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double bond is shown in the ball-and-stick figure model as two cylinders joining the atoms rather than one. The double bond gives the O the neon closed shell configuration as in the O2 molecule. The C needs to share two additional electrons to obtain the neon closed shell configuration, which it does by single bonding to two H atoms. We will discuss double bonds in detail to see how they can be formed from atomic orbitals. Here, we only need to recognize that a double bond concentrates two pairs of electrons between the C and the O. Because of the extra electron density, a double bond is fatter than a single bond. The fatter C=O double bond pushes the C—H single bonds away from it and toward each other. The angles are shown in Figure 14.3. The result is that formaldehyde is still a planar trigonal molecule, but it is not a perfect equilateral triangle.

FIGURE 14.3. Left: BH3. The atoms lie in a plane. The HB bonds are single, and the hydrogens form a perfect equilateral triangle. Each HBH bond angle is 120. Right: H2CO (formaldehyde). The atoms lie in a plane. The CO bond is a double bond. The angles are unequal.

PROMOTING ELECTRONS

Returning to methane, the question is how does methane make four tetrahedrally configured bonds? In Chapter 11, we discussed the electronic configuration of the atoms (see Figure 11.1). Carbon has six electrons, two in the 1s, two in the 2s, and two in 2p orbitals. The valence electrons, the electrons used in bonding, are the 2s and 2p electrons. The top portion of Figure 14.4 shows the atomic orbital energy levels with the four valence electrons filled in. The 1s electrons are not shown. As discussed in Chapter 11 and earlier in this chapter, carbon will form four bonds. In methane, it forms four electron pair sharing bonds to four hydrogen atoms. Each H atom contributes one electron. So the carbon must have four unpaired electrons to form bonds. Each carbon unpaired electron can join with one electron from an H atom to yield an electron pair bond. To have four unpaired electrons, carbon “promotes” a 2s electron to a 2p orbital, as shown in the bottom part of Figure 14.4. For an isolated carbon atom, the configuration shown in the bottom of the figure would not occur unless a lot of energy was pumped into the atom. For a carbon atom, putting a 2s electron into a 2p orbital is a high energy configuration. However, when atoms form molecules, the electrons and nuclei of the different atoms affect each other. Imagine bringing four H atoms toward a C atom. Now, the system wants to assume the lowest energy configuration for all five atoms. Forming the four bonds lowers the energy more than putting a 2s electron into a 2p orbital raises the energy.

FIGURE 14.4. Top: Atomic carbon valence orbitals with the four valence electrons. Bottom: When bonding, a carbon atom “promotes” a 2s to a 2p electron to give four unpaired electrons used to form four bonds to other atoms.

HYBRID ATOMIC ORBITALS—LINEAR MOLECULES

So we see how carbon can make four bonds to yield methane, but why is the shape tetrahedral? The three 2p orbitals are px, py, and pz. These three orbitals are perpendicular to each other, with a 90° angle between any pair. If three H atoms bonded to the three 2p orbitals, the bond angles would be 90°. Furthermore, the 2s orbital is spherical. The fourth H atom’s 1s orbital would have to combine with the carbon 2s orbital. Without something else happening, it is pretty clear that using the carbon 2s and the three 2p orbitals is not going to give methane four identical C-H bonds in a tetrahedral configuration. In addition, how does carbon make the trigonal molecule formaldehyde, or the linear molecule carbon dioxide (O = C = O). In each case, tetrahedral, trigonal, or linear, carbon bonding involves the same 2s and 2p orbitals.

Formaldehyde and carbon dioxide involve double bonds, which we will get to shortly. To bring out the important features of the nature of atomic orbitals that can give linear, trigonal, and tetrahedral shapes, we will examine bonding in BeH2, BH3,