CHM 105 & 106 MO1 UNIT THREE, LECTURE TWELVE 1 IN OUR PREVIOUS MEETING WE WERE TALKING ABOUT DRAWING LEWIS STRUCTURES. WE

Transcription

1 CHM 105 & 106 MO1 UNIT THREE, LECTURE TWELVE 1 CHM 105/106 Program 29: Unit 3 Lecture 12 IN OUR PREVIOUS MEETING WE WERE TALKING ABOUT DRAWING LEWIS STRUCTURES. WE LOOKED AT THE SERIES OF RULES FOR THE PROCESS OF DRAWING LEWIS STRUCTURES AND THE PURPOSE OF DRAWING THE LEWIS STRUCTURES WAS TO ALLOW US THEN TO PREDICT THE GEOMETRIC SHAPE OF EITHER THE MOLECULE OR THE ION SPECIES THAT WE MIGHT BE WORKING WITH. THE LAST ONE THAT WE LOOKED AT WAS THE NITRITE ION AS IT IS CALLED, NO2, AND WE DID THAT RIGHT AT THE END OF THE PERIOD AND KIND OF HURRIEDLY. SO LET ME GO AHEAD AND AT LEAST PUT HIS BACK TOGETHER WITH THE WAY IT LOOKED AND THEN TALK A LITTLE BIT MORE ABOUT THE GEOMETRIC SHAPE. SO WE ENDED UP WHEN WE DID THE LEWIS STRUCTURE HERE WITH A SPECIES THAT HAD TO HAVE NOW A DOUBLE BOND IN IT AND THE OXYGENS THEN WOULD HAVE ENDED UP EACH WITH THESE EXTRA PAIRS AND THE NITROGEN WOULD HAVE ONE PAIR ON IT AND THAT WAS THE LEWIS STRUCTURE, EXCEPT THAT I SHOULD INDICATE HERE THAT IT IS A ION, A MINUS ION THEN THAT WE RE DEALING WITH. NOW AS FAR AS THE GEOMETRY WAS CONCERNED AND SOMEBODY ASKED RIGHT AFTER CLASS HOW I CAME UP WITH THE GEOMETRY OF A BENT MOLECULE. THEY SAID SHOULDN T IT HAVE BEEN TRIANGULAR BECAUSE WE HAD THREE ELECTRON PAIRS AROUND THE CENTRAL PART. WELL LET S LOOK AT THIS AGAIN. WE DO HAVE THREE EFFECTIVE ELECTRON PAIRS AROUND THE CENTRAL GROUP. ONE, TWO, AND THE DOUBLE BOND COUNTS AS ONLY ONE AS FAR AS GEOMETRY, BUT WE DO HAVE THREE. SO IF WE WERE GOING TO DRAW THIS UP WE WOULD DRAW NITROGEN AND THEN A PAIR OF ELECTRONS, BUT THOSE ARE THE NON-BONDING ELECTRONS. WE VE MENTIONED THOSE BEFORE. NONBONDING ELECTRONS. THEN WE WOULD HAVE THIS DOUBLE BOND OXYGEN AND WE WOULD HAVE A SINGLE BOND OXYGEN. SO THE THREE GROUPS, ONE, TWO, THREE, ARE IN FACT INA TRIANGULAR SHAPE. HOWEVER, WE CANNOT SEE THE NON-BONDING ELECTRONS, AND IF WE TAKE THAT PART OF IT AWAY THEN OF COURSE WE RE ONLY LEFT WITH THE MOLECULE THAT JUST LOOKS, OR AN ION IN THIS CASE, THAT LOOKS BENT. YOU CAN PROBABLY DEMONSTRATE THAT USING THIS MODEL HERE AS WELL. WE HAVE FOUR ELECTRON PAIRS ESSENTIALLY HERE AROUND THE NITROGEN. IT HAS

2 CHM 105 & 106 MO1 UNIT THREE, LECTURE TWELVE 2 IT S OCTET SO WE HAVE PUT IT IN FORMS OF THE TETRAHEDRAL. NOW TO ONE OF THESE PAIRS OF ELECTRONS WE RE GOING TO PUT AN OXYGEN ATOM. SO WE LL PUT AN OXYGEN ATOM ON THERE. TO TWO PAIRS NOW WE RE GOING TO PUT ONE OXYGEN ON TWO PAIRS. SO I WILL TAKE SPRINGS HERE SO THAT I CAN HAVE SOME FLEXIBILITY, AND I WILL ATTEMPT TO PUT IT TOGETHER. SO NOW WE HAVE THE OXYGEN WITH THE DOUBLE BONDS OF THE NITROGEN AND THE OXYGEN WITH THE SINGLE BONDS OF THE NITROGEN AND THEN THE PAIR OF NON- BONDING ELECTRONS. WELL IF I CAN T SEE THAT PAIR OF NON-BONDING ELECTRONS NOW YOU SEE THEN THE SHAPE OF THE ION IS AS WE WOULD INDICATE AN ANGULAR OR A BENT SHAPE. ALRIGHT, SO THAT S THE WAY WE CAME UP WITH THE GEOMETRY FOR THAT PARTICULAR SPECIES. LET S GO ON AND TAKE A LOOK AT ONE MORE HERE. WE LL DO THE LEWIS STRUCTURE FIRST AND THEN PREDICT THE GEOMETRIC SHAPE. THE MOLECULE CH2O. NOW AGAIN, THE STEPS OF LEWIS STRUCTURE, WE ADD UP ALL THE VALENCE ELECTRONS, SO WE HAVE FOUR FOR THE CARBON, TWO FOR THE TWO HYDROGENS, SIX FOR THE OXYGEN AND WE END UP THEN WITH TWELVE, OR SIX ELECTRON PAIRS THAT WE HAVE TO WORK WITH. NOW THE FIRST THING WE DO THEN IS WE ATTACH EACH SPECIES TO THE CENTRAL ATOM WITH A SINGLE BOND. THE CENTRAL ATOM IS THE CARBON HERE SO WE LL HAVE CARBON, AN OXYGEN, A HYDROGEN, AND A HYDROGEN. NOW WE VE USED UP THREE ELECTRON PAIRS TO MAKE OUR SINGLE BONDS. SO WE HAVE THREE ELECTRON PAIRS LEFT. THE QUESTION IS HOW MANY ELECTRON PAIRS WOULD WE NEED NOW TO GIVE EVERYTHING A NOBLE GAS-LIKE STRUCTURE? WELL WE WOULD NEED THREE MORE PAIRS ON THIS OXYGEN BECAUSE IT ONLY HAS ONE. THAT WOULD GIVE IT AN OCTET AND WE D NEED ONE MORE PAIR ON THIS CARBON THAT WOULD GIVE IT AN OCTET. SO WE NEED A TOTAL OF FOUR, BUT WE HAVE ONLY THREE, AND OF COURSE THE RULE IS THAT FOR EACH ELECTRON PAIR THAT WE ARE SHORT, IF POSSIBLE, WE WILL PUT IN A MULTIPLE BOND. BY THAT I MEAN THAT THERE ARE ONLY THOSE SIX OR SEVEN ELEMENTS THAT CAN FORM MULTIPLE BONDS. ONCE AGAIN, CARBON, NITROGEN, OXYGEN, AND THREE ELEMENTS UNDER THOSE SILICON, PHOSPHORUS, SULFUR, AND UNDER SULFUR SELENIUM TO A SMALL EXTENT, ARE THE SIX OR SEVEN ELEMENTS THAT CAN FORM MULTIPLE BONDS. WELL WE SEE BOTH CARBON AND OXYGEN ARE ONE OF THOSE

