: Organic and Biological Chemistry Tutorial Answers for gjr s Section Sheet 1
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1 : rganic and Biological hemistry Tutorial Answers for gjr s ection heet 1 Question 1. Draw the Lewis structures (dot & cross diagrams) for the following UR molecules: l 2, Pl 3, 3 (Me) 2 & 3 l 2 l l l l traight in with a nasty one! Both oygen and sulfur are group 16 so 6 valence electrons & chlorine is group 17 so 7 valence electrons. or oygen to complete its octet it needs to form a double bond to the sulfur. f we then form single bonds from sulfur to both the chlorine atoms we complete the chlorine s octet. At this point we still have 2 of sulfur s electrons left, so it must have a lone pair. But hear you cry that makes it have 10 electrons! Yup, it is in the third row so can epand the octet and happily have more Alternatively you could have drawn the Lewis structure as: l l l l This is equally acceptable, all atoms obey the octet rule but you MUT have the formal charges! h yeah, notice that thionyl chloride is tetrahedral and not trigonal planar. Probably a molecule that contains all the tricks in the lecturer s repertoire! Pl 3 l P l l P ll l l P l l P ll Eactly the same ecept phosphorus is group 15 so 5 valence electrons and hence it requires the etra chlorine atom. Just to make sure continue to confuse you, whilst an organic chemist will happily draw either of the resonance forms of l 2 we never draw the single bond resonance l 1
2 structure of phosphine oides. The reason for this is that phosphorus is considered to be etremely oaphilic and will form very strong bonds to oygen. 3 (Me) 2 opefully that was an easy one. 3 Question 2. Draw the Lewis structures (dot & cross diagrams) for the following UR molecules: 3, 5, 3 2, [ 4 ] The difficult one! 1. ount the number of valence electrons. = 7 & = 7. Total = 7 + (5 7) = Draw fluorines around central iodine 2
3 now add two electrons to bond each fluorine to the iodine 3. ow fill all the outer atoms (fluorine) so that they obey the octet rule 4. ave only used 40 of 42 electrons so have one lone pair. Place this on the central atom 3 2 [ 4 ] 2 Another hard one! Remember to add 2 etra electrons for the 2 charge! The formal charge on each oygen is 1 & the formal charge on sulfur is +2. o overall the compound s charge is (4 1) + +2 = 2. y y Question 3. 3
4 Give the formal charge (if one eists) on each atom of the following: B ulfur: atomic number = 16 so it has 16 protons (16+). t has a full 1s orbital (2-), a full 2s (2-), 3 full 2p orbitals (6-). Then we look at the valence shell; 1 lone pair (2-) & 3 single bonds (so 1/2 of 6 electrons = 3-). Add all the electrons (or negative charges) together ( )=15-. The difference between 16+ & 15- is 1+ so sulfur has a formal charge of +1. ygen: atomic number 8 so 8+. as full 1s (2-), then 3 lone pairs (6-) and one bond (1/2 2 = 1-). Total negative charge is 9- and the difference between 8+ & 9- is 1-. Therefore the oygen has a formal charge of 1. verall the molecule is neutral oygen = 8+ 1s = 2-1lp = 2-3bonds = 1/2 6- = 3- verall = oygen = s = 2-3lp = 6-1bond = 1/2 2- = 1-3 verall = 1-3 nitrogen = s = lp = 0-3 4bonds = 1/2 8- = verall = 1+ 4
5 boron = 5+ B 1s = 2-0 lp = 0-4bonds = 1/2 8- = 4- B verall = 1- Question 4. Draw the molecular orbitals, label the orbitals and give the bond angles for the following: 3, 3, 3 +, ( 3 ) 2 = 2 3 sp 3 lone pair sp 3, 109.5! (sp 3 +sp 3 ) sp 3, sp 3 lone pair sp 3, sp 2, 120 empty p orbital ( 3 ) 2 = 2 5
6 ! (sp 3 +sp 2 ) sp 2, 120 sp 3, 109.5! (sp 2 ()+sp 2 ()) sp 2 lone pair sp 3 lone pair sp 3, " (p z +p z ) sp 2, 120! (sp 3 +sp 2 ) Question 5. Draw the molecular orbitals, label the orbitals and give the bond angles for the following: the molecular orbitals, label the orbitals and give the bond angles for the following. sp 3, sp 2, 120! (sp 3 +sp 2 ) asty one! Will accept:! (sp 3 +sp 3 ) sp 3, 109.5! (sp 2 ()+sp 2 ()) sp 3 " (p z +p z ) sp 2 lone pairs sp 2, 120 6
7 ! (sp 3 +sp 3 )! (sp 2 ()+s () ) sp 2, 120! (sp 2 ()+sp 2 ()) sp 2 lone pairs sp 3, sp 2, 120 sp 3, " (p z +p z ) But reality closer to:!* p lone pair R 1 R R sp 2 et two are also a little tricky p & " overlap (parallel) sp 3, " (p z +p z )! (sp 3 +sp 2 ) and sp 3, sp 2, 120! (sp 2 ()+sp 2 ()) empty p orbital 90 to bonds sp 2, 120 sp 3, sp 2, 120! (sp 2 ()+sp 2 ()) p & " overlap (parallel) " (p z +p z ) full p orbital 90 to bonds sp 3, 109.5! (sp 3 +sp 2 ) sp 2, 120 7
8 sp 2, 120! (sp 2 ()+sp 2 ()) " (p z() +p z() ) sp 2 lone pair! (sp 2 ()+sp 2 ())! (sp 2 ()+1s () ) " (p z() +p z() ) Answers are available at: 8
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