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1 Chapter 9 Orbitals and Covalent Bond. 2 Atomic Orbitals Dont Work n to explain molecular geometry. n In methane, CH 4, the shape is tetrahedral. n The.

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Presentation on theme: "1 Chapter 9 Orbitals and Covalent Bond. 2 Atomic Orbitals Dont Work n to explain molecular geometry. n In methane, CH 4, the shape is tetrahedral. n The."— Presentation transcript:

1 1 Chapter 9 Orbitals and Covalent Bond

2 2 Atomic Orbitals Dont Work n to explain molecular geometry. n In methane, CH 4, the shape is tetrahedral. n The valence electrons of carbon should be two in s, and two in p. n The p orbitals would have to be at right angles. n The atomic orbitals change when making a molecule.

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

4 4 In terms of energy Energy 2p 2s Hybridizationsp 3

5 5 How we get to hybridization - CH 4 n We know the geometry from experiment. Four bonds of equal length and strength. n We know the orbitals of the central atom. Hybridizing atomic orbitals can explain the geometry. n So if the geometry requires a tetrahedral shape, it is sp 3 hybridized. n This includes bent and trigonal pyramidal molecules because one of the sp 3 lobes holds the lone pair.

6 6 (YDVD)

7 7 sp 2 hybridization n C 2 H 4 Trigonal planar. 120° n Double bond acts as one pair. This results in 3 effective pairs surrounding the carbon atoms. n One s and two p orbitals hybridize into 3 identical orbitals of equal length and energy to make sp 2 orbitals. n This leaves one p orbital unhybridized.

8 8 In terms of energy Energy 2p 2s sp 2 Hybridization 2p

9 9 Two types of Bonds Sigma bonds ( ) form from the overlap of orbitals along the internclear axis. Sigma bonds ( ) form from the overlap of orbitals along the internclear axis. Pi bond ( ) occupies the space above and below internclear axis. Pi bond ( ) occupies the space above and below internclear axis. n Between adjacent unhybridized p orbitals. The double bond always consists of one bond and one bond. The double bond always consists of one bond and one bond. n C-C double bond (BDVD)

10 10 and bonds and bonds (YDVD)

11 11 sp hybridization n CO 2 n Each carbon has two hybridized orbitals 180º apart. Also 2 unhybridized p orbitals. n p orbitals are at right angles (Fig. 9.17) n Makes room for two p bonds and two sigma bonds.

12 12 In terms of energy Energy 2p 2s Hybridization sp 2p

13 13 (YDVD)

14 14 CO 2 n C can make two s and two p n O can make one s and one p n (Fig. 9.19) COO

15 15 dsp 3 n PCl 5 Five pairs of electrons around the central atom. Trigonal bypyramidal. Only bonds no bonds. Five pairs of electrons around the central atom. Trigonal bypyramidal. Only bonds no bonds. n The model predicts that we must use the d orbitals. n Five electron pairs require dsp 3 hybridization. (Fig. 9.21) n There is some controversy about how involved the d orbitals are.

16 16 d 2 sp 3 n SF 6 n Six pairs of electrons around the central atom. n Octahedral shape. (Fig. 9.23)

17 17 How do we figure this out? Use the Localized Electron Model. Use the Localized Electron Model. n Draw the Lewis structure(s). n Determine the arrangement of electron pairs (VSEPR model). n Specify the necessary hybrid orbitals based upon the pairs of electrons around the central atom.


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