3 CHM 105 & 106 MO1 UNIT THREE, LECTURE TWELVE 3 SIX, SO IN FACT WE CAN FORMA MULTIPLE BOND. WE NEEDED FOUR, WE HAD ONLY THREE. WE RE A BOND SHORT, WE NEED ONE MULTIPLE BOND. WELL THE ONLY PLACE WE CAN PUT IT WOULD BE HERE, AND THEN OF COURSE THAT CARBON NOW HAS AN OCTET SO WE WOULD GIVE THIS OXYGEN TWO MORE PAIR, ADDING UP A TOTAL NUMBER OF ELECTRON PAIRS NOW ONE, TWO, THREE, FOUR, FIVE, SIX. WE STARTED WITH SIX AND SO WE HAVE THE CORRECT NUMBER OF ELECTRONS INDICATED. NOW WHAT ABOUT THE GEOMETRIC SHAPE FOR THIS SPECIES? WELL AGAIN THEN WE COUNT THE NUMBER OF ELECTRON PAIRS AROUND THE CENTRAL ATOM. MULTIPLE BONDS COUNT THE SAME AS SINGLES SO WE HAVE ONE, TWO, THREE. SO THAT TELLS US THAT THIS HERE SHOULD IN FACT BE TRIANGULAR IN GEOMETRY. IT SHOULD BE TRIANGULAR AND IT SHOULD BE THEN PLANAR AS WELL. SO IN OTHER WORDS, WE WOULD HAVE CARBON DOUBLE BOND OXYGEN, 120 DEGREES FROM THAT WOULD BE HYDROGEN AND 120 DEGREES WOULD BE THE OTHER HYDROGEN. WE CAN GO BACK TO OUR SAME SPECIES THAT WE STARTED WITH A MOMENT AGO, THE ONLY DIFFERENCE IS HERE NOW, WE RE GOING TO HAVE A COUPLE OF HYDROGENS THAT WE PUT ON. SO WE START WITH CARBON, FOUR ELECTRON PAIRS AROUND IT AS WE VE SHOWN. WE HAVE THE OXYGEN WHICH IS GOING TO FORM THE DOUBLE BOND WITH THE CARBON. SO NOW WE VE GOT THE OXYGEN DOUBLE BONDED TOT HE CARBON AND NOW WE RE GOING TO PUT ON TWO HYDROGENS AND SO IN FACT WE SEE THEN THAT THE MOLECULE IS TRIANGULAR. AS A MATTER OF FACT, IT S EQUILATERAL TRIANGULAR AND IT IS PLANAR. IN OTHER WORDS, EVERYTHING IS IN A TWO-DIMENSIONAL RATHER THAN A THREE DIMENSIONAL STRUCTURE. SO WE COULD SHOW THE GEOMETRIC SHAPE THEN OF MOLECULE INVOLVING THE MULTIPLE BONDS OR SINGLE BONDS BY USING THIS METHOD. ALRIGHT, ANY QUESTIONS ON ANY OF THOSE? NOW WE HAVE LOOKED AT, AND WE LL JUST LOOK AT ONE MORE HERE QUICKLY, WE HAVE LOOKED AT THOSE LEWIS STRUCTURES IN WHICH EVERYTHING DIDN T SATISFY ITS OCTET. BCL3 FOR INSTANCE, WE HAD BORON, THREE ELECTRONS, AND FOR CHLORINE THREE TIMES SEVEN, 21. SO WE HAD 24 OR 12 ELECTRON PAIR. AND WE STARTED THEN WE HOOKED EACH CHLORINE TO THE BORON. SO WE USED UP THREE ELECTRON PAIR AND THAT LEFT US WITH NINE ELECTRON PAIR TO WORK WITH. NOW, LOOKING AT THE STRUCTURE THERE WE CAN SEE THAT WE WOULD NEED THREE MORE

4 CHM 105 & 106 MO1 UNIT THREE, LECTURE TWELVE 4 PAIRS ON THIS CHLORINE, THREE ON THIS ONE, THREE ON THIS ONE AND ONE MORE ON THE BORON TO GIVE EVERYTHING AN OCTET. THAT WOULD BE THREE PLUS THREE PLUS THREE PLUS ONE FOR TEN. WE HAVE ONLY NINE, SO WE RE ONE ELECTRON PAIR SHORT, WHICH WOULD PREDICT THAT WE SHOULD HAVE A MULTIPLE BOND, BUT AGAIN, ONLY IF IT S ONE OF THOSE SEVEN ELEMENTS. NEITHER OF THESE ARE ONE OF THOSE SEVEN ELEMENTS SO THEREFORE WE CANNOT FORM A MULTIPLE BOND. THAT MEANS THAT SOMETHING IS NOT GOING TO BE ABLE TO SATISFY ITS OCTET. THE QUESTION IS WHICH ONE OF THESE SPECIES IS THE ONE THAT S NOT GOING TO HAVE ITS OCTET FILLED? WELL THE RULE IS THAT THE ELEMENT FURTHEST TO THE LEFT WILL BE THE ONES THAT WILL BE LEAST LIKELY TO ATTRACT ELECTRONS. REMEMBER ELECTRONEGATIVITY. CHLORINE OVER HERE HAS A HIGHER ELECTRONEGATIVITY THAN DOES BORON. IN OTHER WORDS, IT HAS A GREATER ATTRACTION FOR ELECTRONS AND THEREFORE WE WOULD GIVE THE EIGHT ELECTRONS TO EACH OF THE CHLORINES. WE D SATISFY ITS OCTET AND THE BORON WOULD HAVE LESS ATTRACTION FOR ELECTRONS, WOULD BE LEFT THEN ELECTRON DEFICIENT. SO WE WOULD FINISH THE STRUCTURE THEN BY GIVING EACH OF THE CHLORINES NOW THOSE PAIRS OF ELECTRONS AND WE VE NOW USED UP EVERYTHING AND THAT IS THE CORRECT LEWIS STRUCTURE. GEOMETRICALLY, ONCE AGAIN WE HAVE THREE ELECTRON PAIRS AROUND THE BORON THEN AND OF COURSE THIS MOLECULE THEN WOULD BE TRIANGULAR IN SHAPE. ALRIGHT, THAT S ONE TYPE OF EXCEPTION TO THE OCTET. BUT WE ALSO HAVE ANOTHER TYPE OF EXCEPTION TO THE OCTET AND THAT IS FOR A FEW. FOR A FEW ELEMENTS WE FIND THAT THEY ACTUALLY EXCEED THE OCTET OF ELECTRONS. THIS IS NOT SOMETHING WE D NECESSARILY PREDICT, BUT IT S SOMETHING THAT WE DO FIND. IN OTHER WORDS, IN NATURE THERE IS A MOLECULE WITH A CHEMICAL FORMULA PCL5. NOW NEITHER PHOSPHORUS NOR CHLORINE CAN FORM MULTIPLE BONDS, AND SO IF I M GOING TO PUT THESE TOGETHER, THIS PCL5 NOW, THE ONLY WAY THAT I CAN PUT THAT PCL5 TOGETHER, EXCUSE ME, IS TO ATTACH EACH OF THE CHLORINES NOW TO THE PHOSPHORUS. IF WE ADD UP ALL THE ELECTRONS HERE NOW, PHOSPHORUS IS IN THE FIFTH COLUMN NOW SO WE HAVE FIVE AND THEN CHLORINE OR FIVE TIMES SEVEN WOULD BE 35. SO WE HAVE 40 ELECTRONS OR 20 ELECTRON PAIR. WE VE USED UP FIVE, OKAY, AND THEN WE WOULD NEED 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,

