Sp1 orbital

Nickel carbonyl, Ni(CO) , boils at 43°C, and uses the sp1 orbitals of Ni for bonding. Give reasons to justify the following ... 

R3N — C5H5N — RC=N , the unshared pairs are in sp3, sp2 and sp1 orbitals, respectively, and the declining basicity is reflected in the two pKa values quoted above, and in the fact that the basicity of alkyl cyanides is very small indeed (MeCN, pR = -4-3). 

Qualitatively, the resonance picture is often used to describe the structure of molecules, but quantitative valence-bond calculations become much more difficult as the structures become more complicated (e.g., naphthalene, pyridine, etc.). Therefore the molecular-orbital method is used much more often for the solution of wave equations.5 If we look at benzene by this method (qualitatively), we see that each carbon atom, being connected to three other atoms, uses sp1 orbitals to form a bonds, so that all 12 atoms are in one plane. Each carbon has a p orbital (containing one electron) remaining and each of these can overlap equally with the two adjacent p orbitals. This overlap of six orbitals (see Figure 2.1) produces six new orbitals, three of which (shown) are bonding. These three (called it orbitals) all occupy approximately the same space.6 One of the three is of lower energy than... 

The pair of electrons in the sp1 orbital may be donated to a metal lo form a u bond, and an empty p. orbital is present to accept it electron density. Filled d orbitals of (he metal may donate electrons to the p. orbital, to give a metal-carbon double bond, and electrons from filled p orbitals of the oxygen atom may also be donated to form a carbon-oxygen double bond (Fig. 15.20). Resonance form 15.20b appears to be dominant and, although the M—C bond is shorter than expected fora single bond, it is too long for an M—C double bond, leading to the conclusion that the bond order is between one and two. 

The basis set used is Dunning s cc-pVDZ [16]. The CASSCF wave function was obtained with CAS(4,4). The geometries of the equilibrium and TS structures were determined with this basis set and active space. The orbitals were then localized in the active orbital space. The orbitals were transformed so as to have maximum overlap with two carbon sp2 orbitals and hydrogen Is orbitals. The sp1 orbitals were used with the fixed hybridization ratio of 2s to 2p orbitals (1 2) and with a fixed angle of 120° relative to the CO axis... 

The electronic descriptions of pyridine and benzene are very similar. The pyridine ring is formed by the a overlap of carbon and nitrogen sp1 orbitals. In addition, six p orbitals, perpendicular to the plane of the ring, hold six electrons. These six p orbitals form six % molecular orbitals that allow electrons to be delocalized over the tc system of the pyridine ring. The lone pair of nitrogen electrons occupies an sp2 orbital that lies in the plane of the ring. 

The hybridization of the s, p , and py atomic orbitals results in the formation of three sp1 orbitals centered in thexy plane. The large lobes of the orbitals lie in the plane at angles of 120 degrees and point toward the corners of a triangle. 

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. 

Noting that the bond angle of an sp hybridized atom is 109.5° and that of an sp1 hybridized atom is 120°, do you expect the bond angle between two hybrid orbitals to increase or decrease as the s-character of the hybrids is increased ... 

Similar, but different, redeployment is envisaged when a carbon atom combines with three other atoms, e.g. in ethene (ethylene) (p. 8) three sp2 hybrid atomic orbitals disposed at 120° to each other in the same plane (plane trigonal hybridisation) are then employed. Finally, when carbon combines with two other atoms, e.g. in ethyne (acetylene) (p. 9) two sp1 hybrid atomic orbitals disposed at 180° to each other (idigonal hybridisation) are employed. In each case the s orbital is always involved as it is the one of lowest energy level. 

These are all valid ways of deploying one 2s and three 2p atomic orbitals—in the case of sp2 hybridisation there will be one unhybridised p orbital also available (p. 8), and in the case of sp1 hybridisation there will be two (p. 10). Other, equally valid, modes of hybridisation are also possible in which the hybrid orbitals are not necessarily identical with each other, e.g. those used in CH2C12 compared with the ones used in CC14 and CH4. Hybridisation takes place so that the atom concerned can form as strong bonds as possible, and so that the other atoms thus bonded (and the electron pairs constituting the bonds) are as far apart from each other as possible, i.e. so that the total intrinsic energy of the resultant compound is at a minimum. 

Arynes present structural features of some interest. They clearly cannot be acetylenic in the usual sense as this would require enormous deformation of the benzene ring in order to accommodate the 180° bond angle required by the sp1 hybridised carbons in an alkyne (p. 9). It seems more likely that the delocalised 7i orbitals of the aromatic system are left largely untouched (aromatic stability thereby being conserved), and that the two available electrons are accommodated in the original sp2 hybrid orbitals (101) ... 

Both carbon atoms in ethene undergo sp2 hybridization. The C-H bonds involve overlap of sp1 carbon orbitals with Is orbitals of the H atoms. The carbon-carbon double bond involves the overlap of sp2 orbitals from each carbon to give the o bond and the side-on overlap of a p orbital from each carbon atom to give the n bond. 

Hybrid orbitals formed two sp three sp1 four spi one d five sp1d two d six sp dl... 

In the boron trimethyl molecule the boron atom is surrounded by three pairs of valence electrons, which are involved in the formation of single covalent bonds to the three carbon atoms of the methyl groups. An electron-diffraction study10 has shown the molecule to be planar (except for the hydrogen atoms), as would be expected for sp1 hybrid orbitals. The —C distance is 1.56 i 0.02 A, which agrees reasonably well with the value 1.54 A calculated, with the electronegativity correction, by the use of 0.81 A for the boron single-bond radius.11... 

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. 

Exercise 1.8. What is the hybridization of the carbon orbitals which form the C—H and C—C bonds of cyclopropane (HCH = 114°) Verify that if the carbon hybrids which are used for the C—H bonds are exactly sp1, then the two equivalent hybrids for the C—C bonds must be sp5 and the interorbital angle is 101.5° ... 

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