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Oct 1, 2018 at 17:02 history edited orthocresol CC BY-SA 4.0
Seems a typo.
Oct 1, 2018 at 16:57 history edited Martin - マーチン CC BY-SA 4.0
truncated to fit question
Mar 20, 2018 at 13:39 vote accept Tyrion Lannister
Feb 4, 2018 at 9:48 comment added Martin - マーチン @GaurangTandon the geometry depends on various factors, as such it is not possible to determine it from the atoms involved a priori. You'll have to use the tools at your disposal, be it experiment or quantum chemical analysis, even VSEPR may give you a good starting point. However, don't expect to much accuracy from the latter.
Feb 4, 2018 at 9:32 comment added Gaurang Tandon @Martin-マーチン Re: "the geometry is reason for a hybridisation concept" You mean to say that we deduce the hybridisation from the bond angle? If so, then, is it possible to predict the bond angle of $\ce{SnCl2}$ - i.e. whether it will be near $90^\circ$, $120^\circ$ or $180^\circ$ (only an approximation) - just from the atoms involved ($\ce{Sn}$ and $\ce{Cl}$)?
Feb 4, 2018 at 9:15 comment added Martin - マーチン @GaurangTandon it applies in the way that the geometry is reason for a hybridisation concept and not the other way around. It doesn't matter which atoms are involved, it only matters how the atoms are arranged. (In this kind of approximation.)
Feb 4, 2018 at 9:09 comment added Gaurang Tandon Re: "See also this answer of me and this answer of Ben Norris." Those answers talk about mainly the hydrides, while this question is about the chloride (of tin). Then, how can you say that the theory - that applies to hydride - will also apply to the chloride? Thanks for the answer!
Apr 13, 2017 at 12:57 history edited CommunityBot
replaced http://chemistry.stackexchange.com/ with https://chemistry.stackexchange.com/
Apr 18, 2016 at 12:10 history answered Martin - マーチン CC BY-SA 3.0