Physical Chemistry Third Edition

21.3 The CH 4 ,NH 3 , and H 2 O Molecules and thesp^3 Hybrid Orbitals 875

EXAMPLE21.2

Show that the angle between alternate diagonals of a cube is 109 degrees, 28 minutes, 16. 39 ...

seconds.

Solution

The diagonal of a face of a cube with sides equal to unity is

√

2 by the Pythagorean theorem.

The diagonal through opposite corners is then equal to

(1^2 +(

√

2)^2 )^1 /^2 

√

3

by the Pythagorean theorem. A line drawn from the center of the cube to the center of a face of

the cube bisects the angle between two corner-to-corner diagonals. Denote the angle between

this line and one diagonal byα:

αarcsin

(

(0.5)(

√

2)

(0.5)(

√

3)

)

arcsin

(√

2

3

)

 54 .735610 degrees

Twice this angle gives the angle between alternate diagonals, 109.471221 degrees

109 degrees, 28 minutes, 16. 39428 ...seconds.

A cross section of the orbital region of a 2sp^3 orbital in a plane containing its axis

of symmetry is shown in Figure 21.5b. For comparison, cross sections of the orbital

regions for the 2sand 2pzunhybridized orbitals are also shown. The orbital region

of the hybrid orbital extends farther in the direction of its symmetry axis than that of

either the 2sor 2porbital, making it possible to form a larger overlap region with an

orbital on another atom by using a 2sp^3 hybrid orbital instead of a 2sora2porbital,

as well as making a bond in the desired direction possible.

To describe the bonding in the methane molecule, we place the hydrogen atoms at

alternate corners of the cube shown in Figure 21.5b. We form four bonding molecular

orbitals, each of which is a linear combination of a 2sp^3 carbon hybrid orbital and a 1s

hydrogen orbital. The bonds point in the tetrahedral directions, toward the hydrogen

atoms. The molecule has 10 electrons. We assign one pair of electrons to the 1scarbon

nonbonding orbital and the other four pairs to the bonding LCAOMOs. The molecule

has four sigma bonds.

The Ammonia Molecule

For the NH 3 molecule, we place the hydrogen atoms on three of the symmetry axes

of the 2sp^3 hybrid orbitals. We form three sigma bonding orbitals with these hybrid

orbitals and 1sorbitals on the hydrogen atoms. There are 10 electrons. We place two

electrons in the 1sorbital of the N atom (a nonbonding orbital), two in each of the

bonding orbitals, and a lone (nonbonding) pair in the fourth hybrid orbital. This gives

bond angles of 109.5◦, in rough agreement with the experimental bond angles of 107◦.

The Water Molecule

y

x

z

Oxygen

nucleus

Hydrogen

nucleus B

Hydrogen

nucleus A

Figure 21.6 The Positions of the

Nuclei in the Description of the

Water Molecule Usingsp^3 Hybrid

Orbitals.

Figure 21.6 shows the placement of the two hydrogen atoms of the H 2 O molecule to

obtain maximum overlap with two of the 2sp^3 hybrid orbitals. This corresponds to an