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

Oct 2, 2017

Only to a first approximation?

Explanation:

In $\text{carbon tet}$, $\angle C l - C - C l = {109.5}^{\circ}$, as we would predict by vesper....

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

Both molecules are formally $s {p}^{3} - \text{hybridized}$.....

How would you describe the geometry of ${H}_{3} {O}^{+}$, $\text{hydronium ion}$?

Oct 2, 2017

Bond angles are different between ${H}_{2} O$(104.5 deg) and $C$$C {l}_{4}$(109.5 deg).

Explanation:

Both molecule have $s {p}^{3}$ hybrid orbital, but the bond angles are different.

Carbon tetrachloride($C$$C {l}_{4}$) molecule have a perfect tetrahedral structure and its bond angle is about $109.5$ degree.

In contrast, water(${H}_{2} O$) 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)