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Review for Exam Three Principles of Chemistry 1 Section 38 Dr K VSEPR GEOMETRY REVIEW Geometry is dictated by the VSEPR Rules 1 2 Molecule must be arranged so that the repulsion of electrons is minimal Forces between electron pairs follow these orders Lone pair Lone pair Repulsion a b Lone pair Bonded pair Repulsion 1 c Bonded pair Bonded pair Repulsion 3 If there is more than one central atom the geometry of each central atom must be determined separately Example 1 IF 2 First draw the Lewis structure Next determine the SN SN of lone pairs on central atom of atoms around central atom SN 3 2 5 Now determine the shape SN 5 Trigonal bipyramidal However around the central atom I there are 3 lone pairs not bonds so there is a larger repulsion resulting in an alternate SN 5 form In the row the molecular geometry for a SN 5 structure with 3 lone pairs is called Linear The electron geometry would still be Trigonal bipyramidal because electron geometry is asking the electron pair number around the central atom which is 5 Trigonal bipyramidal However the molecular geometry is asking for the molecule geometry which accounts for the increased repulsion of lone pairs to lone pairs and bonded pairs which changes the angles slightly So although still SN 5 the angles are different and the geometry is linear for the molecular geometry 2 VALENCE BOND THEORY We know the geometry of molecules but why do they behave the way they do Some geometries and bond angles and even bond capacities don t line up with the electron configuration of the central atom This theory explains why the theoretical geometry bond angle or capacity don t match the actual The theory basically states that because COVALENT bonds share electrons they are actually sharing their partially filled VALENCE electron orbitals by overlapping them Hybridization when atomic orbitals mix and form new identical hybrid orbitals Only occurs between valence electron orbitals Leads to a better overlap electron electron resistance is minimized Creates an orientation that is consistent with the actual shape of the molecule STEPS Promote valence electron from the ground state configuration to a higher energy orbital 4 or less valence electrons have their own orbital Lone pairs kept in single orbital Hybridize the appropriate valence electron orbitals to make the correct of hybrid orbitals for the appropriate valence electron geometry Must have same number of orbitals before and after hybridization Example 1 CH 4 SN 4 3 Geometry Tetrahedral and therefore has a bond angle of 109 5 Carbon Bonding Capacity 4 However looking at the electron configuration carbon only needs to share 2 electrons to be stable so why does it bond 4 times To explain this carbon hybridizes its valence orbitals which in this case is 2s and 2p So now the s and p orbitals mix and create sp3 Now carbon must form 4 bonds to be stable which explains why it bonds 4 times not 2 SN 4 of lone pair on central atom 2 of atoms around the central atom 2 Electron Geometry Tetrahedral Molecular Geometry Bent Oxygen Bonding Capacity 2 Electron Configuration Example 2 H 2 O Hybridized 4 The orbitals are combined to be 2sp 3 Now how do we know it s a 2sp or sp or etc orbital combined What does this mean Well the superscripts in the 2sp 3 don t indicate the amount of electrons for that orbital but rather how many orbitals there are S only has one orbital so it will be s 1 P has three orbitals if all three are filled it will be p 3 If it had only 2 p orbitals it would be p 2 They need to add up to the steric number Combined orbitals can be predicted based on steric number SN So if we have a tetrahedral like CH4 we know carbon has a SN of 4 so the hybrid orbital will be sp 3 What about BF 3 SN 3 Hybrid Orbitals sp 2 So now that we have hybridized our orbitals for the atoms how do the lewis diagram bonds match up with the geometry SIGMA PI BONDS There are two types of bonds 1 2 Sigma bonds Pi Bonds Sigma Bonds are bonds formed head to head This indicates that the overlap of the shared electrons in these covalent bonds are on the head of each orbital Here we see that the heads or top of the orbitals are overlapping horizontally creating a sigma bond 5 Also because of the orientation of this bond there are no rotation restrictions with sigma bonds Because sigma bonds are horizontal and have more direct overlap they are stronger than Pi Bonds which orient side to side Pi Bonds are formed with parallel orbitals in a side to side orientation The blue area is the overlap of the green orbitals which are parallel to each other Notice the decreased overlap and the fact that there are 2 bonds Pi Bonds for this reason are weaker than sigma bonds Also if you look the pi bonds are parallel which would create a limited bond rotation unless the pi bonds were broken Now how do we know which is which All single bonds and hybridized orbitals form sigma bonds Double Bonds and triple bonds form pi bonds But if pi bonds are weaker why are double and triple bonds stronger than sigma bonds Because double and triple bonds have one sigma bond and all other bonds are pi bonds Triple Bond Double Bond Single Bond Example 1 Acetylene 6 C SN 2 which means a sp hybridization refer to table in hybrid section In this case both carbons are sp but not always Here we see that when a molecule has an sp hybridized orbital there will be two unhybridized ones To explain When the s and p orbitals combine they will create sigma bonds The unhybridized creates pi bonds Since carbon has a triple bond this orientation with 2 sp and 2 p orbitals allows the creation of the triple bond sp 3 has 4 hybridized orbitals s 1 and there are 3 p orbitals so 4 and therefore has no unhybridized Sp 2 have 3 orbitals and therefore has one unhybridized and so on If you look at the electron configuration of s and p s will always have 1 orbital and p will always have 3 So these orbitals should add up to four whether hybridized or unhybridized The Bond and geometry of this molecule looks like this H and C are single bonds so they form sigma bonds C and C have a triple bond So they form a sigma bond on their sp orbitals and then form pi bonds with the unhybridized p orbitals Why does it look like 4 Because the pi bonds are formed on top and bottom of p orbitals That is one bond 7 Now that we know the basics of pi and sigma bonds let s simplify this Example 2 How many sigma and pi bonds are in this molecule 12 single bonds 1 per double triple bond Sigma …

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KSU CHEM 1211 - Exam 3 Chemistry

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