Tetrahedral arrangement hybrid orbitals

A tetrahedral arrangement of orbitals indicates sp hybridization (steric number 4). A compound s geometry depends on the number of lone pairs and can be tetrahedral, trigonal pyramidal, or bent. 

FIGURE 3.14 Each C H bond in methane is formed by the pairing of an electron in a hydrogen U-orbital and an electron in one of the four sp hybrid orbitals of carbon. Therefore, valence-bond theory predicts four equivalent cr-bonds in a tetrahedral arrangement, which is consistent with experimental results. 

So far, we have not considered whether terminal atoms, such as the Cl atoms in PC15, are hybridized. Because they are bonded to only one other atom, we cannot use bond angles to predict a hybridization scheme. However, spectroscopic data and calculation suggest that both s- and p-orbitals of terminal atoms take part in bond formation, and so it is reasonable to suppose that their orbitals are hybridized. The simplest model is to suppose that the three lone pairs and the bonding pair are arranged tetrahedrally and therefore that the chlorine atoms bond to the phosphorus atom by using sp hybrid orbitals. 

Four equivalent ip -hybridized orbitals achieve maximum distance from one another when they arrange in a tetrahedral structure ... 

Similarly, in H2O, the oxygen atom is sp hybridized. So all four orbitals are in a tetrahedral arrangement, just as we would expect for an sp hybridized atom. But only two of the orbitals are being used for bonds. So if we look just at the atoms that are connected, we do not see a tetrahedron. Rather, we see a bent arrangement ... 

The methyl carbanion, CH3, has bond angles close to that in a tetrahedral arrangement of atoms, 109°, indicating 4 electron groups around the central C atom and sp3 hybridization. The lone pair of electrons exerts significant repulsive force on the electrons in bonding orbitals and must be counted as an electron group. 

We have already explained. In terms of hybridisation, how a carbon atom can form four sp hybrid orbitals (see p. 47). We can apply this concept to explain the bonding in alkanes. Ethane is taken as an example of a typical alkane. The four sp hybrid orbitals on each carbon atom will overlap end-on with four other orbitals three hydrogen Is orbitals and one sp hybrid orbital on the other carbon atom. Four cr bonds will be formed and they will adopt a tetrahedral arrangement. This is illustrated for ethane in the diagram. 

These new hybrid orbitals are then all equivalent, and spaced to minimize any interaction this is a tetrahedral array, the best way of arranging four groups around a central point. Each hybrid orbital can now accommodate one electron. 

With atoms such as carbon and silicon, the valence-state electronic configuration to form four covalent bonds has to be (s)1(px)1(py)1(ps)1. Repulsion between the electron pairs and between the attached nuclei will be minimized by formation of a tetrahedral arrangement of the bonds. The same geometry is predicted from hybridization of one s and three p orbitals, which gives four sp3-hybrid orbitals directed at angles of 109.5° to each other. The predicted relative overlapping power of s -hybrid orbitals is 2.00 (Figure 6-10). 

Some molecules, such as ammonia, have a lone pair of electrons on the central atom. This electron pair occupies an orbital confined to the central atom. According to the VSEPR model, the four electron pairs in NH3 take up a tetrahedral electron arrangement, so we describe the nitrogen atom in terms of four sp3 hybrid orbitals. Because nitrogen has five valence electrons, one of these hybrid orbitals is already doubly occupied (45). The ls-electrons of the three hydrogen atoms pair with the... 

Many organic compounds contain carbon atoms that are bonded to four other atoms. When four bonds are oriented as far apart as possible, they form a regular tetrahedron (109.5° bond angles), as pictured in Figure 2-15. This tetrahedral arrangement can be explained by combining the s orbital with all three p orbitals. The resulting four orbitals are called sp hybrid orbitals because they are composed of one s and three p orbitals. 

The four electron pairs around the nitrogen atom require a tetrahedral arrangement. We have seen that a tetrahedral set of sp3 hybrid orbitals is obtained by combining the 2s and three 2p orbitals. In the NH3 molecule three of the sp3 orbitals are used to form bonds to the three hydrogen atoms, and the fourth spJ orbital is used to hold the lone pair, as shown in Fig. 14.7. 

Carbon occurs in the allotropes (different forms) diamond, graphite, and the fullerenes. The fullerenes are molecular solids (see Section 16.6), but diamond and graphite are typically network solids. In diamond, the hardest naturally occurring substance, each carbon atom is surrounded by a tetrahedral arrangement of other carbon atoms, as shown in Fig. 16.26(a). This structure is stabilized by covalent bonds, which, in terms of the localized electron model, are formed by the overlap of sp3 hybridized atomic orbitals on each carbon atom. 

The simplest member of the saturated hydrocarbons, which are also called the alkanes, is methane (CH4). As discussed in Section 14.1, methane has a tetrahedral structure and can be described in terms of a carbon atom using an sp-J hybrid set of orbitals to bond to the four hydrogen atoms (see Fig. 22.1). The next alkane, the one containing two carbon atoms, is ethane (C2H6), as shown in Fig. 22.2. Each carbon in ethane is surrounded by four atoms and thus adopts a tetrahedral arrangement and sp3 hybridization, as predicted by the localized electron model. 

The quasi-localized orbitals of CH4 imply sp hybrid orbitals in so far as the molecular geometry is tetrahedral. The same occurs for ethane (Fig. 8.10) and all saturated unstrained hydrocarbons. Thus it is found that, as a result of an essentially tetrahedral arrangement of four atoms around each C, all C-H bonds in these compounds can be described in terms of identical localized molecular orbitals . Although these cannot be directly related to electron energies, this description parallels the observation that C-H bond... 

Even though the valence would be correct after promotion, the structure still would be wrong. Beryllium hydride would have two different kinds of bonds, and methane would have three identical bonds formed by overlap of H(ls) with the C(2p) orbitals and a different bond formed by H(ls) and C(2s). Pauling proposed that new orbitals with the proper symmetry for bond formation could be formed by hybridization of 2s and 2p orbitals after promotion. The Be(2s) and Be(2pz) orbitals would combine to form two equivalent hybrid orbitals oriented 180° apart. The C(2s) would hybridize with the three C 2p) orbitals to give four equivalent new orbitals in a tetrahedral arrangement around the carbon atom. 

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