5 CHM 105 & 106 MO1 UNIT THREE, LECTURE TWELVE 5 11, 12, 13, 14, 15 MORE AND OF COURSE THAT S WHAT WE HAVE AND SO EVERYTHING THERE, ALL THE CHLORINES HAVE OCTETS, THEY ALL HAVE FOUR PAIRS OF ELECTRONS BUT NOTICE THE PHOSPHORUS NOW HAS HOW MANY PAIRS OF ELECTRONS AROUND IT? FIVE, FOR A TOTAL OF TEN. SO IT S OCTET HAS BEEN EXCEEDED. AND YES, THIS MOLECULE EXISTS SO THIS IS A WAY WE HAVE TO DRAW THE STRUCTURE THEN TO REPRESENT IT. WHAT ABOUT THE GEOMETRIC SHAPE? WELL WE HAVEN T REALLY SAID ANYTHING ABOUT THE GEOMETRIC SHAPE OF THESE THAT EXCEED THE OCTET. THE SHAPES THAT WE HAD IN THE TABLE OF THE LEWIS STRUCTURES WERE ONLY FOR THOSE THAT WERE LESS THAN OR EQUAL TO AN OCTET OF ELECTRONS. WELL IT TURNS OUT THEN THAT THE GEOMETRIC SHAPE FOR THAT PARTICULAR SPECIES IS LIKE A PYRAMID EXCEPT THAT WE HAVE A DOUBLE PYRAMID. SO IN OTHER WORDS WE HAVE A PYRAMID STACKED ON TOP OF A PYRAMID UPSIDE DOWN, THE TWO BASES TOGETHER. THIS WOULD BE ONE PYRAMID AND THIS WOULD BE THE OTHER, THREE SIDED PYRAMID, AND SO WE SPECIFICALLY REFER TO THIS SHAPE AS BI-PYRAMIDAL. BI-PYRAMIDAL STRUCTURE, A DOUBLE PYRAMID STACKED BASE TO BASE. ALRIGHT, AND WE SEE THAT ON THE EQUATOR I GUESS WE COULD CALL IT, ON THE EQUATOR THE ATOMS ARE 120 DEGREES FROM EACH OTHER. OF COURSE OBVIOUSLY THESE TWO ARE 180 DEGREES FROM EACH OTHER AND THESE ARE 90 DEGREES FROM EACH OTHER. IT S IMPOSSIBLE TO PUT FIVE THINGS ON AND GET THEM ALL EQUAL DISTANT FROM EACH OTHER AND SO THIS IS THE GEOMETRIC SHAPE THAT HAS BEEN DETERMINED BY X-RAY DIFFRACTION WITH THIS TYPE OF STRUCTURE. ALRIGHT, SO WE HAVE THEN A BI-PYRAMIDAL SHAPE. WE ALSO HAVE ONE OTHER ONE. SULFUR HEXAFLUORIDE. MATTER OF FACT, SULFUR HEXAFLUORIDE HAS BEEN USED FOR MANY YEARS AS AN INSULATING MATERIAL, AS A NON-ELECTROLYTE IN TRANSFORMERS. IN TRANSFORMERS WHERE YOU HAVE VERY HIGH ELECTRICAL CHARGES YOU DON T WANT ANY KIND OF INTERNAL ELECTRICITY FLOW OR SPARKING TO OCCUR AND THEY USE THEN A CHEMICAL IN THERE THAT S A NON-ELECTROLYTE TO INSULATE IT, A LIQUID. AND THE LIQUID THAT WAS USED FOR MANY YEARS WAS SULFUR HEXAFLUORIDE. NOW WITH OUR CONCERN ABOUT FLUORINE AND THE OZONE LAYER THEN THIS COMPOUND HAS BEEN REMOVED AS FAR AS BEING AN INSULATOR. AS TRANSFORMERS WORE OUT, AS TRANSFORMERS WERE REPLACED

6 CHM 105 & 106 MO1 UNIT THREE, LECTURE TWELVE 6 THEY WERE THEN REPLACED WITH TRANSFORMERS THAT NO LONGER CONTAIN THIS HEXAFLUORO COMPOUND. BUT AGAIN, IF WE LOOK AT THIS, FLUORINE CAN ONLY FORM SINGLE BONDS. NOW SULFUR CAN FORM MULTIPLE BONDS BUT THAT DOESN T HELP US WITH FLUORINE ONLY BEING ABLE TO FORM SINGLE BONDS. WE HAVE SIX FROM THE SULFUR, SIX TIMES SEVEN, THAT WOULD BE 42 FROM THE FLUORINE AND SO WE HAVE A TOTAL OF 48 ELECTRONS OR 24 ELECTRON PAIRS. WELL IF WE GO AHEAD THEN AND START WITH THE SULFUR A ND WE HOOK IN NOW THE SIX FLUORINES LIKE SUCH. WE HAVE USED UP SIX ELECTRON PAIRS WHICH MEANS WE HAVE 18 ELECTRON PAIRS LEFT 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, AND SO ALL OF THE FLUORINES HAVE THEIR OCTET BUT THE SULFUR HAS EXCEED THE OCTET. IT ACTUALLY HAS NOW 12 ELECTRONS AROUND THAT NUCLEUS. AGAIN, GEOMETRICALLY WE HAVEN T SAID ANYTHING ABOUT THAT SHAPE, BUT THE SHAPE OF IT LOOKS LIKE THIS. SEE THAT WE HAVE A SQUARE SHAPE HERE OR A SQUARE SHAPE THERE. NO MATTER WHICH WAY YOU TURN THIS YOU HAVE A SQUARE SHAPE SO IT S KIND OF LIKE A SQUARE PYRAMID STACKED ON TOP OF A SQUARE PYRAMID. I GUESS WE COULD CALL IT THAT WAY, BUT ACTUALLY THE WAY THEY NAME IT IS BASED ON THE NUMBER OF FACES. WE HAVE 1, 2, 3, 4 FACES AND OF COURSE WE D HAVE 1, 2, 34, FACES BELOW, SO WE HAVE A TOTAL OF 8 FACES ON THAT STRUCTURE AND OF COURSE THE PREFIX FOR 8 IS WHAT? THE PREFIX FOR 8? OH CERTAINLY SOMEBODY KNOWS? PREFIX FOR 8? OCTA. SO THIS IS CALLED OCTA, AND OF COURSE THE WORD THAT WE USE FOR FACE OCTAHEDRAL. SO THAT S AN OCTAHEDRAL STRUCTURE FOR THAT PARTICULAR MOLECULE. WHOOPS, AND THAT WAS A WEAK FLUORINE SULFUR BOND THAT WE SAW THERE THAT BROKE. OKAY. ALRIGHT, SO NEITHER OF THESE ARE WE GOING TO WORRY ABOUT IN TERMS OF THAT WE WILL HAVE TO NECESSARILY NAME THAT STRUCTURE, BUT THE FACT IS THAT THAT IS THE GEOMETRY THAT WE WOULD GET. NOW, IN THE BONDING THAT WE TALKED ABOUT SO FAR WE TALKED ABOUT TWO ATOMS, EACH CONTRIBUTING ONE ELECTRON FROM ONE OF ITS ORBITALS, ONE OF ITS VALENCE ORBITALS GOES TO FORM A BOND. BUT WE CAN ALSO HAVE A TYPE OF COVALENT SHARING PROCESS IN WHICH ONE OF THE SPECIES PROVIDES BOTH ELECTRONS AND THE OTHER PROVIDES NONE, AND THIS IS REFERRED TO AS COORDINATE COVALENT BONDING AND HERE S AN EXAMPLE OF A

