CHM 105 & 106 MO1 UNIT THREE, LECTURE ELEVEN 1 IN OUR LAST LECTURE WE WERE LOOKING AT THE SHARING PROCESS IN CHEMICAL BONDING

Transcription

1 CHM 105 & 106 MO1 UNIT THREE, LECTURE ELEVEN 1 CHM 105/106 Program 28: Unit 3 Lecture 11 IN OUR LAST LECTURE WE WERE LOOKING AT THE SHARING PROCESS IN CHEMICAL BONDING AND TALKED ABOUT THE FACT THAT WE MAY HAVE EQUAL SHARING OR UNEQUAL SHARING, AND AS A MATTER OF FACT, IONIC BONDING IS ACTUALLY AN EXTREME CASE OF UNEQUAL SHARING, AND THE QUESTION IS HOW CAN WE DETERMINE THAN WHETHER WHEN TWO ATOMS COME TOGETHER THAT THEY FORM AN IONIC BOND OR A SHARING BOND OR IF IT IS A SHARING BOND IT S EQUAL OR NOT EQUAL SHARING. AND WE DO HAVE A NUMERIC VALUE CA LLED ELECTRONEGATIVITY WHICH WE DEFINED IN THE PREVIOUS LECTURE AS THE ATTRACTION OF TWO NUCLEI FOR THE PAIR OF SHARED ELECTRONS. ELECTRONEGATIVITY. AND LINUS PAULING, A WELL KNOWN AMERICAN CHEMIST WHO DID A LOT OF THE EARLY WORK IN CHEMICAL BONDING DEVISED A NUMERIC WAY OF EVALUATING OR RATING THIS ELECTRONEGATIVITY. THE ELEMENTS THAT HAVE THE STRONGEST ATTRACTION FOR PAIRS OF SHARED ELECTRONS ARE LOCATED ON THE FAR RIGHT SIDE OF THE PERIODIC TABLE, AND THE HIGHEST IS LOCATED AT THE UPPER RIGHT CORNER. THE LOWEST ELECTRONEGATIVITY, THE LEAST TENDENCY TO WANT TO SHARE A PAIR OF ELECTRONS ARE THE ELEMENTS ON THE LOWER LEFT-HAND CORNER. AGAIN, AS WE SAW IN IONIZATION ENERGY AS WE WENT FROM LEFT TO RIGHT ON THE CHART, IONIZATION ENERGY THE ENERGY NEEDED TO TAKE AN ELECTRON AWAY FROM A SINGLE ATOM INCREASE. IT BECAME MORE DIFFICULT TO TAKE THE ELECTRON AWAY AS WE MOVED FROM LEFT TO RIGHT, AND IT BECAME EASIER TO TAKE IT AWAY AS WE WENT FROM TOP TO BOTTOM. WE SEE THAT ELECTRONEGATIVITY, ALTHOUGH NOT EXACTLY THE SAME THING, FOLLOWS THE SAME PATTERN. NOTICE AS WE COME DOWN A COLUMN THE ELECTRONEGATIVITY DECREASES. HERE WE HAVE LITHIUM, NUMERIC VALUE 1.0, FRANCIUM.7, AND IF WE LOOK AS WE GO FROM LEFT TO RIGHT HERE AS WE GO FROM LITHIUM AT 1.0 AND WE GET OVER HERE TO FLUORINE WE NOTICE THAT THE ELECTRONEGATIVITY IS 4.0. NOW YOU MIGHT SAY WELL WHAT HAPPENED TO THE HALOGENS, I LL MOVE THIS OVER A LITTLE BIT THERE, WHAT HAPPENED TO THE, I MEAN EXCUSE ME, WHAT HAPPENED TO THE NOBLE GASSES? WHY DON T WE SEE NUMBERS THERE? WELL REMEMBER NOBLE GASSES DO

