10.4^10.4. Hybridization of Atomic Orbitals^336^345^,,^16001^16218%
The concept of atomic orbital overlap should apply also to polyatomic molecules. However, a satisfactory bonding scheme must account for molecular geometry. We will discuss three examples of VB treatment of bonding in polyatomic molecules.
Consider the CH4 molecule. Focusing only on the valence electrons, we can represent the orbital diagram of C as
Because the carbon atom has two unpaired electrons (one in each of the two 2p orbitals), it can form only two bonds with hydrogen in its ground state. Although the species CH2 is known, it is very unstable. To account for the four C—H bonds in methane, we can try to promote (that is, energetically excite) an electron from the 2s orbital to the 2p orbital:
Now there are four unpaired electrons on C that could form four C—H bonds. However, the geometry is wrong, because three of the HCH bond angles would have to be 90° (remember that the three 2p orbitals on carbon are mutually perpendicular), and yet all HCH angles are 109.5°.
To explain the bonding in methane, VB theory uses hypothetical hybrid orbitals, which are atomic orbitals obtained when two or more nonequivalent orbitals of the same atom combine in preparation for covalent bond formation. Hybridization is the term applied to the mixing of atomic orbitals in an atom (usually a central atom) to generate a set of hybrid orbitals. We can generate four equivalent hybrid orbitals for carbon by mixing the 2s orbital and the three 2p orbitals:
Because the new orbitals are formed from one s and three p orbitals, they are called sp3 hybrid orbitals. Figure 10.6 shows the shape and orientations of the sp3 orbitals. These four hybrid orbitals are directed toward the four corners of a regular tetrahedron. Figure 10.7 shows the formation of four covalent bonds between the carbon sp3 hybrid orbitals and the hydrogen 1s orbitals in CH4. Thus, CH4 has a tetrahedral shape, and all the HCH angles are 109.5°. Note that although energy is required to bring about hybridization, this input is more than compensated for by the energy released upon the formation of C—H bonds. (Recall that bond formation is an exothermic process.)
sp3 is pronounced “s-p three.”
The following analogy is useful for understanding hybridization. Suppose that we have a beaker of a red solution and three beakers of blue solutions and that the volume of each is 50 mL. The red solution corresponds to one 2s orbital, the blue solutions represent three 2p orbitals, and the four equal volumes symbolize four separate orbitals. By mixing the solutions we obtain 200 mL of a purple solution, which can be divided into four 50-mL portions (that is, the hybridization process generates four sp3 orbitals). Just as the purple color is made up of the red and blue components of the original solutions, the sp3 hybrid orbitals possess both s and p orbital characteristics.
Another example of sp3 hybridization is ammonia (NH3). Table 10.1 shows that the arrangement of four electron pairs is tetrahedral, so that the bonding in NH3 can be explained by assuming that N, like C in CH4, is sp3-hybridized. The ground-state electron configuration of N is 1s22s22p3, so that the orbital diagram for the sp3 hybridized N atom is
Three of the four hybrid orbitals form covalent N—H bonds, and the fourth hybrid orbital accommodates the lone pair on nitrogen (Figure 10.8). Repulsion between the lone-pair electrons and electrons in the bonding orbitals decreases the HNH bond angles from 109.5° to 107.3°.
It is important to understand the relationship between hybridization and the VSEPR model. We use hybridization to describe the bonding scheme only when the arrangement of electron pairs has been predicted using VSEPR. If the VSEPR model predicts a tetrahedral arrangement of electron pairs, then we assume that one s and three p orbitals are hybridized to form four sp3 hybrid orbitals. The following are examples of other types of hybridization.
The beryllium chloride (BeCl2) molecule is predicted to be linear by VSEPR. The orbital diagram for the valence electrons in Be is
We know that in its ground state Be does not form covalent bonds with Cl because its electrons are paired in the 2s orbital. So we turn to hybridization for an explanation of Be's bonding behavior. First, we promote a 2s electron to a 2p orbital, resulting in
Now there are two Be orbitals available for bonding, the 2s and 2p. However, if two Cl atoms were to combine with Be in this excited state, one Cl atom would share a 2s electron and the other Cl would share a 2p electron, making two nonequivalent BeCl bonds. This scheme contradicts experimental evidence. In the actual BeCl2 molecule, the two BeCl bonds are identical in every respect. Thus, the 2s and 2p orbitals must be mixed, or hybridized, to form two equivalent sp hybrid orbitals:
Figure 10.9 shows the shape and orientation of the sp orbitals. These two hybrid orbitals lie on the same line, the x-axis, so that the angle between them is 180°. Each of the BeCl bonds is then formed by the overlap of a Be sp hybrid orbital and a Cl 3p orbital, and the resulting BeCl2 molecule has a linear geometry (Figure 10.10).
