Question

Are the bond angles in water, and carbon tetrachloride the same?

Answer 1

Only to a first approximation?

Explanation:

In `"carbon tet"`, `/_Cl-C-Cl=109.5^@`, as we would predict by vesper....

In water, the equivalent `/_H-O-H` would be `109.5^@`, however, the oxygen lone pairs, which are not bound to another atom and thus lie closer to the oxygen atom, thus tend to compress `/_H-O-H` to give angles of `104-5^@`, and thus we describe water as a bent molecule.

Both molecules are formally `sp^3-"hybridized"`.....

How would you describe the geometry of `H_3O^+`, `"hydronium ion"`?

Answer 2

Bond angles are different between `H_2O`(104.5 deg) and `C` `Cl_4`(109.5 deg).

Explanation:

Both molecule have `sp^3` hybrid orbital, but the bond angles are different.

Carbon tetrachloride(`C` `Cl_4`) molecule have a perfect tetrahedral structure and its bond angle is about `109.5` degree.

In contrast, water(`H_2O`) molecule has two covalent bonds and two lone(unshared) electron pairs. Lone pairs are very close to the center atom(`O`), and the repulsive force between lone pairs is stronger than that between covalent bonds.

Therefore, angle between lone pairs must be larger than `109.5` degree, and bond angle becomes smaller than that.

Bond angle between two `O-H` bonds is `104.5` degree.

(The picture is cited from a Japanese site, "水の話 ～化学の鉄人小林映章が「水」を斬る！～ " http://www.con-pro.net/readings/water/doc0002.html)