2 CHM 105 & 106 MO1 UNIT THREE, LECTURE ELEVEN 2 NOT TEND TO CHEMICALLY BOND UNDER NORMAL CONDITIONS, AND SO HE DID NOT DETERMINE NUMERIC VALUES THEN FOR THEIR CHEMICAL SHARING PROCESS. ALRIGHT, SO WE CAN USE THIS TABLE OF ELECTRONEGATIVITIES THEN TO DETERMINE FIRST OF ALL WHAT TYPE OF BOND WE RE GOING TO HAVE. IS IT IONIC OR IS IT PRIMARILY COVALENT, AND WHETHER OR NOT IT IS OR IS NOT EQUAL SHARING. IONIC THEN WE DEFINE AS ANYTIME TWO ATOMS BOND AND THE ELECTRONEGATIVITY DIFFERENCE, THAT S WHAT THIS DELTA M EANS DIFFERENCE IN ELECTRONEGATIVITY IF THE DIFFERENCE IN ELECTRONEGATIVITY IS GREATER THAN 1.6 IT IS GOING TO END UP AS AN IONIC BOND, INITIALLY, ALL CHEMICAL BONDS START AS A SHARING PROCESS BUT THEY CAN END UP AS AN IONIC PROCESS. FOR INSTANCE, IF I BRING SODIUM ONE ELECTRON, CHLORINE SEVEN ELECTRONS. WE CAN PUT THEM TOGETHER BY THIS PAIR OF ELECTRONS HERE THAT WE COULD SAY INITIALLY IS BEING SHARED, BUT IF WE LOOK AT THE ELECTRONEGATIVITY WE FIND THAT FOR SODIUM THE VALUE IS 0.9 AND FOR CHLORINE 3.0, AND SO THEREFORE THE DELTA EL IN THIS CASE IS 2.1 WHICH MEANS THAT OF COURSE IT S GREATER THAN THE 1.6 AND SO WE SAY THAT THE BONDING IN SODIUM CHLORIDE IS PRIMARILY IONIC. IN OTHER WORDS WE SAID A GIVE AND TAKE PROCESS. THE ELECTRON IS GIVEN BY THE SODIUM TO THE CHLORINE. REALLY AND TRULY IT STARTS AS A SHARING AND THEN THE ELECTRONS SEPARATE. SO WE END UP WITH AN NA + AS WE SHOW IT AND A CL - BECAUSE WE RE SAYING THAT THE CHLORINE HAS THE ELECTRON MORE THAN 50% OF THE TIME. NOW IN THE COVALENT WHICH WE RE SHARING PROCESS, NOT A GIVE AND TAKE, COVALENT BONDING OCCURS THEN WHEN THE DELTA ELECTRONEGATIVITY IS LESS THAN 1.6. WE HAVE THEN WHAT WE CALL COVALENT OR SHARING. WE TALKED ABOUT PURE COVALENT WHERE WE HAVE EQUAL SHARING. WHEN WE HAVE EQUAL SHARING THE DIFFERENCE IN THE ELECTRONEGATIVITY OF THE TWO ELEMENTS IS ZERO. SO QUITE OBVIOUSLY IF I WERE TO COMBINE HERE A HYDROGEN WITH ANOTHER HYDROGEN, EACH ONE GIVING ONE ELECTRON, EACH OF THESE HAS AN ELECTRONEGATIVITY OF 2.1. THE DIFFERENCE IS ZERO. THEY ARE GOING TO EQUALLY SHARE THAT PAIR OF ELECTRONS AND WE CALL THAT THEN PURE COVALENT BONDING. ALRIGHT, NOW, IF ITS POLAR BUT IT S NOT EQUAL TO ZERO, IN OTHER WORDS IF THE ELECTRONEGATIVITY IS LESS THAN 1.6 BUT GREATER THAN ZERO THEN WE

