Unhybridized p orbitals

Section 2 20 Carbon is sp hybridized in ethylene and the double bond has a ct com ponent and a rr component The sp hybridization state is derived by mix mg the 2s and two of the three 2p orbitals Three equivalent sp orbitals result and their axes are coplanar Overlap of an sp orbital of one car bon with an sp orbital of another produces a ct bond between them Each carbon still has one unhybridized p orbital available for bonding and side by side overlap of the p orbitals of adjacent carbons gives a rr bond between them... 

FIGURE 14 3 (a) The unshared electron pair occupies an sp hybridized orbital in dichlorocarbene There are no electrons in the unhybridized p orbital (b) An electrostatic potential map of dichlorocarbene shows negative charge is concentrated in the region of the unshared pair and positive charge above and below the carbon... 

Figure 1.13 An sp hybridized carbon. The three equivalent sp2 hybrid orbitals (green) lie in a plane at angles of 120° to one another, and a single unhybridized p orbital (red/blue) is perpendicular to the sp2 plane.

When two sp2-hybridized carbons approach each other, they form a cr bond by sp2-sp2 head-on overlap. At the same time, the unhybridized p orbitals approach with the correct geometry for sideways overlap, leading to the formation of what is called a pi (ir) bond. The combination of an >p2-sp2 a bond and a 2p-2p 77 bond results iii the sharing of four electrons and the formation of a carbon-carbon double bond (Figure 1.14). Note that the electrons in then-bond occupy the region centered between nuclei, while the electrons in the 77 bond occupy regions on either side of a line drawn between nuclei. 

Figure 1.14 The structure of ethylene. Orbital overlap of two sp hybridized carbons forms a carbon-carbon double bond. One part of the double bond results from a (head-on) overlap of sp2 orbitals (green), and the other part results from (sideways) overlap of unhybridized p orbitals (red/blue). The ir bond has regions of electron density on either side of a line drawn between nuclei.

STRATEGY Use the VSEPR model to identify the shape of the molecule and then assign the hybridization consistent with that shape. All single bonds are cr-bonds and multiple i bonds are composed of a cr-bond and one or more TT-bonds. Because the C atom is attached to three atoms, we anticipate that its hybridization scheme is sp1 and that one unhybridized p-orbital remains. Finally, we form cr- and Tr-bonds by allowing the 1 orbitals to overlap. 

Multiple bonds are formed when an atom forms a tr-bond by using an sp or sp2 hybrid orbital and one or more ir-bonds by using unhybridized p-orbitals. The side-by-side overlap that forms a ir-bond makes a molecule resistant to twisting, results in bonds weaker than tr-bonds, and prevents atoms with large radii from forming multiple bonds. 

We can understand the differences in properties between the carbon allotropes by comparing their structures. Graphite consists of planar sheets of sp2 hybridized carbon atoms in a hexagonal network (Fig. 14.29). Electrons are free to move from one carbon atom to another through a delocalized Tr-network formed by the overlap of unhybridized p-orbitals on each carbon atom. This network spreads across the entire plane. Because of the electron delocalization, graphite is a black, lustrous, electrically conducting solid indeed, graphite is used as an electrical conductor in industry and as electrodes in electrochemical cells and batteries. Its... 

Use the four-step procedure described earlier. Begin with the Lewis structure for the molecule, and then identify the appropriate hybrid orbitals. Construct a a bond framework, and complete the bonding picture by assembling the bonds from the unhybridized p orbitals. 

Figure 1.5 The shapes of some s and p orbitals. Pure, unhybridized p orbitals are almost-touching spheres. The p orbitals in hybridized atoms are lobe-shaped (Section 1.14).

The planar framework has a bonds as just shown, which involve sp2 hybrid orbitals on the boron atoms. This leaves one unhybridized p orbital that is perpendicular to the plane. The B2H6 molecule can be considered being made by adding two H+ ions to a hypothetical B21142 ion that is isoelectronic with C2H4 because each carbon atom has one more electron than does a boron atom. In the B2I l42 ion, the two additional electrons reside in a tt bond that lies above and below the plane of the structure just shown. When two H+ ions are added, they become attached to the lobes of the n bond to produce a structure, the details of which can be shown as... 

