S Orbitals hybridization

The concept of hybridization explains how carbon forms four equivalent tetrahedral bonds but not why it does so. I he shape of the liybrid orbital suggests the answer. When an s orbital hybridizes with three p orbitals, the resultant sp hybrid orbitals are unsymmetrical about the nucleus. One of the hvo... 

The peculiar shape of sp hybrid orbitals turn out to have an important consequence. Since most of the electron density in an sp hybrid orbital lies to one side of a carbon atom, overlap with a half-filled 1 orbital of hydrogen, for example, on that side produces a stronger bond than would result otherwise. If the electron probabilities were equal on both sides of the nucleus, as it would be in a p orbital, half of the time the electron would be remote from the region between the bonded atoms, and the bond would be weaker. Thus, not only does Pauling s orbital hybridization proposal account for carbon forming four bonds rather than two, these bonds are also stronger than they would be otherwise. 

On the basis of the above interpretation NQI can be unequivocally represented in terms of the population of orbitals. Actually, NQR spectroscopy allows the determination, at best, of two parameters (e Qqh and rj) and in many cases (I = j) only e Qqh can be obtained. However, the above-mentioned Equation [1] contains four parameters (s, d, /g, which cannot be determined from one or two experimental parameters. It is therefore necessary to include approximations, which neglect the d orbital participation in hybridization, and also, in some cases, p bonding and to consider that the s orbital hybridization is a small contribution and remains constant in a series of compounds. Thus changes of e Qqh are directly related to bond ionic character or p electron charge transfer in the case of a hydrogen bond. 

In reality, there are four primary bonds associated with each Si atom. This can happen by hybridization between s and p wave functions. The s orbital hybridizes with all the three p orbitals to formsp hybrid orbitals (Figure 8.5b). Hybrid orbitals possess both s and p character, and directionally reaches out in space as lobes in a tetrahedral arrangement with a bond angle of 109° (Figure 8.5c). Each orbital is populated by one electron (Figure 8.5b). Now, each Si can bond to four other Si atoms or any other four atoms to give three-dimensional structures. 

This sum describes the polarization of the Is orbital in terms of functions that have PO symmetry by combining an s orbital and po orbitals, one can form a hybrid-like orbital that is nothing but a distorted Is orbital. The appearance of the excited npo orbitals has... 

For example, in formaldehyde, H2CO, one forms sp hybrids on the C atom on the O atom, either sp hybrids (with one p orbital "reserved" for use in forming the n and 7i orbitals and another p orbital to be used as a non-bonding orbital lying in the plane of the molecule) or sp hybrids (with the remaining p orbital reserved for the n and 7i orbitals) can be used. The H atoms use their 1 s orbitals since hybridization is not feasible for them. The C atom clearly uses its sp2 hybrids to form two CH and one CO a bondingantibonding orbital pairs. 

FIGURE 2 8 sp Hybridization (a) Electron configuration of carbon in its most stable state (b) Mixing the s orbital with the three p orbitals generates four sp hybrid orbitals The four sp hybrid orbitals are of equal energy therefore the four valence electrons are distributed evenly among them The axes of the four sp orbitals are directed toward the corners of a tetrahedron... 

Because each carbon m acetylene is bonded to two other atoms the orbital hybridization model requires each carbon to have two equivalent orbitals available for CT bonds as outlined m Figure 2 19 According to this model the carbon 2s orbital and one of Its 2p orbitals combine to generate two sp hybrid orbitals each of which has 50% s character and 50% p character These two sp orbitals share a common axis but their major lobes are oriented at an angle of 180° to each other Two of the original 2p orbitals remain unhybridized... 

Conformational analysis is far simpler m cyclopropane than m any other cycloalkane Cyclopropane s three carbon atoms are of geometric necessity coplanar and rotation about Its carbon-carbon bonds is impossible You saw m Section 3 4 how angle strain m cyclopropane leads to an abnormally large heat of combustion Let s now look at cyclopropane m more detail to see how our orbital hybridization bonding model may be adapted to molecules of unusual geometry... 

All of these trends can be accommodated by the orbital hybridization model The bond angles are characteristic for the sp sp and sp hybridization states of carbon and don t require additional comment The bond distances bond strengths and acidities are related to the s character m the orbitals used for bonding s Character is a simple concept being nothing more than the percentage of the hybrid orbital contributed by an s orbital Thus an sp orbital has one quarter s character and three quarters p an sp orbital has one third s and two thirds p and an sp orbital one half s and one half p We then use this information to analyze how various qualities of the hybrid orbital reflect those of its s and p contributors... 

An orbital hybridization description of bonding in methylamine is shown in Figure 22.2. Nitrogen and carbon are both s/r -hybridized and are joined by a a bond. The unshared electron pair on nitrogen occupies an s/r -hybridized orbital. This lone pair-is involved in reactions in which fflnines act as bases or nucleophiles. The graphic that opened this chapter is an electrostatic potential map that clearly shows the concentration of electron density at nitrogen in rnethylanine. 

An answer was provided in 1931 by Linus Pauling, who showed how an s orbital and three p orbitals on an atom can combine mathematically, or hybridize, to form four equivalent atomic orbitals with tetrahedral orientation. Shown in Figure 1.10, these tetrahedrally oriented orbitals are called sp3 hybrids. Note that the superscript 3 in the name sp3 tells how many of each type of atomic orbital combine to form the hybrid, not how many electrons occupy it. 

The asymmetry of sp3 orbitals arises because, as noted previously, the two lobes of a p orbital have different algebraic signs, + and -. Thus, when a p orbital hybridizes with an s orbital, the positive p lobe adds to the s orbital but the negative p lobe subtracts from the s orbital. The resultant hybrid orbital is therefore unsymmetrical about the nucleus and is strongly oriented in one direction. 

What accounts for the stability of conjugated dienes According to valence bond theory (Sections 1.5 and 1.8), the stability is due to orbital hybridization. Typical C—C bonds like those in alkanes result from a overlap of 5p3 orbitals on both carbons. In a conjugated diene, however, the central C—C bond results from conjugated diene results in part from the greater amount of s character in the orbitals forming the C-C bond. 

The carbon-oxygen double bond of a carbonyl group is similar in many respects to the carbon-carbon double bond of an alkene. The carbonyl carbon atom is s/ 2-hybridized and forms three valence electron remains in a carbon p orbital and forms a tt bond to oxygen by overlap with an oxygen p orbital. The oxygen atom also has two nonbonding pairs of electrons, w hich occupy its remaining two orbitals. 

Hybrid orbital (Section 1.6) An orbital derived from a combination of atomic orbitals. Hybrid orbitals, such as the sp3, s/J2, and sp hybrids of carbon, are strongly directed and form stronger bonds than atomic orbitals do. 

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