Next we will look at the BF3 (boron trifluoride) molecule, known to have planar geometry based on VSEPR. Considering only the valence electrons, the orbital diagram of B is
First, we promote a 2s electron to an empty 2p orbital:
sp2 is pronounced “s-p two.”
Mixing the 2s orbital with the two 2p orbitals generates three sp2 hybrid orbitals:
These three sp2 orbitals lie in the same plane, and the angle between any two of them is 120° (Figure 10.11). Each of the BF bonds is formed by the overlap of a boron sp2 hybrid orbital and a fluorine 2p orbital (Figure 10.12). The BF3 molecule is planar with all the FBF angles equal to 120°. This result conforms to experimental findings and also to VSEPR predictions.
You may have noticed an interesting connection between hybridization and the octet rule. Regardless of the type of hybridization, an atom starting with one s and three p orbitals would still possess four orbitals, enough to accommodate a total of eight electrons in a compound. For elements in the second period of the periodic table, eight is the maximum number of electrons that an atom of any of these elements can accommodate in the valence shell. This is the reason that the octet rule is usually obeyed by the second-period elements.
The situation is different for an atom of a third-period element. If we use only the 3s and 3p orbitals of the atom to form hybrid orbitals in a molecule, then the octet rule applies. However, in some molecules the same atom may use one or more 3d orbitals, in addition to the 3s and 3p orbitals, to form hybrid orbitals. In these cases, the octet rule does not hold. We will see specific examples of the participation of the 3d orbital in hybridization shortly.
To summarize our discussion of hybridization, we note that
The concept of hybridization is not applied to isolated atoms. It is a theoretical model used only to explain covalent bonding.
Hybridization is the mixing of at least two nonequivalent atomic orbitals, for example, s and p orbitals. Therefore, a hybrid orbital is not a pure atomic orbital. Hybrid orbitals and pure atomic orbitals have very different shapes.
The number of hybrid orbitals generated is equal to the number of pure atomic orbitals that participate in the hybridization process.
Hybridization requires an input of energy; however, the system more than recovers this energy during bond formation.
Covalent bonds in polyatomic molecules and ions are formed by the overlap of hybrid orbitals, or of hybrid orbitals with unhybridized ones. Therefore, the hybridization bonding scheme is still within the framework of valence bond theory; electrons in a molecule are assumed to occupy hybrid orbitals of the individual atoms.
Table 10.4 summarizes sp, sp2, and sp3 hybridization (as well as other types that we will discuss shortly).
|Pure Atomic Orbitals of the Central Atom||Hybridization of the Central Atom||Number of Hybrid Orbitals||Shape of Hybrid Orbitals||Examples|
|s, p, p||sp2||3||BF3|
|s, p, p, p||sp3||4||CH4,|
|s, p, p, p, d||sp3d||5||PCl5|
|s, p, p, p, d, d||sp3d2||6||SF6|
Before going on to discuss the hybridization of d orbitals, let us specify what we need to know to apply hybridization to bonding in polyatomic molecules in general. In essence, hybridization simply extends Lewis theory and the VSEPR model. To assign a suitable state of hybridization to the central atom in a molecule, we must have some idea about the geometry of the molecule. The steps are as follows:
Draw the Lewis structure of the molecule.
Predict the overall arrangement of the electron pairs (both bonding pairs and lone pairs) using the VSEPR model (see Table 10.1).
Deduce the hybridization of the central atom by matching the arrangement of the electron pairs with those of the hybrid orbitals shown in Table 10.4.
Determine the hybridization state of the central (underlined) atom in each of the following molecules: (a) BeH2, (b) AlI3, and (c) PF3. Describe the hybridization process and determine the molecular geometry in each case.
The steps for determining the hybridization of the central atom in a molecule are:
The ground-state electron configuration of Be is 1s22s2 and the Be atom has two valence electrons. The Lewis structure of BeH2 is
There are two bonding pairs around Be; therefore, the electron pair arrangement is linear. We conclude that Be uses sp hybrid orbitals in bonding with H, because sp orbitals have a linear arrangement (see Table 10.4). The hybridization process can be imagined as follows. First we draw the orbital diagram for the ground state of Be:
By promoting a 2s electron to the 2p orbital, we get the excited state:
The 2s and 2p orbitals then mix to form two hybrid orbitals:
The two Be—H bonds are formed by the overlap of the Be sp orbitals with the 1s orbitals of the H atoms. Thus, BeH2 is a linear molecule.
