Chemistry 102 Summary July 2 nd
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1 Chemistry 102 Summary July 2 nd - So far in our discussion of molecules we have focused on the arrangement of valence electrons. - Today we are going to concentrate on the atomic orbitals that hold these valence electrons and more specifically what happens to these atomic orbitals when bonds form. Example: H2 Question: Why does bonding represent a lower energy state? Example: F2 1s 2 2s 2 2p 5 p z p z! bond - Electron density is on the internuclear axis (designated as z). - End-end or head to head overlap of orbitals, ex. pz+pz or 1s +1s as in H2. - Sigma (σ) bonds can be rotated since orbitals still overlap when atoms rotate. Example: O2 1s 2 2s 2 2p 4 - We have two bonds to account for. - Obvious choice of orbitals is the two 2p atomic orbitals with unpaired electrons. - Assume the z axis is the internuclear axis, then the 2pz orbitals overlap to form the sigma bond. - The second bond is formed from side to side overlap of the 2px orbitals.
2 z-axis p x p x! bond - Electron density in a π (pi) bond is above and below the internuclear axis. - Pi bonds are formed through side to side overlap of parallel p atomic orbitals (px + px or py + py). - Pi bonds cannot be rotated without breaking the bond. - All multiple bonds contain at least one pi bond. - A single bond consists of one sigma bond. - A double bond consists of one sigma bond and one pi bond (px+px or py+py). - A triple bond consists of one sigma bond and two pi bonds (px+px and py+py). Example: N2 Bonding in more complicated molecules: Example: CH4 -Simple overlap of atomic orbitals can t explain the degree bond angles of CH4. - if s,p and d orbitals overlapped to form all bonds then all bond angles would be ninety and 180 degrees and this is not the case! Hybrid Orbitals used to explain the geometry seen in polyatomic molecules. This is a human construct which mixes atomic orbitals to form hybrid orbitals. - For the central carbon atom in CH4 we combine the carbon 2 s orbital and three carbon 2 p orbitals to result in four sp 3 hybrid orbitals.
3 - These sp 3 hybrid orbitals are at bond angles of degrees and overlap with the hydrogen 1s orbitals to form four equivalent C-H bonds in CH4.. - Figures 9.3 and After overlap the electrons in the bonds are on average localized between the two nuclei. Examples: (i) NH3 (ii) H2O (iii) C2H4 Question: What do all these molecules have in common? Example: C2H4 Question: What is the geometry around each carbon? - Molecules which exhibit trigonal planar geometry exhibit sp 2 hybridization. - In sp 2 hybridization, the 2s orbital is combined with two of the three 2 p atomic orbitals to form 3 equivalent sp 2 hybrid oribitals, one p orbital is left unhybridized. Example: C2H2 Question: What is the geometry around each carbon? - In linear geometry we mix one s orbital with one p orbital to form two sp hybrid orbitals and leave two p orbitals unhybridized. Examples: (i) CO2 (ii) O2 (iii) N2
4 Requisite Skills - Know how to predict hybridization around central atoms. - Know difference between sigma and pi bonds. - Be able to count the number of sigma and pi bonds. - Be able to predict which orbitals overlap to form sigma and pi bonds. Example: Fill in the following table and also list the sigma and pi bonds and the orbitals that overlap to form them. O H H C 1 C 2 C 3 N H Atom Geometry Bond Angles Hybrid Orbitals C1 C2 C3 N O Example: Is allene (C3H4) a planar molecule?
5 Example: PCl5 Question: What is the geometry about phosphorous? - In trigonal bipyramidal geometry dsp 3 hybrids are employed they are formed through mixing one s orbital + one d orbital + three p orbitals to give five dsp 3 hybrid orbtials. Example: SF6 Question: What is the geometry about sulfur? - In octahedral geometry d 2 sp 3 hybrid orbitals are employed they are formed through mixing one s orbital + two d orbitals + three p orbitals to give six d 2 sp 3 hybrid oribtials. Delocalization Example: NO2 - - Experiment tells us that NO bonds are equal in length and strength. - We explain equal bond lengths in molecules that exhibit resonance by saying the pi electrons are delocalized over the entire surface of the molecule/ion. - One unhybridized p orbital from nitrogen and one from each of the two oxygen atoms all overlap together to give a pi bond delocalized above and below the entire plane of the NO2 - ion. - There is an extra stability associated with the greater number of resonance structures that can be drawn for a molecule because electron-electron repulsions are minimized. Example: Benzene (C6H6)
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