7 CHM 105 & 106 MO1 UNIT THREE, LECTURE TWELVE 7 COORDINATE COVALENT BONDING CONDITION. AMMONIA REACTING WITH A HYDROGEN ION TO FORM AMMONIUM. NOW IF WE LOOK AT THE STRUCTURE OF THE AMMONIA HERE IT WOULD LOOK LIKE THIS, WOULD BE A PYRAMIDAL STRUCTURE AND IT HAS A PAIR OF ELECTRONS. HYDROGEN ION IS MERELY AN EM PTY S ORBITAL, I MEAN HYDROGEN ONLY HAS ONE ELECTRON AND IF WE HAVE A HYDROGEN ION IT IS NOTHING MORE THAN A PROTON WITH AN EMPTY S ORBITAL. SO YOU SEE THIS HYDROGEN THEN COULD ACTUALLY SIT DOWN ON TOP OF THE PAIR OF ELECTRONS, ITS EMPTY S ORBITAL, AND FORM THEN A COVALENT BOND AND WE LL END UP WITH A MOLECULE NOW THAT LOOKS LIKE THIS. STILL WE LL HAVE NOW A TETRAHEDRAL SHAPE, EXCEPT THAT IT IS AN ION. NOW, ONCE IT IS FORMED I WOULD NOT BE ABLE TO TELL WHICH ONE OF THOSE FOUR BONDS IN THERE WAS THE ONE THAT WAS THE COORDINATE COVALENT BOND. THEY RE ALL STILL THE SAME BOND ENERGY BUT THE PAIR OF ELECTRONS YOU USE TO FORM THE BOND INITIALLY CAME FROM ONLY ONE SPECIES AND THAT S CALLED COORDINATE COVALENT BONDING. NOW NOTICE THAT THE AMMONIA, THE AMMONIA HAS ACCEPTED A PROTON. HMMM. AND WHAT DID WE DEFINE SOMETHING THAT ACCEPTED A PROTON? CHAPTER FIVE, BETTER KNOW THIS BECAUSE THIS IS GOING TO BE ON ONE OF THE EXAMS. WHAT DID WE DEFINE SOMETHING THAT COULD ACCEPT A PROTON? BRONSTED -LOWRY DEFINITION? WHAT? I HEARD SOMEBODY SAY? AN ACID. NOPE. GOT A 50/50 CHANCE SO ALRIGHT WE LL TAKE THE BASE THEN. SOMETHING THAT CAN ACCEPT A PROTON, REMEMBER BY DEFINITION, AN ACID IA A PROTON DONOR. BRONSTED-LOWRY DEFINITION. A BASE IS A PROTON ACCEPTOR. SO THE AMMONIA HERE HAS ACCEPTED A PROTON TO FORM THIS AMMONIUM ION. NOW LEWIS WHO IS THE ONE OF COURSE THAT WE RE TALKING ABOUT LEWIS STRUCTURES, LEWIS THEN SAID WHAT IS IT THAT S REALLY INVOLVED IN ACID/BASE REACTIONS? IT ISN T THE PROTON, IT S THE ELECTRON. ELECTRONS FORM BONDS. WHY DON T WE DEFINE ACIDS AND BASES THEN IN TERMS OF BONDS, IN TERMS OF ELECTRON PAIRS, AND SO LEWIS THEN SAID WE CAN DEFINE A BASE THEN AS ANYTHING THAT HAS AN ELECTRON PAIR TO DONATE. SO A BASE IS AN ELECTRON PAIR DONOR. IN OTHER WORDS, IT HAS A PAIR OF NON-BONDING ELECTRON THAT IT CAN DONATE OR SHARE TO FORM A CHEMICAL BOND, AND OBVIOUSLY IF WE CALL A BASE AN ELECTRON PAIR DONOR THEN AN

8 CHM 105 & 106 MO1 UNIT THREE, LECTURE TWELVE 8 ACID BECOMES AN ELECTRON PAIR ACCEPTOR. SO LEWIS THEN GAVE US A NEW DEFINITION FOR THE ACID A ND THE BASE, AND ACTUALLY LEWIS DEFINITION OF AN ACID AND A BASE NOW ADD SOME THINGS TO ACID CATEGORIES THAT PREVIOUSLY WERE NOT INCLUDED IN ACIDS BECAUSE BEFORE, ANYTHING THAT WAS AN ACID HAD TO HAVE ESSENTIALLY AN IONIZABLE HYDROGEN, A PROTON TO GIVE AWAY. NOW WE RE SAYING THE ONLY THING YOU HAVE TO DO TO BE AN ACID IS TO BE ABLE TO ACCEPT A PAIR OF ELECTRONS, AND THERE ARE MANY COMPOUNDS THAT FIT INTO THAT CATEGORY. IF YOU GO ON TO TAKE A COURSE IN ORGANIC CHEMISTRY FOR INSTANCE YOU LL PROBABLY TALK QUITE A BIT ABOUT LEWIS ACIDS AND LEWIS BASES. THEY RE VERY WEAK ACIDS, BUT THEY RE INVOLVED IN ORGANIC CHEMICAL REACTIONS QUITE EXTENSIVELY. NOW ONE OF THE PLACES THAT WE RE GOING TO RUN INTO THESE LEWIS BASES THEN ARE IN A GROUP OF ORGANIC COMPOUNDS CALLED HETEROCYCLIC AMINES. NOW WE VE ALREADY TALKED ABOUT THE GROUP AMINES IN THE CHAPTER ON ACID BASE. RECALL THAT AN AMINE HAS A GENERAL FORMULA THAT LOOKS LIKE THAT. IT HAS A NITROGEN IN IT, AND THIS NITROGEN DOES HAVE A PAIR OF ELECTRONS. AS A MATTER OF FACT, IT S THAT PAIR OF ELECTRONS ON THE NITROGEN THAT MAKES THE MOLECULE A BASE. ALRIGHT, THOSE ARE THE ORGANIC BASES. THESE ARE AMINES, THEY RE CALLED CYCLIC BECAUSE OBVIOUSLY THEY RE HOOKED TOGETHER IN A CYCLE AND THEY RE CALLED HETEROCYCLIC BECAUSE IN THE RING STRUCTURE THEY HAVE AN ATOM OTHER THAN A CARBON. NOW IF ALL OF THESE WERE CARBONS, IF WE TALKED ABOUT ONE OF THOSE, WE TALKED ABOUT IT THE OTHER DAY. AS A MATTER OF FACT WE HAD THIS ONE HERE IN WHICH WE HAD THE SIC CARBONS. WE DON T HAVE A LOT OF ROOM TO DRAW HERE. THE SIX CARBONS WITH ALTERNATING DOUBLE BONDS AND REMEMBER WE CALLED THAT BENZENE. BENZENE WAS A CYCLIC HYDROCARBON. IT IS REFERRED TO AS A HOMOCYCLIC BECAUSE ALL OF THE ATOMS IN THE RING ARE THE SAME TYPE. THEY RE ALL CARBONS THAT S A HOMOCYCLIC. HETEROCYCLIC MEANS AT LEAST ONE OF THE ATOMS IN THE RING, IN THE CYCLE, IS NOT A CARBON. THAT S WHAT HETEROCYCLIC MEANS, AND WE SEE EACH OF THESE HAS AT LEAST ONE NITROGEN IN ITS STRUCTURE AND EACH OF THESE NITROGENS HAS A PAIR OF ELECTRONS, NON-BONDING ELECTRONS. THIS PARTICULAR ONE HAS TWO NON-BONDING