The ground-state electron configuration of Al is [Ne]3s23p1. Therefore, the Al atom has three valence electrons. The Lewis structure of AlI3 is
There are three pairs of electrons around Al; therefore, the electron pair arrangement is trigonal planar. We conclude that Al uses sp2 hybrid orbitals in bonding with I because sp2 orbitals have a trigonal planar arrangement (see Table 10.4). The orbital diagram of the ground-state Al atom is
By promoting a 3s electron into the 3p orbital we obtain the following excited state:
The 3s and two 3p orbitals then mix to form three sp2 hybrid orbitals:
The sp2 hybrid orbitals overlap with the 5p orbitals of I to form three covalent Al—I bonds. We predict that the AlI3 molecule is trigonal planar and all the IAlI angles are 120°.
The ground-state electron configuration of P is [Ne]3s23p3. Therefore, the P atom has five valence electrons. The Lewis structure of PF3 is
There are four pairs of electrons around P; therefore, the electron pair arrangement is tetrahedral. We conclude that P uses sp3 hybrid orbitals in bonding to F, because sp3 orbitals have a tetrahedral arrangement (see Table 10.4). The hybridization process can be imagined to take place as follows. The orbital diagram of the ground-state P atom is
By mixing the 3s and 3p orbitals, we obtain four sp3 hybrid orbitals.
As in the case of NH3, one of the sp3 hybrid orbitals is used to accommodate the lone pair on P. The other three sp3 hybrid orbitals form covalent P—F bonds with the 2p orbitals of F. We predict the geometry of the molecule to be trigonal pyramidal; the FPF angle should be somewhat less than 109.5°.
Determine the hybridization state of the underlined atoms in the following compounds: (a) SiBr4 and (b) BCl3.
We have seen that hybridization neatly explains bonding that involves s and p orbitals. For elements in the third period and beyond, however, we cannot always account for molecular geometry by assuming that only s and p orbitals hybridize. To understand the formation of molecules with trigonal bipyramidal and octahedral geometries, for instance, we must include d orbitals in the hybridization concept.
Consider the SF6 molecule as an example. In Section 10.1 we saw that this molecule has octahedral geometry, which is also the arrangement of the six electron pairs. Table 10.4 shows that the S atom is sp3d2-hybridized in SF6. The ground-state electron configuration of S is [Ne]3s23p4:
Because the 3d level is quite close in energy to the 3s and 3p levels, we can promote 3s and 3p electrons to two of the 3d orbitals:
sp3d2 is pronounced “s-p three d two.”
Mixing the 3s, three 3p, and two 3d orbitals generates six sp3d2 hybrid orbitals:
The six S—F bonds are formed by the overlap of the hybrid orbitals of the S atom with the 2p orbitals of the F atoms. Because there are 12 electrons around the S atom, the octet rule is violated. The use of d orbitals in addition to s and p orbitals to form an expanded octet (see Section 9.9) is an example of valence-shell expansion. Second-period elements, unlike third-period elements, do not have 2d energy levels, so they can never expand their valence shells. (Recall that when n = 2, l = 0 and 1. Thus, we can only have 2s and 2p orbitals.) Hence, atoms of second-period elements can never be surrounded by more than eight electrons in any of their compounds.
Describe the hybridization state of phosphorus in phosphorus pentabromide (PBr5).
Follow the same procedure shown in Example 10.3.
The ground-state electron configuration of P is [Ne]3s23p3. Therefore, the P atom has five valence electrons. The Lewis structure of PBr5 is
There are five pairs of electrons around P; therefore, the electron pair arrangement is trigonal bipyramidal. We conclude that P uses sp3d hybrid orbitals in bonding to Br, because sp3d hybrid orbitals have a trigonal bipyramidal arrangement (see Table 10.4). The hybridization process can be imagined as follows. The orbital diagram of the ground-state P atom is
Promoting a 3s electron into a 3d orbital results in the following excited state:
Mixing the one 3s, three 3p, and one 3d orbitals generates five sp3d hybrid orbitals:
These hybrid orbitals overlap the 4p orbitals of Br to form five covalent P—Br bonds. Because there are no lone pairs on the P atom, the geometry of PBr5 is trigonal bipyramidal.
Describe the hybridization state of Se in SeF6.
Similar problem: 10.40.
Why is it not possible to generate sp4 hybrid orbitals?