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In the MO diagram for $O_{2}$ :

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Shouldn't there be 3 different $3\sigma_{g}$ Bonding MO's as each p orbital can form a $\sigma_{g}$ Bonding MO along its respective axes?

In which case the $3\sigma_{g}$ has 6 electrons rather than 2.

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  • $\begingroup$ Do you know what the definition of a $\sigma$ orbital is? Also, a single orbital can never have 6 electrons in it. It can only have a maximum of 2. $\endgroup$ – bon Dec 5 '15 at 10:00
  • $\begingroup$ Yes I think its an orbital which has no nodal planes as its rotated about its inter-nuclear axis. Its invariant under cylindrical rotation about the nucleus. I understand that an orbital can never have more than 2 electrons, but doesn't the p orbital along the x axis form a different $\sigma_{g}$ bonding MO in comparison the p orbital along the y and z axes? $\endgroup$ – J.Gudal Dec 5 '15 at 10:05
  • $\begingroup$ Look at the orbitals formed by overlap of two $p_x$ orbitals (taking z as the internuclear axis, by convention) and tell me whether they have $\sigma$ symmetry. $\endgroup$ – bon Dec 5 '15 at 10:08
  • $\begingroup$ No they don't. Oh right thanks I get it now, it makes sense. $\endgroup$ – J.Gudal Dec 5 '15 at 10:10
  • $\begingroup$ Feel free to answer your own question so that it can be of use to future users. $\endgroup$ – bon Dec 5 '15 at 10:13
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Thanks to bon in the comments I understand that since the z-axis is taken by convention as the inter-nuclear axis, the $p_x$ and $p_y$ orbitals are unable to form molecular orbitals that are invariant under cylindrical rotation about the z axis. Consequently there is only one $3\sigma_g$ and one $3\sigma_u$ orbital formed from $p_z$ orbitals.

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  • $\begingroup$ The convention is not important. You can only create one sigma-bond with the p-orbitals. I can give you the reason why in an answer but I have not time to do it right now. $\endgroup$ – ParaH2 Dec 5 '15 at 12:34

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