9 CHM 105 & 106 MO1 UNIT THREE, LECTURE TWELVE 9 ELECTRONS THAT CAN BE USED AS A BASE, AS AN ELECTRON PAIR DONOR TO SOMETHING ELSE. THESE WILL BECOME VERY IMPORTANT TO US IN CHAPTER 14 AS WE BEGIN TO TALK ABOUT NUCLEIC ACIDS. NUCLEIC ACIDS ALL INVOLVE ONE OF THESE HETEROCYCLIC AMINE STRUCTURES AS PART OF THE OVERALL MOLECULAR STRUCTURE. SO WE LL COME BACK AND LOOK AT THAT, BUT AGAIN THIS IS PAIR OF ELECTRON THAT ALLOWS IT TO BE BASE. AS A MATTER OF FACT, EVERYTHING THAT SERVES AS A BASE, WHETHER IT BE THIS OR AMMONIA OR EVEN JUST THE HYDROXIDE ION, THE REASON THE HYDROXIDE ION SERVES AS BASE, THAT S THE LEWIS STRUCTURE FOR THE HYDROXIDE ION, THE REASON THE HYDROXIDE ION IS A BASE IS IT HAS A PAIR OF ELECTRONS THAT IT CAN DONATE TO A HYDROGEN ION TO MAKE A WATER MOLECULE. AN ACID-BASE REACTION. ALRIGHT, NOW ONE OF THE KEY FUNCTIONS OF DRAWING LEWIS STRUCTURES IS TO BE ABLE TO TALK ABOUT MOLECULAR POLARITY. NOW POLARITY OF MOLECULES, NOT POLARITY OF BONDS. WE VE ALREADY TALKED ABOUT POLARITY OF BONDS. POLARITY OF BONDS MEANS THAT THERE S AN ELECTRONEGATIVITY DIFFERENCE BETWEEN THE TWO ATOMS THAT ARE ATTACHED. THAT S ALL. IF THEY HAVE AN ELECTRONEGATIVITY DIFFERENCE THEY ARE A POLAR BOND. BUT A POLAR MOLECULE THEN INVOLVES NOT ONLY POLAR BONDS BUT ALSO THE GEOMETRY OF THE MOLECULE. NOW TO BE A POLAR MOLECULE, NUMBER ONE, YOU MUST HAVE POLAR BONDS. CAN T BE A POLAR MOLECULE OF YOU DON T HAVE A POLAR BOND. BUT NUMBER TWO IS YOU MUST HAVE THE CORRECT GEOMETRIC SHAPE. AS A MATTER OF FACT, WE SOMETIMES SAY THAT MOLECULAR POLARITY IS THE VECTOR SUM OF ELECTRONEGATIVITY DIFFERENCES, DELTA ( ). THE VECTOR SUM OF THE ELECTRONEGATIVITY DIFFERENCES. IN OTHER WORDS, VECTORS INDICATE DIRECTION AND MAGNITUDE OF ELECTRONEGATIVITY DIFFERENCE. LET S TAKE FOR EXAMPLE WATER HERE AS OUR MOLECULE AND WE VE DOME THE LEWIS STRUCTURE NOW AND WHEN WE DO THE LEWIS STRUCTURE FOR THE WATER OF COURSE WE WOULD HAVE A COUPLE OF NON- BONDING PAIRS THERE. WE DO THE LEWIS STRUCTURE AND THEN THE GEOMETRY WOULD TELL US THAT THIS IS A BENT MOLECULE. AS A MATTER OF FACT THE BOND ANGLES IN HERE ARE ABOUT 109 DEGREES. NOW WHEN I TALK ABOUT THE VECTOR SUM I LL START DOWN HERE THEN NOW. I LL PUT THIS AS THE OXYGEN CENTER. NOW I SEE THAT I HAVE TWO VECTORS.

10 CHM 105 & 106 MO1 UNIT THREE, LECTURE TWELVE 10 OXYGEN HAS AN ELECTRONEGATIVITY, I LL PUT THIS DOWN IN HERE SO WE CAN SEE IT A LITTLE BETTER, 3.5. HYDROGEN 2.1. SO THERE S AN ELECTRONEGATIVITY DIFFERENCE BETWEEN THOSE TWO OF 1.4 AND OXYGEN IS THE ONE THAT IS MORE STRONGLY ATTRACTING SO WE WOULD DRAW THE VECTOR THEN IN THE MOLECULE HERE AS ACTING TOWARD THE OXYGEN. SO THE OXYGEN WOULD BE NEGATIVE, THE HYDROGENS WOULD BE POSITIVE. NOW LET S TALK ABOUT THE TOTAL VECTOR IN MOLECULES HOWEVER. SO IF I START RIGHT HERE THEN AND I M GOING TO TAKE THE FIRST VECTOR WHICH WAS A DIFFERENCE OF 1.4 AND IT WAS ANGLED UP THEN AT A DIRECTION UP IN THIS WAY. SO WE LL CALL EACH BLOCK ABOUT.2 SO WE LL GO UP ABOUT 7 BLOCKS HERE. 1, 2, 3, 4, 5, 6, 7, SOMETHING LIKE THAT. THAT S MY FIRST VECTOR. THAT WAS 1.4 UNITS FROM THE OXYGEN. WE RE SAYING THAT WE VE DISPLACED THE ELECTRON CHARGE THAT MUCH. NOW WE D START THERE AND THE OTHER VECTOR WE D SEE NOW IS GOING TO BE ABOUT 109 DEGREES, SO NOT QUITE 90 BUT ABOUT 109, A LITTLE BIT FURTHER, AND IT S ALSO 1.4, AND SO WE GO UP HERE NOW ABOUT 7 BLOCKS AGAIN AND WE SHOULD END UP ABOUT RIGHT THERE. NOW WHEN WE ADD THESE TWO TOGETHER GRAPHICALLY, THEN WE END UP WITH A NET VECTOR. THIS IS THE VECTOR SUM RIGHT THERE. SO WE RE SAYING THAT WE HAVE A NET VECTOR ACTING IN THE MOLECULE THEN IN THIS DIRECTION THROUGH THE MOLECULE OF A CERTAIN MAGNITUDE. WE DON T CARE WHAT IT LOOKS LIKE IN THIS CASE IT WOULD PROBABLY BE A MAGNITUDE OF ABOUT 2.0 OR SOMETHING LIKE THAT. BUT W HAVE A DEFINITE VECTOR. SO THIS SIDE OF THE MOLECULE IS DEFINITELY NEGATIVE AND THIS SIDE OF THE MOLECULE IS DEFINITELY POSITIVE, AND SO IF I AGAIN I LL JUST DRAW THE WATER OVER HERE. THEN I M SAYING THAT MY VECTOR IS LIKE THAT. THIS IS THE NEGATIVE SIDE, THIS IS THE POSITIVE SIDE AND SO WE DEFINITELY HAVE A POLAR MOLECULE. WE HAVE POLAR BONDS, GEOMETRIC SHAPE ALSO GIVES US THEN A VECTOR THAT DOESN T ADD TO ZERO. IF THE VECTOR ADDS TO ZERO IT WOULD MEAN THAT WE WOULD HAVE A NON-POLAR MOLECULE. IF THE VECTOR ADDS TO ANY NUMERIC VALUE WE HAVE A POLAR MOLECULE. NOW WHAT S THE IMPORTANCE OF POLARITY? WELL ONE OF THE THINGS WHEN WE TALKED IN CHAPTER FOUR ABOUT DISSOLVING IS WE SAID THAT LIKES DISSOLVE LIKES. WHY DOES WATER DISSOLVE ETHYL ALCOHOL AND CARBON TETRACHLORIDE NOT? BECAUSE

