S-p hybridization

The valence atomic orbitals which are available to form the orbitals of a CC single bond, directed along the x axis, are the 2s and 2px atomic orbitals on each carbon atom. Their admixture—in proportions which depend on the number of neighbors at each carbon and on the subsequent hybridization—creates two (s, p ) hybrids on each atom. One of these hybrids points away from the other atom and can be used for bonding to additional atoms. The pair of hybrids which point at each other overlap and interact in the conventional fashion [we symbolize the non-interacting orbitals by an interruption of the bond axis (Fig. 1)]. The two bond orbitals which are formed in this manner both have [Pg.3]

Verification of the shape of the curve has been obtained with the collaboration of Dr. J. Sherman by the theoretical treatment of a somewhat similar problem (the effect of s-p hybridization of bond orbitals on interatomic distance). 

Fig. 4.—Calculated one-electron bond energy-values (D/k, full curves) and squared bond strength values (S, dashed curves) for s-p hybridization.

James also carried out a treatment of the lithium molecule-ion with use of 25 orbitals and explicit consideration of the four K electrons, obtaining the energy value —0.304 e. v. at r = 2.98 A. (this being not necessarily the minimum point of the curve). Our curve for an 5 bond similarly gives a very small energy value, — 0.22 e. v., at this value of r. The calculations described above indicate that the principal source of inaccuracy in these treatments is the assumption that the bonds are 5 bonds, and that the consideration of s-p hybridization makes a great improvement. 

Theoretical energy curves for one-electron bonds between two atoms are calculated for bond orbitals formed by hybridization of 2s and 2p orbitals, 35 and 3 orbitals, and 35, 3p, and 3d orbitals, the same radial part being used for the orbitals in a set. It is found that for s-p hybridization the bond energy is closely proportional to S3, with 5 the magnitude of the angular part of the bond orbital in the bond direction. This relation is less satisfactorily approximated in the case of s-p-d hybridization. 

The curve of single-bond metallic radii for the elements of the first long period has a characteristic appearance (Fig. 3) which must be attributed in the main to variation in the type of bond orbital. The rapid decrease from potassium to chromium results from increase in bond strength due to increasing s-p and d-s—p hybridization. The linear section of the curve from chromium to nickel substantiates the assumption that the same bonding orbitals (hybrids of 2.56 3d orbitals, one 4s orbital, and 2.22 4p orbitals) are effective throughout this series. The increase in radius from nickel to copper is attributed not... 

This view somehow seems dubious in the case of heavier elements like 6 row metals. The high energy separation, as well as the very different spatial distribution of the 6s/6p wavefunctions, which are found for these elements because of the strong influence of relativity, stand against an efficient s-p hybridization. The first excited state of Th (in the gas phase), s p lies 7.4 eV above the... 

In summary, one can state that s-p-hybridization on the heavier main group metals is not responsible for the stereochemical activity of a lone pair. Instead, the general conclusion can be drawn that anti-bonding metal ns-ligand np interactions lead to structural distortions in order to minimize these unfavorable interactions. 

Any hybrid orbital is named from the atomic valence orbitals from which It Is constmcted. To match the geometry of methane, we need four orbitals that point at the comers of a tetrahedron. We construct this set from one s orbital and three p orbitals, so the hybrids are called s p hybrid orbitais. Figure 10-8a shows the detailed shape of an s p hybrid orbital. For the sake of convenience and to keep our figures as uncluttered as possible, we use the stylized view of hybrid orbitals shown in Figure 10-8Z). In this representation, we omit the small backside lobe, and we slim down the orbital in order to show several orbitals around an atom. Figure 10-8c shows a stylized view of an s p hybridized atom. This part of the figure shows that all four s p hybrids have the same shape, but each points to a different comer of a regular tetrahedron. 

An inner atom with a steric number of 4 has tetrahedral electron group geometry and can be described using S p hybrid orbitals. 

Remember that the molecular shape ignores the lone pair. The hydronium ion has a trigonal pyramidal shape described by the three s p hybrid orbitals that form bonds to hydrogen atoms. 

Both inner atoms have steric numbers of 4 and tetrahedral electron group geometry, so both can be described using s p hybrid orbitals. All four hydrogen atoms occupy outer positions, and these form bonds to the inner atoms through 1 s-s p overlap. The oxygen atom has two lone pairs, one in each of the two hybrid orbitals not used to form O—H bonds. 

You can visualize methanol either as methane with one —H replaced by —OH or as water with one —H replaced by — CH3. Either way of visualizing the molecule reveals a tetrahedral inner atom, for which s p hybridization is appropriate. 

P n s p hybridized atom has three coplanar hybrid orbitals separated by 120° angles. One unchanged p orbital is perpendicular to the plane of the hybrids. 

In triethylaluminum, each A1—C bond can be visualized as an. y p hybrid on aluminum overlapping with an S p hybrid on a carbon atom. Figure 10-13 shows this bonding representation, with three equivalent A1—C bonds and the unused 3 p orbital on the aluminum atom. 

Remember from Chapter 6 that energy is released when a bond forms. Consequently, atoms that form covalent bonds tend to use all their valence s and p orbitals to make as many bonds as possible. We might expect the S p -hybridized aluminum atom to form a fourth bond with its unused 3 p orbital. A fourth bond does not form in A1 (C2 115)3 because the carbon atoms bonded to aluminum have neither orbitals nor electrons available for additional bond formation. The potential to form a fourth bond makes triethylaluminum a very reactive molecule. 

The Zn—C bonds of dimethyizinc are formed by overlap of an sp hybrid orbital from the metal with an s p hybrid orbitai from carbon. 

The carbon and nitrogen atoms have a p orbital left over after construction of the. s p hybrids. These two p orbitals are perpendicular to the plane that contains the five nuclei. Side-by-side overlap of the p orbitals gives a jrbond that completes the bonding description of this molecule ... 

Once the a bonding framework is completed, the only atoms with unused valence orbitals are the s p -hybridized... 

Why does siiicon not form itt bonds as readiiy as carbon, its Group 14 neighbor The answer can be found by examining the sizes of the vaience orbitais of siiicon and carbon. Figure 10-23 compares side-by-side p-orbital overiap for carbon atoms and siiicon atoms. Notice that for the larger silicon atom, the 3 orbitals are too far apart for strong side-by-side overiap. The resuit is a very weak ttbond. Consequently, silicon makes more effective use of its vaience orbitais to form four a bonds, which can be described using s p hybrid orbitals. 

Ozone, which has 18 valehce electrohs, exemplifies bent molecules. Another example is the hitrite ahioh, the subject of Extra Practice Exercise. The bohdihg of NO2 can be represented using s p hybrid orbitals for the inner nitrogen atom and one set of delocalized n orbitals. 

All the atoms of butadiene lie in a plane defined by the s p hybrid orbitals. Each carbon atom has one remaining p orbital that points perpendicular to the plane, in perfect position for side-by-side overlap. Figure 10-42 shows that all four p orbitals interact to form four delocalized molecular orbitals two are bonding MOs and two are antibonding. The four remaining valence electrons fill the orbitals, leaving the two p orbitals empty. 

The Lewis structure shows that methyl methaciylate has the formula C5 Hg O2, with 40 valence electrons. You should be able to verily that the two CH3 groups have. s -hybridized carbons, the inner oxygen atom is s hybridized, the outer oxygen atom uses 2 p atomic orbitals, and the three double-bonded carbons are s p hybridized. These assignments lead to the following inventory of a bonds and inner-atom lone pairs ... 

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