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I want to know the exact mechanism for how silver would start to grow during a mirroring with Tollen's reagent. I've made two pictures to illustrate how I see the nucleation/growth process of silver on silica:

nucleation

growth

As you can see, I didn't draw any chemical bonds, because I think that the silver adherence is due to Van Der Waals forces. Am I correct? If that is the case, then I believe a place with high permanent polarisation would be better suited for the nucleation of silver.

Now, let's follow the logic. We have a greater electronegativity difference between $\ce{O}$ and $\ce{Si}$ ($3.44 - 1.90 = 1.54$) than between $\ce{O}$ and $\ce{H}$ ($3.44 - 2.2 = 1.24$).

For the upper oxygen, assuming an angle of 109 degrees, you get a total dipole moment of: $2(1.54)\cos\left(\frac{109}{2}\right) = 1.79$

For the lower oxygen with the bonded hydrogen, assuming an angle of 120 degrees, you would get a total dipole moment of: $1.54 + 1.24 = 2.78$ (please tell me if I'm wrong to assume a straight line!)

This would mean that you have a greater accumulation of electrons at the bottom oxygen, which would favor Van Der Waals attraction (Debye interaction, I guess, since you have a permanent dipole and an induced dipole). This, in turn, would mean that it would be a better nucleation site.

If you make a surface treatment using silane groups, I believe you promote the presence of $\ce{OH}$ groups at your silica surface, though I'm unsure about this. I do know that it makes better silver mirroring though. I've made another drawing:

silanisation

Following my logic, this one is wrong in many ways. First of all, I drew angles on the $\ce{Si-O-H}$ bonds, which I didn't draw in my first pictures. Secondly, the silver is not even near the oxygen atoms, which I believe are the nucleation points.

I'm confused, could you please point out to me where I'm right or wrong in my logic/knowledge?

Mainly:

  • Is there a bond forming between the silica surface and the silver? Or is there only Van Der Waals (Debye) attraction?
  • Is my depiction of nucleation & growth correct (for the first two pictures)?
  • Is my logic regarding electronegativity correct?
  • Is my logic of " higher total electronegativity difference, hence better nucleation site" correct? (assuming I'm right about the Van Der Waals Forces)
  • Is my depiction of silanisation correct?
  • Am I right to say that the higher abundance of surface $\ce{OH}$ groups by silanisation implies better nucleation?
  • Could I do the same when using a strong acid on my glass surface?
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    $\begingroup$ Note that the $\ce{Si-O-H}$ groups are acidic and the Tollens reagent is alkaline. Thus, there are $\ce{Si-O- }$ groups, which are missing in your drawings. $\endgroup$
    – user7951
    May 29, 2015 at 17:16
  • $\begingroup$ Then, would you say that we get an ionic bond between the positive silver and the negative oxygen? Without the need to reduce the silver? $\endgroup$
    – victorbg
    May 29, 2015 at 17:24
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    $\begingroup$ I'd say that your representation of the surface of glass is naive. There would be all sorts of surface irregularities, and all sorts of impurities. $\endgroup$
    – MaxW
    Oct 22, 2015 at 7:00
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    $\begingroup$ I am also not very sure about it but perhaps there are only Van Der Walls forces because when we tilt the tube the silver deposited on surface leaves off the surface and suspends back into the solution. This phenomenon favors the presence of weak bonds. I have also read that some of silver forms a thin layer of a non-stoichiometric compound on glass surface which allows silver to be deposited on it but it is not confirmed. $\endgroup$
    – Hamza
    Jun 4, 2016 at 18:11
  • $\begingroup$ Maybe because the pH is quite high (you add sodium hydroxide), and you heat the actual surface of the test tube, for example, some of the Si-OH groups get deprotonated and actually coordinate to silver ions, and so they get reduced at the very surface of the glass. That would be my guess about the actual formation mechasm. I would guess that VDW forces hold silver at the glass surface, as it can very easily be scraped off. $\endgroup$
    – Uros
    Jun 27, 2016 at 1:18

2 Answers 2

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You are on the right track, and the question is somewhat answerable. But, there are some problems with the proposed mechanisms of nucleation and—to an extent—the fundamental nature of the question to address first.

The first to regard is the nature of the Tollen's reagent: this deposition onto glass occurs in solution and only occurs after the reduction of silver(I) in silver nitrate to silver(0) by some aldehyde. Before the surface even becomes a factor to consider, chemical processes are occurring. Silver(0) can form dimers (Ag$_2$), trimers (Ag$_3$), etc. These byproducts can also be oxidized, encapsulate silver(I) ions, et cetera, et cetera to form polyatomic ions in solution. There exists a whole menagerie of silver nanoclusters and microscopic allotropes that live in this solution from the moment reduction begins, and they only continue react with each other to form other compounds.

This is not unique to silver(0), but it's prominent enough with silver that silver is a major subject of nanoparticle formation—a large part of nanoparticle research goes into taming these processes to design microscopic particles with unique properties. By the time silver(0) has a chance to reach the surface, it could be in a host of configurations, nanocluster morphologies, etc. It would be rare if all or even most of the deposition on glass came from single-atom nucleation. This does not matter so much in the bulk if you are just trying to get a silver mirror, but the more heterogeneity that exists, the rougher and thus more dull the mirror.

Single-atom nucleation on surfaces is possible—many techniques like molecular beam epitaxy and whatnot can be used to study those processes, and silver thin-films have loads of applications in crystal engineering and surface chemistry. Those processes just won't be insanely relevant to mirrors formed from the Tollen's reagent.

Second is glass as the surface—silica glass comes in just too many forms for there to be one answer. It is also harder to answer the question with glass because it is an amorphous solid, which means characteristics about the surface may only be understood from a broad, quasi-statistical perspective. You are correct to note that hydroxylation has an impact, however again your proposed mechanisms rely on single-atom interactions, and so there will never be just one interaction that occurs.

This paper comptuationally simulates the action of both single silver atoms and silver nanoparticles on different forms of silica surface. They determine using Monte Carlo simulation that for hydroxylation levels and temperatures relevant to lab chemistry, the silver adherence to the surface is indeed van der Waals-like. Modeling a silver nanodroplet on glass melting and then freezing, they found that sometimes the silver may adopt an icosahedral structure, but usually it will adopt a face-centered cubic truncated octahedron. They also found in the case of single-atom silver(0) on the surface, interactions with oxygen dominated (they use a Lennard-Jones potential to plot this). Both atomic distances were well below the sum of their van der Waal radii, although if you use oxygen's ionic radius it's less pronounced. They compare some of their results to experimental precedent where possible.

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"If you make a surface treatment using silane groups, I think that you promote the presence of OH groups at your silica surface". This is NOT correct. You siliconize glass to cap the free SiOH groups

2 silica_Si-OH + n SiCl2Me2 + HO-Si_silica -> silica_Si-[O(SiMe2)]$_n$-O-Si_silica. As a result you decrease the number of active OH groups and make it highly non-polar.

Before siliconization silica is surrounded by a hydration sphere. Ag$^+$ can bind it well, but we need to precipitate neutral Ag. In case of non-siliconized glass we have two effects:

(1) It is easier for silver to grow on the surface that already has a silver nano-crystal
(2) Before treatment glass have inhomogeneities, so nucliation in some places is heavily favored.

Both effects are negated by siliconization.


Note that I didn't account for surface charges which change as a result of siliconization and might play a role.

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