Why is atomic size mismatch a sign of poor bond strength? Wouldn't orbital overlap be proportional whether the two atoms are small or the two atoms are big?

For example I heard someone comment that an argument that the C-I bond is weaker than the C-F bond based on orbital overlap made no sense to them, because why would the size of the individual orbitals matter in overlap? Small atom, big atom and small atom, small atom - shouldn't their overlaps be proportional? Or something along the lines of that. I'll have to pull up the post again ...

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    $\begingroup$ Atomic orbitals of similar energy lead to a better overlap than atomic orbitals of dissimilar energy. The valence 5p orbital involved in bonding for iodine is much higher in energy than the valence 2p orbital. $\endgroup$ Oct 22, 2014 at 14:14

1 Answer 1


Small atom, big atom and small atom, small atom - shouldn't their overlaps be proportional

No. Look at the figure below and let's consider s-orbital overlap, just to keep it simple. Remember that these are spheres with the electron density distributed throughout the entire volume. In the "small-small" case there is effective overlap of the electron density in both 1s orbitals; we would expect the resultant bond to be strong. In the "small-big" case, the small 1s orbital involves a significant fraction of its electron density in overlap, but not much of the electron density in the large s orbital is involved in overlap. In the "small-big" case the overlap of electron clouds is ineffective and any resultant bond would be weak.

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  • $\begingroup$ what would happen when the overlap percentage is the same, but orbitals have different sizes? how do you compare the bond strength of 1s-1s, 2s-2s, and 3s-3s overlaps? Do we need to take the spherical nodes into play? $\endgroup$
    – Chetan
    May 24 at 6:32

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