11 CHM 105 & 106 MO1 UNIT THREE, LECTURE TWELVE 11 OF THE MOLECULAR POLARITY. IN CHAPTER 10 WE LL TALK MORE EXTENSIVELY ABOUT LIQUIDS AND SOLIDS AND SOME OF THE PHYSICAL PROPERTIES. MELTING POINTS, BOILING POINTS, DENSITY, AND IT ALL WILL TIE BACK INTO THIS IDEA OF MOLECULAR POLARITY, OR AT LEAST THAT WILL BE ONE OF THE MAJOR CONTRIBUTING FORCES THAT WE LL HAVE TO TAKE INTO CONSIDERATION. WELL LET S TRY ANOTHER ONE HERE QUICKLY. JUST A MOMENT AGO WE DREW THE LEWIS STRUCTURE FOR BCL3, AND WE CAME UP WITH THIS FOR ITS GEOMETRIC SHAPE. NOW ELECTRONEGATIVITIES, BORON HAS AN ELECTRONEGATIVITY OF 2.0, AND CHLORINE HAS AN ELECTRONEGATIVITY OF 3.0. SO YES WE DO HAVE POLAR BONDS. WE WOULD HAVE A VECTOR GOING THAT WAY AND THIS WAY AND THIS WAY AND EACH OF THOSE WOULD HAVE AN ELECTRONEGATIVITY DIFFERENCE OF 1.0. WELL LET S TRY TO DEPICT THIS NOW ON THE GRAPH. SO ONCE AGAIN WE LL START TIGHT HERE. THAT S THE CENTER BEFORE WE LOOK AT ANY OF THE VECTORS, AND LET S TAE THEN OUR FIRST VECTOR UP HERE AND WE WILL GO UP AGAIN WE LL USE.2 PER BLOCK SO WE RE GONNA GO UP FIVE BLOCKS. 1, 2, 3, 4, 5. SO WE RE GONNA MAKE OUR FIRST VECTOR UP TO THERE. THEN WE LL TAKE THIS VECTOR 1.0. IT S GOING TO START AT THIS POINT AND IT S GOING TO GO 5 BLOCKS NOW TO THE RIGHT AND DOWN. SO IT S GOING TO GO LIKE THIS. FIVE BLOCKS AND THEN WE RE GOING TO TAKE THIS VECTOR WHICH IS ALSO 1.0 AND IT S GOING TO GO 120 DEGREES IN THIS DIRECTION 5 BLOCKS AND WHERE DID WE END UP? WE END UP WHERE WE STARTED. WHAT S THE NET VECTOR WHEN I ADD ALL THOSE UP? ZERO. I M RIGHT BACK WHERE I STARTED. THERE IS NO CHARGE DISPLA CEMENT IN THAT MOLECULE. THERE IS NO POSITIVE AND NEGATIVE SIDE. AS A MATTER OF FACT, ALL OF THE CHLORINES ARE PARTIALLY NEGATIVE IN THAT STRUCTURE. THERE ISN T A POSITIVE END AND A NEGATIVE END OR POSITIVE SIDE AND A NEGATIVE SIDE. IT S ALL SURROUNDED BY NEGATIVE CHLORINES IN THAT CASE. SO THEREFORE BCL3 IS A NON- POLAR MOLECULE. ALRIGHT, YOU HAVE ANY QUESTIONS RAISE YOUR HANDS OR WE LL JUST KEEP BLASTING FORWARD. CARBON DIOXIDE. IF I DO THE LEWIS STRUCTURE FOR CARBON DIOXIDE WE LL HAVE THIS FOR THE LEWIS STRUCTURE. AND IF WE WERE TO LOOK AT THE CARBON HERE NOW WITH ESSENTIALLY TWO ELECTRON PAIRS AROUND IT, AROUND THE CARBON, WHAT WOULD BE THE GEOMETRIC SHAPE OF CARBON DIOXIDE? LINEAR.

12 CHM 105 & 106 MO1 UNIT THREE, LECTURE TWELVE 12 ABSOLUTELY RIGHT. LINEAR. ALRIGHT, SO THAT MEANS THAT BOTH OXYGENS ARE ON OPPOSITE SIDES. CARBON, 2.5. OXYGEN, 3.5. SO WE HAVE TWO VECTORS, ONE GOING THAT DIRECTION AND ONE GOING THAT DIRECTION AND AT 1.0 MAGNITUDE. AGAIN IF WE TAKE OUR CHART HERE WE START HERE AND THAT WILL TAKE THE FIRST VECTOR UP THERE, 1.0. WE GO OVER THERE FIVE SQUARES, AGAIN USING EACH SQUARE COUNTING AS.2, AND THEN I TAKE THIS VECTOR AND START WHERE I WAS AND I GO THIS DIRECTION, 1.0, WHERE AM I? RIGHT BACK WHERE I WAS. THERE IS ZERO CHARGE DISPLACEMENT IN THE MOLECULE. CARBON DIOXIDE IS A NON-POLAR LINEAR MOLECULE. SO THE LEWIS STRUCTURE GAVE US THE ABILITY TO PREDICT THE GEOMETRIC SHAPE AND THEN KNOWING ELECTRONEGATIVITY DIFFERENCES WE CAN DETERMINE THE ACTUAL POLARITY. NOW, DOES CARBON DIOXIDE DISSOLVE VERY EXTENSIVELY IN WATER? NO. HARDLY AT ALL. MATTER OF FACT, IT S VERY DIFFICULT TO GET CARBON DIOXIDE IN WATER. THAT S WHY WHEN YOU TAKE THE CAP OFF A CARBONATED BEVERAGE ALL THE CARBON DIOXIDE COMES RUSHING OUT. IT DOESN T WANT TO STAY DISSOLVED. THE ONLY WAY WE GET IT IN IS TO PUT IT UNDER HIGH PRESSURE IN THE FIRST PLACE. WHY WOULDN T IT BE SOLUBLE IN WATER? WELL WHAT KIND OF MOLECULES ARE WATER MOLECULES? JUST DREW IT A MOMENT AGO, POLAR. WHAT KIND OF MOLECULE IS CARBON DIOXIDE? NON-POLAR. NON-POLARS DO NOT DISSOLVE IN POLAR SOLVENTS AND SO THEREFORE WE WOULD EXPECT CO2 DOES NOT HAVE VERY MUCH SOLUBILITY. QUESTION? DID YOU HAVE A QUESTION? OKAY. ALRIGHT, THIS IS FORMALDEHYDE. AN ORGANIC COMPOUND, IT S A GROUP, A TYPE OF COMPOUNDS THAT WE HAVEN T TALKED ABOUT AS FA R AS FUNCTIONAL GROUPS BUT THIS IS FORMALDEHYDE AND WE WILL TALK ABOUT THE ALDEHYDES AND KETONES, TWO OTHER ORGANIC GROUPS A LITTLE BIT LATER. IF WE DO THE LEWIS STRUCTURE AGAIN WE COME UP WITH THIS. WE JUST DID THE LEWIS STRUCTURE FOR THIS ONE A LITTLE BIT EARLIER IN THIS LECTURE TODAY, AND WE CAME UP WITH THEN A GEOMETRIC SHAPE THAT THIS WILL BE THEN TRIANGULAR. BECAUSE A DOUBLE BOND COUNTS AS ONE AND THEN THE OTHER TWO SO WE HAVE THREE SO IT S PLANAR AND IT S TRIANGULAR. NOW ELECTRONEGATIVITY WISE, HYDROGENS ARE 2.1. CARBON IS 2.5. OXYGEN IS 3.5. THOSE ARE OUR ELECTRONEGATIVITIES THAT WE RE GOING TO WORK WITH. SO WE HAVE A VECTOR

13 CHM 105 & 106 MO1 UNIT THREE, LECTURE TWELVE 13 ACTING IN THIS DIRECTION WITH A MAGNITUDE OF.4. WE HAVE A VECTOR ACTING IN THAT DIRECTION WITH A MAGNITUDE OF.4, AND THEN WE HAVE A VECTOR ACTING IN THIS DIRECTION WITH A MAGNITUDE OF 1.0. NOW IF WE CAN GO DOWN TO OUR GRAPH PAPER HERE AND WE START HERE IN THE CENTER WE TAKE OUR FIRST VECTOR WHICH WAS THE HYDROGEN CARBON VECTOR,.4, AND SO WE WILL USE WELL LET S SEE THIS TIME I GUESS WE CAN USE, YEAH WE LL USE.2 AGAIN PER BLOCK. SO WE GO OVER ABOUT TWO BLOCKS DOWN HERE IN THIS GENERAL DIRECTION FOR OUR FIRST VECTOR. OUR SECOND VECTOR STARTS WHERE THE FIRST ONE LEFT OFF AND WE RE GOING TO GO ABOUT TWO BLOCKS WHICH WOULD BE.4 UNITS AT 109 DEGREES AND WE RE BACK THERE. AND THEN WE RE GOING TO TAKE OUR LAST VECTOR WHICH IS 1.10 STARTING HERE AND GOING OVER THEN FIVE BLOCKS AND WE RE RIGHT THERE. IF WE ADD ALL THESE TOGETHER TO GET OUR NET VECTOR HERE S WHERE WE ENDED UP, HERE S WHERE WE STARTED, OUR NET VECTOR WOULD BE THAT. SO IS THIS MOLECULE POLAR? WHY DEFINITELY SO. THIS END IS THE NEGATIVE END, THIS END IS THE POSITIVE END. THE POINT OF THE ARROW IS ALWAYS THE END WHERE THE ELECTRONS ARE MOVING TO AND SO IF WE W ERE TO GO BACK UP HERE AND REDRAW THIS QUICKLY LIKE SUCH WE WOULD SAY THIS SIDE IS PARTIALLY POSITIVE AND THIS SIDE IS PARTIALLY NEGATIVE, AND FORMALDEHYDE IS A POLAR MOLECULE AND WOULD I PREDICT THAT FORMALDEHYDE SHOULD DISSOLVE IN WATER? YES. BECAUSE POLAR SOLVENTS LIKE POLAR SOLVENTS. NON-POLAR SOLVENTS LIKE NON- POLAR SOLVENTS. WE TALKED ABOUT THAT IN CHAPTER FOUR A LITTLE BIT. ALRIGHT, LET S LOOK AT PHOSPHORUS AND LET S LOOK AT PH3 FIRST OF ALL. AND WE LL ALSO LOOK AT PCL3. NOW IF WE LOOK AT PH3 IT S GOING TO HAVE THIS SAME GEOMETRIC SHAPE, IT S PYRAMIDAL, THERE S THAT PAIR OF NON-BONDING ELECTRONS AS FAR AS GEOMETRY IS CONCERNED. THEN WE D HAVE PHOSPHORUS WITH A HYDROGEN AND A HYDROGEN AND A HYDROGEN. QUESTION IS THAT POLAR OR NON-POLAR? GETS TO BE A LITTLE BIT DIFFICULT WHEN YOU GO TO THREE DIMENSIONAL MOLECULES BECAUSE NOW YOU HAVE THINGS MOVING INTO AND OUT OF THE TWO DIMENSIONS. SO IT S A LITTLE BIT MORE DIFFICULT, BUT WE CAN MAKE SOME ROUGH ESTIMATES AND WE LL SEE. BUT FIRST OF ALL IN THE CASE OF PH3 WE LOOK UP HYDROGEN 2.1 AND IF WE LOOK AT THE ELECTRONEGATIVITY TABLE WHICH I FORGOT TO BRING ALONG I

