molecular geometries of covalent molecules: Lewis structures and VSEPR theory?

Answer 1

Vague, but that is not what the V is for.

V is force the Valence of valence shell elecron pair repulsion.

Were you ready for that?

Here is the idea. Electron all have the same charge. -1, and so they repell each other.

But they all share the same space around the nucleus of an atom.

For some reason they need to do this in two close, two close then six farther away. That means two for helium. But eight for the next shell.

Look at the periodic table. This is on the inside cover of every chem text. Evenly space electrons around the name. One dot on each side for carbon. and it looks nothing like the 104 degree space around an actual carbon atom. But you can add 4 hydrogens on every side to even it out.

These are the shapes of the shells in the VSEPR chapter. Same book previously referenced. It looks round then 4 lobed and then later 6 lobed later.

Good luck. This is all I had time for.

Answer 2

First off, if you know the name of the compound, you can look to the periodic chart and using the valence of each element determine whether or not there will be lone electrons (making the compound negative) or whether there will be additional protons (making the compound positive); in either case, the compound will be polar (having charge) and will dissolve in water, although the particular solubility limit will differ. Secondly, most compounds that are neutral, but are capable of breaking into ions, will dissolve into water. However, compounds that do not break into ions sometimes dissolve into water. The important point will be whether there are titratable groups, i.e. if the compound has -OH groups on it. If the compound does not have any groups that can hydrogen bond with water, then it will not dissolve into water.

Answer 3

Which ones ???????