A n bond is formed from the side by side overlap of unhybridized p orbitals, p orbital + p orbital => 1 n bond (2 separate electron clouds)... 

A k bond can not be formed alone. It can be formed after the formation of a o bond, if any unhybridized p-orbitals of atoms remain. In another words, to form a it bond, two atoms must form a a bond first. 

When a carbon atom undergoes sp2 hybridization, sp2 hybrid orbitals form n bonds, but the unhybridized p orbital forms a pi bond. 

Both carbon atoms in ethylene molecule undergo sp2 hybridization and form three identical sp2 hybrid orbitals. One p orbital remains unhybridized. Two sp2 hybrid orbitals from each carbon atom overlap end to end with the Is orbital of a hydrogen atom and four C — Ho bonds are formed in total. Also, between the two carbon atoms, a C — Co bond is formed as a result of the overlap between two sp2 hybrid orbitals. So, in the C2H4 molecule in total there are five o bonds. Meanwhile, the unhybridized p orbitals of the two carbon atoms overlap side by side and form a rt bond. So between the two carbon atoms in the C2H4 molecule there is one o bond, formed by the overlapping of sp2 hybrid orbitals and one n bond, formed by the side by side overlapping of the unhybridized p orbitals. In total, two bonds are formed, hence a double bond exists between the two carbon atoms. 

When carbon undergoes sp hybridization the sp hybrid orbitals form s bonds, and the unhybridized p orbitals form n bonds. 

In the C2H4 molecule, unhybridized p orbitals overlap in side by side and form a n bond. 

Meanwhile, the unhybridized p orbitals of two carbon atoms overlap side by side and form two C — C n bonds. Thus, in the C2H2 molecule between the two carbon atoms, one o bond is formed (by the end to end overlap of sp hybrid orbitals) and two n bonds are formed (by the side by side overlap of the unhybridized p orbitals). 

In the CJ I2 molecule, unhybridized p orbitals form two n bonds by overlapping side by side. 

Each atom in the ring must have an unhybridized p-orbital perpendicular to the plane of the ring. 

Heteroatoms (such as O, S, N) that are double-bonded to other atoms in a ring can t donate lone-pair electrons to the pi system because their p-electrons are already involved in the double bond. Single-bonded heteroatoms can donate a single lone-pair to the pi system but not two, because one lone-pair must be in an unhybridized p orbital orthogonal (at 90 degrees) to the sp ring plane. 

Acetylene, C H, is a linear molecule in which each C uses two sp HO s to form two o- bonds with a 180° angle. The unhybridized p orbitals form two tt bonds. 

The bond angles of the carbon atoms in benzene are 120°. All carbon atoms are s p -hybridized, and each carbon atom has a single unhybridized p orbital perpendicular to the plane of the ring. The carbon p -hybridized orbitals overlap to form the ring of the benzene molecule. Because the C—C bond lengths are 1.39 A, the p orbitals are close enough to overlap efficiently and equally all round the ring. 

The SHMO theory was originally developed to describe planar hydrocarbons with conjugated n bonds. Each center is sp2 hybridized and has one unhybridized p orbital perpendicular to the trigonal sp2 hybrid orbitals. The sp2 hybrid orbitals form a rigid unpolarizable framework of equal C—C bonds. Hydrogen atoms are part of the framework and are not counted. The Hiickel equations (3.3) described in the first part of Chapter 3 apply, namely,... 

Each MO is expanded in terms of the unhybridized p orbitals, one per center. The overlap integral between two parallel p orbitals is small and is approximated to be exactly zero. Thus,... 

The halide (rimers consist or planar six-membered rings (Fig. I6.23).77 The bond angles are consistent with sp1 hybridization of the nitrogen and approximately sp3 hybridization of the phosphorus. Two of the sp3 orbitals of nitrogen, containing one electron each, are used for o bonding and the third contains a lone pair of electrons This leaves one electron left for the unhybridized p. orbital. 

Multiple bonds are formed when an atom forms a a-bond by using a hybrid orbital and one or more rr-bonds by using unhybridized p-orbitals. 

FIGURE 7.10 The combination of one s and one p orbital gives two sp hybrid orbitals oriented 180° apart. Two unhybridized p orbitals remain and are oriented at 90° angles to the sp hybrids. 

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