14 CHM 105 & 106 MO1 UNIT THREE, LECTURE TWELVE 14 BELIEVE I M CORRECT ON THIS, PHOSPHORUS IS 2.1. WELL I REALLY DON T NEED TO GO ANY FURTHER IN THIS CASE. GEOMETRIC SHAPE OF THIS WOULD LOOK LIKE IF I GO BACK HERE TO THE CENTER THAT ALL OF MY VECTORS ARE GOING TO BE EITHER TOWARD THE PHOSPHORUS OR AWAY FROM THE PHOSPHORUS, SO THEY RE GOING TO BE SOMETHING LIKE GOING DOWN THIS WAY THEN GOING DOWN THIS WAY AND THEN COMING OUT AT US. SO IF THE ATTACHED GROUP WAS MORE ELECTRONEGATIVE THAN THE PHOSPHORUS WE D END UP WITH A NET VECTOR LIKE THAT OR IF THE ATTACHED GROUP WAS LESS ELECTRONEGATIVE THEN THE VECTORS WOULD GO TOWARDS IT AND SO WE WOULD END UP THEN WITH SOMETHING GOING LIKE THAT AND THEN LIKE THAT AND THEN LIKE THIS AND WE WOULD END UP WITH THIS AS OUR NET VECTOR. SO IT SHOULD BE A POLAR MOLECULE IF IN FACT THERE ARE ELECTRONEGATIVITY DIFFERENCES. BUT WITH PHOSPHORUS AND HYDROGEN WE HAVE A NON- POLAR BOND AND SO EVEN THOUGH THE GEOMETRY IS SUCH THAT IT SHOULD BE POLAR IT CAN T BE POLAR IF WE DON T HAVE POLAR BONDS. IN CONTRAST, HERE WE WOULD HAVE 2.1 AND 3.0 AND SO THEREFORE WE DO HAVE AN ELECTRONEGATIVITY DIFFERENCE AND THE GEOMETRY WOULD THEN TELL US AGAIN THAT WE SHOULD HAVE A POLAR MOLECULE. NOW IN GENERAL FOR PYRAMIDAL OR TETRAHEDRAL TYPE STRUCTURES WHICH ARE GOING TO BE THREE DIMENSIONAL ONE CAN BASICALLY USE IT IN A TWO-DIMENSION TO GET A GOOD ESTIMATE OF THE VECTORS. NOT QUITE ACCURATE BUT PRETTY CLOSE. IN OTHER WORDS WE COULD JUST SAY THAT WE WERE LOOKING AT IT LIKE THIS SO THAT WE RE LOOKING AT A TWO- DIMENSIONAL AND WE RE GOING TO HAVE ABOUT THE SAME EFFECT. THIS VECTOR WOULD OFFSET THIS VECTOR AND WE WOULD END UP WITH A NET VECTOR HERE SO THAT, YES, IF THE ELECTRONEGATIVITY WAS DIFFERENT WE WOULD HAVE A POLAR MOLECULE. SO ONE CAN FORCE IT BACK INTO TWO DIMENSIONS TO KIND OF LOOK AT MOLECULAR POLARITY. ALRIGHT, ANY QUESTION ON ANY PART OF THAT ONE? ONE OTHER ONE WOULD INVOLVE TETRAHEDRAL GEOMETRY LIKE AROUND THE CARBON CENTER AND AGAIN AS WE SEE THERE WE WOULD HAVE A THREE-DIMENSIONAL MOLECULE TO WORK WITH, BUT WE COULD AGAIN GIVE A PRETTY GOOD ESTIMATE BY LOOKING AT IT IN TERMS OF A SQUARE. NOW LET S SUPPOSE THAT WE RE LOOKING AT METHANE, CH4. METHANE, CARBON IS 2.5 AND HYDROGEN S 2.1. SO YES,

15 CHM 105 & 106 MO1 UNIT THREE, LECTURE TWELVE 15 THERE WOULD BE AN ELECTRONEGATIVITY DIFFERENCE, YES IT WOULD BE A POLAR BOND, BUT IF WE WENT OUT HERE THEN, NOW LOOKING OUR VECTORS WOULD LOOK LIKE THIS. EACH ONE WOULD BE.4. SO WE WOULD GO OVER TWO BLOCKS, DOWN TWO BLOCKS, OVER TWO BLOCKS AN BACK UP TWO BLOCKS AND WHERE ARE WE? NOWHERE. WE RE RIGHT BACK WHERE WE STARTED AND SO METHANE, CH4 WOULD BE A NON-POLAR MOLECULE. CCL4, CARBON TETRACHLORIDE WOULD BE AN NON-POLAR MOLECULE. BUT WHAT ABOUT CH2CL2? WHAT ABOUT IT? WE WILL GET IT IN TERMS OF SQUARE PLANAR CONDITION, NOW WE LL PUT, DOESN T MAKE ANY DIFFERENCE WHERE WE PUT THESE. WE CAN PUT THEN THAT WAY. UH, YES IT DOES MAKE A DIFFERENCE NOW. SO IF WE DRAW IT THAT WAY OR VERSUS DRAWING IT THIS WAY WE WOULD HAVE TWO DIFFERENT EFFECTS, AND SO WHEN WE GET TO MIXED AROUND THE TETRAHEDRAL WE REALLY DO NEED TO PICTURE IT THREE DIMENSIONALLY TO DETERMINE MOLECULAR POLARITY. WELL IN OUR NEXT LECTURE WE LL TAKE A LOOK AT ISOMERIZATION AND ITS EFFECT ON GEOMETRIC SHAPE AND GO ON AND BEGIN A LITTLE BIT OF CHAPTER 8 ON CHEMICAL NOMENCLATURE.

