What is the oxidation number of Si (silicon)?

↳Redirected from "Suppose that I don't have a formula for #g(x)# but I know that #g(1) = 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"
1 Answer
anor277
Jun 26, 2016

Elemental silicon has a formal oxidation state of #0#.

Explanation:

Of course, silicon IN ITS COMPOUNDS, as a carbon analogue, usually has an oxidation state of #+IV#.

I use the formal definition of oxidation state/number: #"the charge left on the central atom, when all the bonding"# #"electrons are removed with the charge retained by the "# #"more electronegative atom"#.

As the element, #Si# has neither donated nor accepted electrons, and therefore it is zerovalent; i.e. it has a formal #0# oxidation state. What is the oxidation state of silicon in #SiF_4#; in #SiH_4#; in #Si(CH_3)_4??#

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