Hybridization of the carbon atom

This section and the next are a review of the ways carbon atoms bond to themselves or to other elements to form solids such as the carbides. 

The Carbon Hybrid si Orbital. In the s[ hybrid configuration, the arrangement of the electrons of the L shell of the atom in the ground state is modified as one of the 2s electrons is promoted (or lifted) to the higher orbital 2p as shown in Fig. 3.4. These new orbitals are called hybrids since they combine the 2 and the 2p orbitals. They are labeled sp since they are formed from one s orbital and threep orbitals. 

In this hybrid sp state, the carbon atom has four 2s[ orbitals, instead of two 25 and two 2p of the ground-state atom. The valence state is raised from two to four and can accept four other electrons from another atom. The calculated sp electron-density contour is shown in Fig. 3.5 and a graphic visualization of the orbital, in the shape of an electron cloud, is shown in Fig. 3.6.1 1 This orbital is asymmetric, with most of it concentrated on one side and with a small tail on the opposite side. The lobes are labeled + or -. These refer to the sign of the wave function and not to any positive or negative charges since an electron is always negatively charged. 

The energy required to accomplish the sp hybridization and raise the carbon atom from the ground state to the corresponding valence state V4 is 230 kJ mol This hybridization is possible only because the required energy is more than compensated by the energy decrease associated with forming bonds with other atoms. 

The hybridized atom is now ready to form a set of bonds with other atoms. It should be stressed that these hybrid orbitals (and indeed all hybrid orbitals) are formed only in the bonding process with other atoms and are not representative of an actual structure of a free carbon atom.l l 

Ground-State Carbon Orbitals. The carbon-atom orbiteils in the ground state can be visualized as shown graphically Fig. 2.6. The wave-function calculations represent the s orbital as a sphere with a blurred or fuzzy edge that is characteristic of all orbital representation. As a sphere, the s orbital is non-directional. The 2p orbital can be represented as an elongated barbell which is symmetrical about its axis and directional. 

The Carbon Hybrid sp Orbital. The s 2s 2p configuration of the carbon atom does not account for the tetrahedral symmetry found in structures such as diamond or methane (CH4) where the carbon atom is bonded to four other carbon atoms in the case of diamond, or to four atoms of hydrogen in the case of methane. In both cases, the four bonds are of equal strength. 

In order to have a electron configuration that would account for this symmetry, the structure of the carbon atom must be altered to a state with four valence electrons instead of two, each in a separate orbital and each 

A graphic visualization of the formation of the sp hybridization is shown in Fig. 2.10. The four hybrid sp orbitals (known as tetragonal hybrids) have identical shape but different spatial orientation. Connecting the end points of these vectors (orientation of maximum probability) forms a regular tetrahedron (i.e., a solid with four plane faces) with equal angles to each other of 109° 28.  

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Commonly used in biology as a tissue preservative, formaldehyde, CIDO, contains a carbon -oxygen double bond. Draw the line-bond structure of formaldehyde, and indicate the hybridization of the carbon atom. 

Predict the hybridization of the carbon atom in each of the following functional groups ... 

Unsaturated organic molecules, such as ethylene, can be chemisorbed on transition metal surfaces in two ways, namely in -coordination or di-o coordination. As shown in Fig. 2.24, the n type of bonding of ethylene involves donation of electron density from the doubly occupied n orbital (which is o-symmetric with respect to the normal to the surface) to the metal ds-hybrid orbitals. Electron density is also backdonated from the px and dM metal orbitals into the lowest unoccupied molecular orbital (LUMO) of the ethylene molecule, which is the empty asymmetric 71 orbital. The corresponding overall interaction is relatively weak, thus the sp2 hybridization of the carbon atoms involved in the ethylene double bond is retained. 

When there is more than one central atom in a molecule, we concentrate on each atom in turn and match the hybridization of each atom to the shape at that atom predicted by VSEPR. For example, in ethane, C2H6 (38), the two carbon atoms are both central atoms. According to the VSEPR model, the four electron pairs around each carbon atom take up a tetrahedral arrangement. This arrangement suggests sp hybridization of the carbon atoms, as shown in Fig. 3.14. Each... 

