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I recently learnt about this equation but I would like to know how NaOH reacts with the 3d structure of SiO2 and gives Na2SiO3 and H2O.

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Can anyone explain with simple diagram how NaOH forms new bond in SiO2 to give the above mentioned products?Like how they replace, displace etc in the 3d structure.

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  • $\begingroup$ Just to clarify, if you are thinking that SiO2 has a square planar structure based on that picture, it is in fact nothing like that at all. $\endgroup$
    – Jon Custer
    Commented Feb 16, 2022 at 15:17

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To use simple terms, $\ce{NaOH}$ breaks the "vertical" bonds $\ce{Si−O}$ without touching the "horizontal" bonds of your picture. As a consequence, sodium silicate is not $\ce{Na2SiO3}$. It has the structure of a long polymer $\ce{Na_{2n}(SiO3)_n}$. The polymeric ion $\ce{(SiO3^{2−})_n}$ is made of long chains of atoms $\ce{-O-Si-O-Si-O-Si-O -}$etc., where each $\ce{Si}$ atom is still attached to two negatively charged oxygen atoms $\ce{O^−}$, one "above" and one "under" the main $\ce{-O-Si-O-Si-O -}$chain. And a positive ion $\ce{Na+}$ is placed near each such oxygen atom.

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  • $\begingroup$ ok so what happens to the OH- which comes out as water? $\endgroup$
    – AJknight
    Commented Feb 16, 2022 at 1:52
  • $\begingroup$ Also the reaction says that 1 molecule of SiO2 reacts with 2 molecules of NaOH to give 1 molecule of Na2SiO3 and 1 molecule of H20.What happens to the extra oxygen as when 2 OH-ions react one molecule of water is obtained? $\endgroup$
    – AJknight
    Commented Feb 16, 2022 at 2:21
  • $\begingroup$ Suppose all vertical Si-O-Si bonds are broken so that the only broken bond is Si-O just under Si (and never above Si). Now H2O gives one OH to the Si atom, and one H to the O atom. This produces long chains Si-O-Si-O-Si-O- where each Si atom is attached to 2 groups OH. This makes two groups $\ce{OH}$ attached to each silicium atom of the chain. Each such -OH group can react with NaOH producing $\ce{-O^-}$ attached to the silicium atom, one $\ce{Na+}$ ion and one water molecule $\endgroup$
    – Maurice
    Commented Feb 16, 2022 at 16:39

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