Orbitals tetrahedral

The element before carbon in Period 2, boron, has one electron less than carbon, and forms many covalent compounds of type BX3 where X is a monovalent atom or group. In these, the boron uses three sp hybrid orbitals to form three trigonal planar bonds, like carbon in ethene, but the unhybridised 2p orbital is vacant, i.e. it contains no electrons. In the nitrogen atom (one more electron than carbon) one orbital must contain two electrons—the lone pair hence sp hybridisation will give four tetrahedral orbitals, one containing this lone pair. Oxygen similarly hybridised will have two orbitals occupied by lone pairs, and fluorine, three. Hence the hydrides of the elements from carbon to fluorine have the structures... 

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-... 

The formation of a Si crystal is shown in Fig. 1.10. Aside from the core, each Si atom has four valence electrons two 3s electrons and two 3p electrons. To form a Si crystal, one of the 3s electrons is excited to the 3p orbital. The four valence electrons form four sp hybrid orbitals, each points to a vertex of a tetrahedron, as shown in Fig. 1.10. Thpse four sp orbitals are unpaired, that is, each orbital is occupied by one electron. Since the electron has spin, each orbital can be occupied by two electrons with opposite spins. To satisfy this, each of the directional sp orbitals is bonded with an sp orbital of a neighboring Si atom to form electron pairs, or a valence bond. Such a valence bonding of all Si atoms in a crystal form a structure shown in (b) of Fig. 1.10, the so-called diamond structure. As seen, it is a cubic crystal. Because all those tetrahedral orbitals are fully occupied, there is no free electron. Thus, similar to diamond, silicon is not a metal. 

Herzberg (Nobel prize for Chemistry, 1971) commented on the two distinct photoionizations from methane that this observation illustrates the rather drastic nature of the approximation made in the valence bond treatment of CH4, in which the 2s and 2p electrons of the carbon atom are considered as degenerate and where this degeneracy is used to form tetrahedral orbitals representing mixtures of 2s and 2p atomic orbitals. The molecular orbital treatment does not have this difficulty". 

Fig. 4-2.—The angular dependence of a tetrahedral orbital with bond direction along the x axis.

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... 

Since tetrahedral orbitals form stronger bonds than other orbitals it might be thought that hybridization to tetrahedral orbitals would always occur in bond formation. The tendency to use the best bond orbitals is, however, resisted in the case of atoms with an unshared pair (or more than one) by the tendency to keep the unshared pair in the 8 orbital, which is more stable than the p orbitals. In OFi, for example, the use of tetrahedral orbitals in bond formation would require that half of the 8 orbital (which is divided equally among the four tetrahedral orbitals) be used for shared pairs and only half for unshared pairs. Since a shared pair counts as only one electron for each atom, this would involve the loss of one-quarter of the extra stability due to a pair of s electrons, and the atom will strive to prevent this. On the... 

Their bond strength is 1.991, only slightly less than that of tetrahedral orbitals. 14 H. A. L5vy and L. O. Brockway, J.A.C.S. 59, 2085 (1937). 

The explanation of the deviation of the value from 126.27° (the tetrahedral value) is that the bond orbitals of the tricovalent nitrogen atom are not tetrahedral orbitals, but are orbitals with only about 5 percent of 8 character, plus small amounts of d and / character (Secs. 4-3 and 4-5). For these orbitals a value for the single-bond double-bond angle intermediate between the values 90° for pure p orbitals and 125.27° for tetrahedral orbitals would be expected. 

They have the bond strength 2.694, much greater than that of sp tetrahedral orbitals (2.000). These four square orbitals are formed with use of only two of the 4p orbitals the other p orbital might accordingly also be used by the nickel atom to form another (rather weak) bond. 

Even in the case of s-p orbitals it is not necessary that all the orbitals be equivalent. Consider the water molecule, in which the H—O—H angle is 104 ", which docs not correspond to any of the hybrids described above, but lies between the I09 ° angle for sp3 and 90° for pure p orbitals. Presumably the two bonding orbitals in water are approximately tetrahedral orbitals but contain a little more p character, which correlates with the tendency of the bond angle to diminish toward the 90° of pure p orbitals. The driving forces for this effect will be discussed in Chapter 6. 

Mathematically, the formation of sp3 or tetrahedral orbitals for methane is more complicated but not basically different. The results are four equivalent hybrid orbitals, each containing one part s to three parts p in each wave function, directed to the corners of a tetrahedron. As in the case of sp hybrids, the hybridization of s and p has... 

To a first approximation, the three fluorine atoms and the single oxygen atom will be bonded to the phosphorus atom with <7 bonds from sp3 tetrahedral orbitals. One of the five 3d orbitals on the phosphorus atom also can overlap with a 2p Orbital on the oxygen atom (Fig. 6. Id) and form a fifth bond, d -p , further stabilizing the molecule. 

Molecules Isoelectronic with Neon.—Another set of molecules whose structure is typical of many standard chemical environments is the set of ten-electron first row hydrides Ne, HF, H20, NH3, and CH4. On p. 281 we saw how the outer electrons of the neon atom could be described either as being in the configuration (2s)2(2p)6 or, alternatively, as occupying four tetrahedral orbitals %2, %s, and y4 orientated relative to one another in a tetrahedral manner, the orientation of the tetrahedron in space being arbitrary. The electronic structures of the other molecules of the series can now be discussed in terms of this basic system if we imagine unit positive charges to be removed successively from the nucleus. 

When carbon is linked with four neighbouring atoms, the four valencies are directed towards the corners of a regular tetrahedron. Such a configuration is favourable for several reasons. Firstly, the hybrid sp tetrahedral orbitals permit greater overlapping of orbitals and hence a greater bond energy than s ot p... 

Another series of tetrahedral orbitals may be obtained starting from the... 

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