16 CHM 105 & 106 MO1 UNIT THREE, LECTURE TWELVE 16 SUPPLEMENTAL SEGMENT DEMO OF MOLECULAR POLARITY IN CHAPTER SEVEN AT THE END OF THE CHAPTER WE VE BEEN TALKING ABOUT MOLECULAR SHAPES, MOLECULAR POLARITY AND WE RE GOING TO TAKE A LOOK AT A COUPLE OF MOLECULES AND SEE HOW WE CAN DEMONSTRATE THIS MOLECULAR POLARITY. THE SHAPE OF A WATER MOLECULE IS SUCH IN THE SHAPE OF ETHYL ALCOHOL MOLECULE IS SUCH AND IF WE WERE TO LOOK AT THEIR VECTOR SUMS WE WOULD FIND THAT IN FACT THEY WERE BOTH POLAR MOLECULES. THE MOLECULE WE HAVE HERE IS CARBON TETRACHLORIDE, AND IT WOULD BE TETRAHEDRAL IN GEOMETRIC SHAPE AND THE VECTOR SUM WOULD BE ZERO. IT WOULD BE A NON-POLAR MOLECULE. AND IODINE, WHICH IS A TWO IODINE ATOMS CONNECTED TOGETHER WHICH WILL ALSO BE A NON-POLAR MOLECULE. NOW IF WE TOOK ONE OF THESE TWO LIQUIDS THAT ARE POLAR AND CONSIDERED THEM TO BE LIKE LITTLE PLUS AND MINUS MAGNETS. IF WE WERE TO BRING A CHARGE TO THE SIDE OF A TUBE OF A LIQUID WE WOULD ANTICIPATE THAT THE MOLECULES WOULD ORIENT THEMSELVES IN SUCH A WAY THAT THE OPPOSITE CHARGE WOULD BE ATTRACTED TO THE EXTERNAL CHARGE THAT WE HAVE BROUGHT ALONG AND SO WE RE GOING TO USE THIS PRINCIPLE TO TRY TO DEMONSTRATE THEN HOW WE CAN SHOW THAT SOMETHING IS POLAR OR NON-POLAR IN THE LIQUID STATE. THEN WE LL CONTINUE BY TAKING A LOOK AT THIS PHENOMENA OF POLAR AND NON-POLAR MOLECULES. NOW I HAVE SET UP A DEMONSTRATION HERE AND I HAVE THREE LIQUIDS. THE FIRST BURETTE CONTAINS WATER. WE KNOW THAT IT S WATER BECAUSE IT S BLUE. WE VE GOT ETHYL ALCOHOL OR SOMETIMES BY THE MOONSHINERS CALLED OL RED EYES. SO THAT S WHY IT S COLORED THERE. IT S ETHYL ALCOHOL WITH JUST A DROP OF FOOD COLORING, AND THE LAST SMALL TUBE WE HAVE IS CARBON TETRACHLORIDE WITH A SMALL AMOUNT OF IODINE IN IT TO GIVE IT TH PURPLE COLOR SO THAT WE COULD SEE IT. ALL THREE LIQUIDS THEMSELVES ARE COLORLESS. THIS IS JUST MERELY THAT WE CAN VISUALIZE THEM BETTER. NOW WE NEED THE CHARGES ROD IN ORDER TO CHECK THE POLARITY AND NON-POLARITY AND SO WHAT I HAVE HERE IS A PIECE OF CAP FUR AND JUST A PLASTIC RULER, AND IF WE DO A LITTLE FRICTION HOPEFULLY WE CAN GET A CHARGE ON THIS AND WE CAN CHECK WITH THE

17 CHM 105 & 106 MO1 UNIT THREE, LECTURE TWELVE 17 PIECES OF PAPER AND WE SEE THAT YES WE DO HAVE A CHARGED SPHERE. ALRIGHT, WE NOW AGAIN WILL CHARGE UP OUR ROD HERE. WE RE GOING TO TURN ON A STREAM OF THE WATER. WE CAN SEE THAT IT S FLOWING DOWN STRAIGHT. IT I BRING THE CHARGED SPHERE UP WE SHOULD NOTICE THEN A BENDING OF TH STREAM. I LL TRY TO BRING IT IN FROM THE OTHER SIDE, AND WITH LUCK YOU CAN MAYBE SEE THAT I M NOW PUTTING IT INTO THE LARGE BEAKER BETWEEN THE TWO BURETTES. THIS IS NOT PHYSICAL CONTACT. THERE S NO LIQUID HERE. IT S MERELY THE FACT THAT THE WATER WHICH IS POLAR IS ATTRACTED TO THE CHARGED ROD. NOW WE LL TRY THE ETHYL ALCOHOL. SO WE LL TURN A STREAM OF THE ETHYL ALCOHOL ON. AGAIN WE SEE THAT IT S GOING STRAIGHT INTO THE BEAKER. IF I BRING THE CHARGE UP WE SEE THAT THE STREAM OF ETHYL ALCOHOL OS BEING CURVED AND IF I BRING UP THIS SIDE I HOPEFULLY MAYBE CAN BRING IT INTO THE LARGE BEAKER. IF I CAN GET IT GOING THE RIGHT DIRECTION HERE WELL WE CAN T QUITE GET IT GOING TO THE LARGE BEAKER. NOW WE RE GONNA TEST THE CARBON TETRACHLORIDE. AGAIN WE LL GET A CHARGE ON HERE QUICKLY. WE SEE THAT IT IS CHARGED. I LL NEED TO SWITCH SIDES HERE. I WILL TURN ON THE STREAM OF CARBON TETRACHLORIDE AND AS I BRING UP THE CHARGED ROD WE NOTICE THAT NOTHING IS HAPPENING TO THE STREAM OF CARBON TETRACHLORIDE, INDICATING IT IS NOT ATTRACTED TO THE POSITIVE, OR TO THE NEGATIVE CHARGE. NOW ONE MIGHT ASK MAYBE IT S BECAUSE THIS THING HAS LOST ITS CHARGE. LET S JUST DOUBLE CHECK QUICKLY. WE SEE THAT IS NOT THE CASE, THAT THE ETHYL ALCOHOL IS BENDING PROFUSELY, SO THEREFORE IT IS CHARGES. WE CAN GO BACK QUICKLY AND RE-SHOW THAT THE CARBON TETRACHLORIDE IS NOT AFFECTED, THUS SHOWING THAT IT IS A NON-POLAR MOLECULE. NOW WE RE GOING TO TAKE A LOOK AT THE COMPATIBILITY OF LIQUIDS. IF YOU RECALL IN CHAPTER FOUR ON SOLUTIONS WE TALKED ABOUT THAT LIKE MATERIALS DISSOLVED LIKE MATERIALS. IN OTHER WORDS, POLAR MOLECULES LIKE POLAR MOLECULES. NON-POLAR MOLECULES LIKE NON-POLAR MOLECULES. WE HAVE THE WATER AND THE ETHYL ALCOHOL, BOTH OF WHICH ARE POLAR, AND IF I POUR THEM TOGETHER WE NOTICE THAT WE GET A COLOR MIXTURE, BUT IT IS A PURE SOLUTION. IT IS NOT SEPARATING INTO INDIVIDUAL LAYERS. SO BOTH LIQUIDS ARE TOTALLY MISSABLE IN EACH OTHER. SO WE SEE THAN AGAIN A

18 CHM 105 & 106 MO1 UNIT THREE, LECTURE TWELVE 18 POLAR-LIKE-POLAR. IF WE WERE TO TAKE SOME OF THE CARBON TETRACHLORIDE, WHICH IS NON-POLAR, AND ADD SOME WATER, WHICH I HAVE DONE IN THIS TUBE, WE SEE THAT THEY SEPARATE INTO TWO SEPARATE LAYERS. THE POLAR WATER MOLECULES NOT LIKE THE NON- POLAR CARBON TET. IF WE SHAKE THIS UP AND LET IT STAND HERE FOR JUST A SECOND WE QUICKLY SEE THAT THE MOLECULES LITERALLY SQUEEZE EACH OTHER OUT. THE NON-POLAR GO BACK TO BEING IN THE NON-POLAR LAYER, AND THE POLAR WATER MOLECULES HAVE SQUEEZED OUT THE CARBON TET AND ARE MAKING ONCE AGAIN A COLORLESS WATER LAYER. SO AGAIN WE SEE A DEMONSTRATION OF THE MISSABILITY AND IMMISSABILITY OF DIFFERENT LIQUIDS DEPENDING ON IF THEY RE POLAR OR NON-POLAR.