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How are the 3p orbitals of chlorine lower in energy than the 1s orbital of hydrogen?enter image description here

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up vote 4 down vote accepted

I'd say there are two important things to consider here: Firstly, $\ce{Cl}$ is more electronegative than $\ce{H}$. And secondly, as you proceed from the left to the right in any row of the periodic table, the s and p orbitals go down in energy. This is a consequence of the fact that the valence electrons do not screen each other effectively. Thus, the addition of one proton to the nucleus and one electron does not cancel; instead, the valence electrons "feel" an increased nuclear charge. The electronegativity difference and the reduced screening among the valence electrons seems to be enough to shift the $3\ce{s}$ and even the $3\ce{p}$ orbital of $\ce{Cl}$ below the energy of the $1\ce{s}$ orbital of $\ce{H}$. The experimental state averaged ionization potentials (IP) could serve as a good guide for the placement of starting orbitals in an orbital interaction diagram. For the $\ce{H}$ $1\ce{s}$ orbital you have an IP of $-13.6 \, \mathrm{eV}$. For the $\ce{Cl}$ $3\ce{s}$ and $3\ce{p}$ orbitals you have IP's of $-25.2 \, \mathrm{eV}$ and $-13.7 \, \mathrm{eV}$, respectively.$^{[1]}$ So, this indicates that the $3\ce{p}$ orbital of $\ce{Cl}$ does indeed lie slightly below the $1\ce{s}$ orbital of $\ce{H}$.


Here is a table of the experimental state averaged ionization potentials, in electron volts, for the main group atoms. However, those $\ce{p}$ atomic orbitals for groups 1 and 2 and for the $\ce{s}$ and $\ce{p}$ orbitals of the sixth row are not experimentally known and hence, calculated values (indicated by an asterisk) have been used.$^{[2,3]}$ The latter include relativistic corrections. The table is taken from the 2nd Edition of the book Orbital Interactions in Chemistry by T. A. Albright, J. K. Burdett, and M.-H. Whangbo.

State averaged ionization potentials, in electron volts

[1] L. C. Allen, J. Am. Chem. Soc. 1989, 111, 9003 ; L. C. Allen, J. Am. Chem. Soc. 1992, 114, 1510; L. C. Allen, Acc. Chem. Res. 1990, 23, 175.

[2] A. Vela, J. L. Gazquez, J. Phys. Chem. 1988, 92, 5688.

[3] J. P. Desclaux, At. Nucl. Data Tables 1973, 12, 311.

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You can also test that theory with the bohr model for hydrogen like atoms. It is nothing really accurate, but it draws the picture quite good: $$E= {1\over 2} m_\mathrm{e} v^2 - {Z k_\mathrm{e} e^2 \over r} = - {Z k_\mathrm{e} e^2 \over 2r}.$$ With $Z$ as nuclear charge and $r$ as the radius, $k_\mathrm{e}$ is a constant and $E$ is the energy: $E\propto -Z$. – Martin - マーチン Aug 13 '14 at 11:15
Thank you very much. Do you have a table with the averaged ionization potentials (IP)? – Marko Aug 13 '14 at 11:28
@Marko It's your lucky day ;) I added one for the main group elements. But unfortunately, I don't have any data for transition metals. – Philipp Aug 13 '14 at 11:44

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