A vague answer I've seen is that it has something do with a greater number of possible excited states, but I don't know what this means for a partially ionized plasma - much less a fully ionized plasma - nor how this would allow the plasma inside a star to absorb more radiation.
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Both free-free and bound-free absorption are strongly dependent on the atomic number $Z$ of nuclei in the gas.
For free-free absorption it is simply that the emissivity per unit volume of electrons accelerating in the field of an ion scales as $Z^2$ (there are $Ze$ electrons accelerating near a charge of $+Ze$). Since by Kirchoff's Law, absorptivity is directly related to emissivity then the absorption coefficient and opacity are also proportional to $Z^2$.
There is a similar argument for bound-free absorption, where since electrons are highly accelerated in the field of a strongly positively charged ion, there is a $Z^4$ dependence of the absorption cross-section.
These very strong dependences translate to weaker dependences on the mass-fraction of heavy elements because the total absorption coefficients must be averaged over all the constituents of the gas - most of which are hydrogen and helium with $Z=1$ and $Z=2$ respectively. Nevertheless there still ends up being a fairly strong, positive correlation between metallicity and opacity.
Even in the outer parts of stars where the opacity is dominated by H$^{-}$ ions, it is still metals that supply almost all the free electrons to make those ions and the opacity ends up being proportional to the metallicity.
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$\begingroup$ Why do metals provide so many more free electrons than hydrogen? The most common metals don't have significantly lower first ionization energies than hydrogen. Is it just because metals have more electrons to give? $\endgroup$ Commented Sep 1, 2021 at 20:47
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$\begingroup$ @zucculent actually hydrogen has a relatively high ionization energy. Most atoms have lower ionization energy than hydrogen. Judging from Wikipedia: Ionization energy only He, N, O, F, Ne, Ar and Kr have higher ionization energy that hydrogen. Li and Na have almost three times less ionization energy. $\endgroup$– PrallaxCommented Sep 1, 2021 at 21:23
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$\begingroup$ @Prallax that's true, but of the elements you mentioned, N, O, and Ne are some of the most abundant metals. Li and Na are much less so. $\endgroup$ Commented Sep 1, 2021 at 21:53
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1$\begingroup$ Nevertheless, in the photosphere it is Na that supplies the free electrons. $\endgroup$– ProfRobCommented Sep 1, 2021 at 22:56
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$\begingroup$ Oh so hydrogen's ionization energy is too high to contribute significantly to the free electrons? $\endgroup$ Commented Sep 2, 2021 at 3:02