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

2 Answers
Oct 2, 2017

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"#?

Oct 2, 2017

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.
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(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)