Cyclopropane, C.H, is a hydrocarbon composed of a three-membered ring of carbon atoms, (a) Determine the hybridization of the carbon atoms, (b) Predict the CCC and HCH bond angles at each carbon atom on the basis of your answer to part (a), (c) What must the real CCC bond angles in cyclopropane be (d) What is the defining characteristic of a cr-bond compared with a ir-bond, for example (e) How do the C—C cr-bonds in cyclopropane extend the definition of conventional o-bonds (f) Draw a picture depicting the molecular orbitals to illustrate your answer. 

H/D exchange of H and Hg protons of sulfone 86 and estimated the difference in the free energies of activation for 79a and 79b to be < 1.2 kcal mol , based on the kjk value of 3 0.5. In the base-catalyzed H/D exchange of 87, kjk = 1.6, where k and k are the rate constants of H/D exchange of H, and H, respectively. Based on the small kjk value. Brown and colleagues suggested that if the carbanion is pyramidal, the steric stabilities of 79a and 79b are almost identical. Meanwhile, based on their C-NMR study Chassaing and Marquet proposed that the hybridization of the carbon atom of the sulfonyl carbanion, PhSOjCHj , would be between sp and sp . 

The mechanism described in Scheme 2 was rejected on the grounds that the steric requirement for the abstraction of a hydrogen atom from -CHD-of species (IV) could not be met. Assuming an atomically flat surface, and sp3 hybridization of the carbon atom bonded to the surface, the plane of the Ce-ring in (IV) is in such a configuration that the hydrogen atom of -CHD- is directed away from the surface, and the deuterium atom toward the surface. Thus, unless the species is adsorbed near a step in the metal lattice, the loss of this hydrogen and the formation of a second carbon-metal bond would require a very considerable distortion of adsorbed species. 

Changes in I strain bring about changes in the hybridization of the carbon atom during the reaction. Thus in the reduction of the ketones the trigonal sp2 hybridized carbon changes to tetrahedral carbon atom. 

The observed very substantial shift of the resonance line in the trimethylcarbonium complex, as compared with the position of the C line in the starting covalent sp -hybridized halide, amounting to 273 p.p.m., is difficult to interpret in any other way than as a direct proof (a) that the state of hybridization of the carbon atom involved is changed in the complex to sp and, at the same time, (b) that the carbon atom carries a substantial positive charge. 

When considering the reaction of carbonyl groups, remember the polarity of the carbon-oxygen bond, the hybridization of the carbon atom (sp ), and the bond angles of the planar group. Figure 10-16 summarizes these features. 

Finally, the hybridization of the carbon atom also has a marked effect on its willingness to attach to the transition metal. Allyl or benzyl halides undergo oxidative addition faster than aromatic or vinyl halides. The least reactive are alkyl halides which require the use of nickel(O)9 complexes or highly active catalyst systems.10 If we start from an optically active substrate, then the oxidative addition usually proceeds in a stereoselective manner. 

The magnitude of 13C— H coupling constants depends on the hybridization of the carbon atom (sp3, sp2, sp) coupled to the proton. This is indicative of a mechanism of spin-spin coupling involving the bonding electrons, as is illustrated in Fig. 1.11 for a group AX with nuclei A and X, each with a total spin quantum number of / =... 

PROBLEM 7.22 Describe the hybridization of the carbon atom in the hydrogen cyanide molecule, H-C=N, and make a rough sketch to show the hybrid orbitals it uses for bonding. 

The X-ray crystallographic analysis of 9 provided experimental structural data of the new corannulene-type aromatic system. The bond alternation found in 9 closely follows the pattern found in an X-ray crystal structure of tetramethyl fused corannulene [38c]. Such agreement of the bond alternation patterns indicates that the fullerene framework and the dibenzo conjugation added to the parent corannulene structure have a small structural effect in spite of the rather large difference in hybridization of the carbon atoms between 9 p-orbital axis vector (POAV) angles [40] = 8.0-12.8° and C (POAV = 0.0-11.0°). 

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