What is the electronic and molecular geometry of #"dimethyl ether"#?

2 Answers
Jun 28, 2017

Answer:

We deal with #"dimethyl ether"#, #H_3C-O-CH_3#. The electronic geometry on each NON-HYDROGEN atom is tetrahedral to a first approx.......

Explanation:

And thus, to a first approx., #/_H-C-H=/_C-O-C# #=/_O-C-H=109.5^@#.

Because oxygen bears 2 lone pairs, which, as non-bonding entities, tend to lie closer to the oxygen atom, the low pairs may tend to compress #/_C-O-C#; the actual values are the preserve of experiment.

Jun 28, 2017

I always think about what molecule looks similar, and remember its molecular and tetrahedral geometry.

  • the central oxygen is analogous to the oxygen in water if the two #"CH"_3# fragments are replaced with #"H"# atoms. Water is bent, with a tetrahedral electron geometry, and thus has a VSEPR-predicted bond angle of #bb(109.5^@)#.

Naturally, that angle is not strictly correct (it is actually #104.4776^@# in water), due to the lone pair crunching the molecule together, but given you are asked for the IDEAL bond angle, that's what you'll have to type in regardless.

  • for one choice of carbon atom in this molecule, if you change the remaining #"OCH"_3# to #"H"#, then it looks just like methane, #"CH"_4#. Therefore, the atom has around it a tetrahedral electron geometry, and a #bb(109.5^@)# IDEAL angle as well...

Or, if you just counted electron groups around both central atoms (#"C/O"#), there are four each, and (if you refer to the table in your book that lists number of electron domains for each geometry) thus we automatically defer to VSEPR predictions of #109.5^@#.

See this for a mathematical proof of the ideal tetrahedral bond angles.