connection between the higher effective nuclear charge leading to more contracted orbitals and ionized CO having a shorter bond length than normal CO. Could you eleborate on this some more, pls?More
I just went through your post again (unfortunately I already upvoted on it) and I noticed, that there might be an important argument missing: Upon ionisation, the nuclear charge in relation to the electronic charge is higher. Hence the orbitals have to become more contracted as the fewer electrons are more attracted. So another reason would be plain and simple electrostatics. (Upon ionisation, the former HOMO of CO also drops below the $\pi$ orbitals in CO(+1) - missing spin pairing, lower electron repulsion, and $\sigma$ is more contracted than $\pi$.)More
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Molecular Orbital theory does NOT predict a longer bond. It correctly predicts a shorter bond. The HOMO for $\ce{CO}$ is a sigma antibonding orbital. When it is ionized, the bond order is increased to 3.5, and the bond length decreases.
Molecular Orbital theory does NOT predict a longer bond. It correctly predicts a shorter bond. The HOMO for $\ce{CO}$ is a sigma antibonding orbital. When it is ionized, the bond order is increased to 3.5, and the bond length decreases.
Unfortunately, the arguments presented by
buckminstandUncle Alaren't completely right. The MO schemes are correct but the HOMO-$\sigma$ orbital ($s_{\sigma}^{*}(5\sigma)$ inbuckminst's diagramm, $\sigma_{3}$ inUncle Al's diagram) is not antibonding but slightly bonding in character because there is some mixing with the $\ce{p}$ atomic orbitals of the right symmetry (in this respect the MO scheme ofUncle Alis better than the one ofbuckminstbecause it indicates this mixing).So, the original problem still persists. One possible explanation for the bond shortening after ionization is that the ionization leads to a shift of the $\ce{CO}$-electron-polarization (on ionization an electron is lost from the mostly $\ce{C}$-centered HOMO-$\sigma$ orbital and this leads to a formation of a positive partial charge on the $\ce{C}$ atom). This strengthens the covalence of the $\ce{CO}$-bond and thus reduces the bond length (since the HOMO-$\sigma$ orbital is only slightly bonding there is not much bonding lost by taking away one of its electrons - the lost bonding is outweighed by the gain in covalency). You can think of this strenghtening of the bond covalence in the following way: Two atomic orbitals can interact better (form stronger bonds) if their energies are close. Without the positive partial charge on $\ce{C}$ the AOs of $\ce{O}$ lie energetically a lot below the AOs of $\ce{C}$ (this is most pronounced for the $\ce{s}$ AOs which form the HOMO-$\sigma*$ orbital). But with the positive partial charge on $\ce{C}$ the AOs of $\ce{C}$ are shifted down in energy and thus their energies are closer to the related AOs of $\ce{O}$ which leads to a stronger interaction when bonds are formed.
Unfortunately, the arguments presented by
buckminstandUncle Alaren't completely right. The MO schemes are correct but the HOMO-$\sigma$ orbital ($s_{\sigma}^{*}(5\sigma)$ inbuckminst's diagramm, $\sigma_{3}$ inUncle Al's diagram) is not antibonding but slightly bonding in character because there is some mixing with the $\ce{p}$ atomic orbitals of the right symmetry (in this respect the MO scheme ofUncle Alis better than the one ofbuckminstbecause it indicates this mixing).So, the original problem still persists.One possible explanation for the bond shortening after ionization is that the ionization leads to a shift of the $\ce{CO}$-electron-polarization (on ionization an electron is lost from the mostly $\ce{C}$-centered HOMO-$\sigma$ orbital and this leads to a formation of a positive partial charge on the $\ce{C}$ atom).This strengthens the covalence of the $\ce{CO}$-bond and thus reduces the bond length (since the HOMO-$\sigma$ orbital is only slightly bonding there is not much bonding lost by taking away one of its electrons - the lost bonding is outweighed by the gain in covalency).You can think of this strenghtening of the bond covalence in the following way: Two atomic orbitals can interact better (form stronger bonds) if their energies are close. Without the positive partial charge on $\ce{C}$ the AOs of $\ce{O}$ lie energetically a lot below the AOs of $\ce{C}$ (this is most pronounced for the $\ce{s}$ AOs which form the HOMO-$\sigma*$ orbital). But with the positive partial charge on $\ce{C}$ the AOs of $\ce{C}$ are shifted down in energy and thus their energies are closer to the related AOs of $\ce{O}$ which leads to a stronger interaction when bonds are formed.
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Molecular Orbital theory does NOT predict a longer bond. It correctly predicts a shorter bond. The HOMO for $\ce{CO}$ is a sigma antibonding orbital. When it is ionized, the bond order is increased to 3.5, and the bond length decreases.
Molecular Orbital theory does NOT predict a longer bond. It correctly predicts a shorter bond. The HOMO for $\ce{CO}$ is a sigma antibonding orbital. When it is ionized, the bond order is increased to 3.5, and the bond length decreases.
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