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I was trying to construct a MO diagram for ClO2, and since I'm not too sure about my final result I thought I'd check it from one of the experts here.

ClO2 is of C2v point group, so I just read off the C2v character table and got the following group orbitals [where z axis is that of the principal axis, and the outer atoms are aligned such that the y-axis points towards the centre atom]:

A1 symmetry, two s orbitals

B1 symmetry, two s orbitals

A1 symmetry, two py orbitals

B1 symmetry, two py orbitals

A1 symmetry, two pz orbitals

B1 symmetry, two pz orbitals

A2 symmetry, two px orbitals

B2 symmetry, two px orbitals

Adding those to the s orbital (A1), px orbital (B2), py orbital (B1) and pz orbital (A1) of the chlorine, we get the following (see follow-up post):

Would my prediction of the 12 MOs be correct?

Thanks!

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Yes, there are 12 orbitals, but they are typically drawn differently from what you have. It is generally assumed that the oxygen $2s$ orbitals lie too low in energy to be involved in bonding, so you have an $a_1$ and $b_1$ orbital at low energy comprised of those two $2s$ orbitals in the same phase ($a_1$) or opposite phase ($b_1$) with no $\ce{Cl}$ orbital involved. You have only included the $b_1$ of this pair. The $a_1$ replaces the $a_1$ orbital you have drawn with oxygen $p_y$ orbitals. Then your $a_1$ pair comprised of $s$ orbitals from all three atoms would instead be comprised of chlorine $3s$ and oxygen $p_y$.

You can read a description of this in Walsh's original paper[1] on AB2 non-hydride molecules.

He also assumes that in bent molecules like $\ce{ClO2}$ there is significant mixing of the $s$ and $p_z$ orbitals on the central atom so that in the $\sigma/\sigma^*$ and $\pi/\pi^*$ $a_1$ groups, the orbital on $\ce{Cl}$ is similar to an $sp$ hybrid in both cases rather than an $s$ orbital in one pair and $p$ in the other.

The relative energy levels of the orbitals for $\ce{ClO2}$ have been reported by Wang and Wang[2].

[1] Walsh, A.D. J. Chem. Soc. (1953) 2266-88.

[2] Wang, X.-B. and Wang, L.-S. J. Chem. Phys. (2000) 113:10928-33.

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  • $\begingroup$ Would it look more like this (before hybridisation)? i.stack.imgur.com/2T1IZ.png $\endgroup$ Jan 23 '19 at 13:37
  • $\begingroup$ Yes, that's right. $\endgroup$
    – Andrew
    Jan 23 '19 at 13:44
  • $\begingroup$ What would the ordering of the MOs look like, if you'd be kind enough to grind through all of that? Thanks a lot! $\endgroup$ Jan 23 '19 at 15:32
  • $\begingroup$ First comment - The convention is to use $\sigma_v'$ for the plane containing the O ligands and $\sigma_v$ for the plane between them. By that convention, your labels of $B_1$ and $B_2$ are reversed. There is no impact on the $A_1$ and $A_2$. $\endgroup$
    – Andrew
    Jan 23 '19 at 18:37
  • $\begingroup$ As for the ordering, I'm not 100% sure of the order. Using your labels, I'd start by assuming the two orbitals from the O2s are lowest, with a1 < b1. Then the two $\sigma$ orbitals, with the a1 < b1. Then b2$\pi$, a1$\pi$, a2 nb (px), b1 nb (pz), a1$\pi^*$, b2$\pi^*$, b1$\sigma^*$, a1$\sigma^*$. But the b1$\sigma$ could go up in energy enough to be higher than the $\pi$ orbitals, and the a1$\pi^*$ is lowered in energy, possibly to the point that it's below the b1 nb(pz) (which is raised up). If I recall correctly, that's consistent with the experimental results. $\endgroup$
    – Andrew
    Jan 23 '19 at 18:42

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