Chemistry Asked on January 2, 2022
Cinnabar($ce{HgS}$) is one of the most common ores of mercury.
Why does it not exist at $ce{HgO}$ or some other such compound?
Is there any reason for why it is so prevalent?
Does it react less with other compounds, or maybe due to some solid structure of HgS having some characteristic property giving it more stability?
I’m seeking an explanation without the mention of free energy, because it doesn’t properly explain the phenomenon for me. I understand that cinnabar is abundant in the crust, and my question is why is $ce{HgS}$ so common?
I am afraid that no one knows why! The superficial rationalizations such as HSAB and ionic size etc are good for decorating general chemistry textbooks.
As scientists, we should accept the fact that all questions which a human mind can generate do not have answers. Unfortunately, as science students, we are trained that every question must have an answer at the back of the book.
There is a whole field of geology, called ore genesis which tries to think of the origin of minerals and ores Ore Genesis on Wikipedia. It is a remarkable wonder and miracle of Nature, as how it deposits ore in certain chemical forms.
More specifically, you would search the geochemistry literature, you would see how complex is the whole concept e.g. if you go through Quicksilver Deposits of Chile Quicksilver deposits you can get a feel.
Answered by M. Farooq on January 2, 2022
Why $ce{HgS}$ and not $ce{HgO}$?
Because of HSAB. F'x has already answered a previous question of why mercury has propensity towards thiols(or in general sulfur). Quoting the relevant information:
In Pearson's HSAB theory (hard and soft acids and bases), the reason the $ce{S-Hg}$ bond is be stronger than the $ce{O-Hg}$ can be explained because $ce{S^2-}$ is a soft basis and $ce{Hg(II)}$ is a soft acid, making a good fit while $ce{O^2-}$ is harder.
If this answer helps, please make sure to upvote F'xs answer.
Answered by Nilay Ghosh on January 2, 2022
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