Tetrahedral Bond Orbitals

Each of the four bonds formed by the carbon atom involves one of the four orbitals of the L shell (Section 6-2). These orbitals are given in Chapter 5 as the 2s orbital and the three 2p orbitals. We might hence ask whether or not the bonds to the four hydrogen atoms are all alike. Would not the 2s electron form a bond of one kind, and the three 2p electrons form bonds of a different kind  

Some molecules are known in which the bond angles are required by the molecular structure to differ greatly from the tetrahedral value. These 

Diagram illustrating (left) (he D orbital in the K shell of the carbon atom, and (right) the four tetrahedral orbitals of the L shell. 

Sometimes two valence bonds of an atom are used in the formation of a double bond with another atom. There is a double bond between two carbon atoms in the molecules of ethylene (ethene), C2H4  

It is interesting to note that the four other bonds that the two carbon atoms in ethylene can form lie in the same plane, at right angles to the plane containing the two bent bonds. 

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Pig. 1. Polar graph showing the dependence on and q> of a tetrahedral bond orbital. The value of y> in cross-section is shown the function is cylindrically symmetrical about the midline of this cross-section. 

Derivation of Results about Tetrahedral Orbitals.—The results about tetrahedral bond orbitals described. above are derived in the following way. We assume that the radial parts of the wave functions fa and faxf faM are so closely similar that their differences can be neglected. The angular parts are... 

An equivalent set of tetrahedral bond orbitals, differing from the e only in orientation, is... 

Fig. 4-6.—Tetrahedral bond orbital with 4 percent d character and 2 percent /.

With this structure the nickel atom lias achieved the krypton electron configuration its outer shell contains five unshared pairs (in the five M orbitals) and five shared pairs (occupying the 4s4p3 tetrahedral bond orbitals). The Ni—C bond length expected for this structure is about 2.16 A, as found by use of the tetrahedral radius 1.39 A obtained by extrapolation from the adjacent values in Table 7-13 (Cu, 1.35 A Zn, 1.31 A). 

Pauling s prescription [169] to specify a set of four equivalent tetrahedral bond orbitals,... 

Photo 37 Linus Pauling lecturing on chemical quantum mechanics in a seminar at Oslo University in 1982. The tetrahedral bond orbitals of the carbon atom can be seen at the left. The lecture appears to have summarized Pauling s development of bond-orbital hybridization in the early 1930 s (SP 5). 

Chemists have made many experiments to answer this question, and have concluded that the four bonds of the carbon atom are identical. A theory of the tetrahedral carbon atom was developed in 1931. According to this theory, the theoiy of hybrid bond orbitals, the 2s orbital and the three 2p orbitals of the carbon atom are hybridized (combined) to form four tetrahedral bond orbitals (or sp - orbitals). They are exactly equivalent to one another, and are directed toward the corners of a regular tetrahedron, as shown in Figure 6-8. Moreover, the nature of s and p orbitals and their hybrids is such that of all possible hybrid orbitals of s and p the tetrahedral orbitals are the best suited for forming strong bonds. Accordingly the tetrahedral arrangement of the bonds is the stable one. 

The tetrahydrides (CH4 to SnH4) have a regular tetrahedral structure, corresponding to the use of sp tetrahedral bond orbitals by the central atom (bond angles 109.5°, Section 6-5). The other hydrides have smaller bond angles, approaching 90°, the value for/ bond orbitals (Section 6-7). 

N, and C hybridize their sp orbit upon reacting with atoms in any phase to create tetrahedral bonding orbits. 

The concept of hybridization explains how carbon forms four equivalent tetrahedral bonds but not why it does so. The shape of the hybrid orbital suggests the answer. When an 5 orbital hybridizes rvith three p orbitals, the resultant sp3 hybrid orbitals are unsyimmetrical about the nucleus. One of the two... 

N, 0, F, P, S, and Cl the bond orbitals for normal valence compounds lead to about the same radii as tetrahedral orbitals, whereas in atoms below these in the periodic system normal valence bonds involve orbitals which approach p-orbitals rather closely, and so lead to weaker bonds, and to radii larger than the tetrahedral radii. This effect should be observed in Br, Se, and As, but not in Ge, and in I, Te, and Sb, but not Sn. For this reason we have added 0.03 A to the tetrahedral radii for As and Se and... 

The internuclear distance at which the best bond orbitals are tetrahedral orbitals is in each case somewhat larger than the equilibrium distance given by the minimum of the energy curve. It is possible that in actual molecules the repul-... 

In this discussion of the transition elements we have considered only the orbitals (n— )d ns np. It seems probable that in some metals use is made also of the nd orbitals in bond formation. In gray tin, with the diamond structure, the four orbitals 5s5p3 are used with four outer electrons in the formation of tetrahedral bonds, the 4d shell being filled with ten electrons. The structure of white tin, in which each atom has six nearest neighbors (four at 3.016A and two at 3.17.5A), becomes reasonable if it is assumed that one of the 4d electrons is promoted to the 5d shell, and that six bonds are formed with use of the orbitals 4dSs5p35d. 

After rising at copper and zinc, the curve of metallic radii approaches those of the normal covalent radii and tetrahedral covalent radii (which themselves differ for arsenic, selenium, and bromine because of the difference in character of the bond orbitals, which approximate p orbitals for normal covalent bonds and sp3 orbitals for tetrahedral bonds). The bond orbitals for gallium are expected to be composed of 0.22 d orbital, one s orbital, and 2.22 p orbitals, and hence to be only slightly stronger than tetrahedral bonds, as is indicated by the fact that R(l) is smaller than the tetrahedral radius. 

It is interesting that a straight line drawn through the tetrahedral radii passes through the metallic radius for calcium this suggests that the metallic bonding orbitals for calcium are sp orbitals, and that those for scandium begin to involve d-orbital hybridization. 

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