Orbitals and Covalent Bond Chapter 9 Orbitals and Covalent Bond

Atomic Orbitals Don’t Work to explain molecular geometry. In methane, CH4, the shape is tetrahedral. The valence electrons of carbon should be two in s, and two in p. The p orbitals would have to be at right angles. The atomic orbitals change when making a molecule.

9.1 Hybridization We blend the s and p orbitals of the valence electrons and end up with the tetrahedral geometry. We combine one s orbital and 3 p orbitals. The atoms are responding as needed to give the minimum energy for the molecule. sp3 hybridization has tetrahedral geometry.

In terms of energy 2p Hybridization sp3 Energy 2s

How we get to hybridization - CH4 We know the geometry from experiment. Four bonds of equal length and strength. We know the orbitals of the central atom. Hybridizing atomic orbitals can explain the geometry. So if the geometry requires a tetrahedral shape, it is sp3 hybridized. This includes bent and trigonal pyramidal molecules because one of the sp3 lobes holds the lone pair.

(YDVD)

sp2 hybridization C2H4 Trigonal planar. 120° Double bond acts as one pair. This results in 3 effective pairs surrounding the carbon atoms. One s and two p orbitals hybridize into 3 identical orbitals of equal length and energy to make sp2 orbitals. This leaves one p orbital unhybridized.

In terms of energy 2p 2p Hybridization sp2 Energy 2s

Two types of Bonds Sigma bonds () form from the overlap of orbitals along the internclear axis. Pi bond () occupies the space above and below internclear axis. Between adjacent unhybridized p orbitals. The double bond always consists of one  bond and one  bond. C-C double bond (BDVD)

 and  bonds (YDVD)

sp hybridization CO2 Each carbon has two hybridized orbitals 180º apart. Also 2 unhybridized p orbitals. p orbitals are at right angles (Fig. 9.17) Makes room for two p bonds and two sigma bonds.

In terms of energy 2p 2p sp Hybridization Energy 2s

(YDVD)

CO2 O C O C can make two s and two p O can make one s and one p (Fig. 9.19) O C O

dsp3 PCl5 Five pairs of electrons around the central atom. Trigonal bypyramidal. Only  bonds no  bonds. The model predicts that we must use the d orbitals. Five electron pairs require dsp3 hybridization. (Fig. 9.21) There is some controversy about how involved the d orbitals are.

d2sp3 SF6 Six pairs of electrons around the central atom. Octahedral shape. (Fig. 9.23)

How do we figure this out? Use the Localized Electron Model. Draw the Lewis structure(s). Determine the arrangement of electron pairs (VSEPR model). Specify the necessary hybrid orbitals based upon the pairs of electrons around the central atom.