3 CHM 105 & 106 MO1 UNIT THREE, LECTURE ELEVEN 3 HAVE UNEQUAL SHARING AND POLAR COVALENT BONDING AND SO WE RIGHT AT THE END YESTERDAY LOOKED AT SPECIFICALLY HYDROGEN AND CHLORINE AND AGAIN IF WE LOOK AT THE ELECTRONEGATIVITIES HERE WE HAVE 2.1 FOR THE HYDROGEN AND 3.0. SO IN FACT WE DO SEE THAT THERE IS A DIFFERENCE. IT S LESS THAN 1.6, GREATER THAN ZERO, WHICH MEANS THAT THE HYDROGEN AND THE CHLORINE ARE NOT GOING TO EQUALLY SHARE THAT PAIR OF ELECTRONS, AND IF THE DON T EQUALLY SHARE THEN WHAT HAPPENS IS THAT THE PAIR OF ELECTRONS, WE LL DRAW A DASH FOR THE BOND HERE THE PAIRED ELECTRON SPENDS A LITTLE MORE TIME TOWARDS THE CHLORINE THAN IT DOES THE HYDROGEN, AND THIS THEN RESULTS IN THIS SIDE OF THE BOND BEING PARTIALLY NEGATIVE AND THIS SIDE OF THE BOND BEING PARTIALLY POSITIVE. SO WE SAY THAT WE HAVE FORMED THEN A POLAR BOND IN THIS CASE. LOOKING AT ONE OF THE QUESTIONS IN THE TEXT THEN PERTAINING TO THIS, THE QUESTION IS WHICH OF THE FOLLOWING MOLECULES HAVE POLAR BONDS? H2S, HCL, BR2 AND OF2. WELL THIS ONE YES BECAUSE WE JUST DID IT. WE JUST LOOKED AT THE ELECTRONEGATIVITY. LET S LOOK AT A HYDROGEN SULFUR THEN AND SEE WHAT IT IS. SO WE HAVE HYDROGEN SULFUR. WE ONLY HAVE TO LOOK AT ONE, THERE ARE TWO OF THOSE HYDROGEN SULFUR BONDS IN THERE, BUT ONE OF THEM, AND WE LOOK AT THE ELECTRONEGATIVITY OF HYDROGEN 2.1. THIS IS FROM THIS TABLE THAT WE JUST HAD UP HERE A MOMENT AGO WHICH IS IN THE TEXT. SO WE LOOK AT HYDROGEN 2.1. WE LOOK FOR SULFUR NOW, WHICH WE FIND TO BE, UH, IF I CAN SEE IT, 2.5. SO YES, THERE IS AN ELECTRONEGATIVITY DIFFERENCE, SO YES THIS WOULD BE A POLAR BOND AND THAT WOULD MEAN THAT THE SULFUR IS GOING TO BE PARTIALLY NEGATIVE AND THIS HYDROGEN IS GOING TO BE PARTIALLY POSITIVE. SO WE CAN SAY YES TO THIS ONE. NOW WHEN WE LOOK AT THE BROMINE DIATOMIC BROMINE I THINK THAT S PRETTY EVIDENT THAT WHATEVER THE ELECTRONEGATIVITY OF ONE IS IT S THE SAME FOR THE OTHER ONE SO THE DIFFERENCE IN ELECTRONEGATIVITY IS OBVIOUSLY ZERO. SO THEREFORE IT IS NON-POLAR. BROMINE, BY THE WAY, IS 2.8. SO BROMINE DOES NOT HAVE POLAR BONDS. FINALLY WE LOOK AT THE MOLECULE OF2, AND WE ONLY HAVE TO LOOK AT AGAIN ONE OF THESE, ONE OF THE OXYGEN FLUORINES. OXYGEN IS 3.5, FLUORINE IS 4.0. SO YES THERE IS A DIFFERENCE. THE FLUORINE

4 CHM 105 & 106 MO1 UNIT THREE, LECTURE ELEVEN 4 WOULD BE PARTIALLY NEGATIVE AND THE OXYGEN WOULD BE PARTIALLY POSITIVE, AND SO YES, OF2 HAS POLAR BONDS. NOW WE RE GOING TO COME BACK AND UTILIZE THIS POLARITY OF ASPECT A LITTLE BIT LATER IN THE CHAPTER AS WE LOOK AT MOLECULAR POLARITY. I WANT TO STRESS, THIS IS TALKING MERELY ABOUT A BOND, ONE SINGLE BOND IN A MOLECULE. IS IT OR IS IT NOT POLAR? THE POLARITY OF THE TOTAL MOLECULE WE HAVE TO TAKE ALL OF THESE BONDS THAT IT HAS AND IN THE PROPER GEOMETRIC SHAPE TO DETERMINE WHETHER OR NOT A MOLECULE IS GOING TO BE POLAR OR NON-POLAR. SO WE LL COME BACK AND TAKE A LOOK AGAIN AT NON-POLARITY AS WE LOOK AT MOLECULAR POLARITY IN JUST A LITTLE BIT. BEFORE WE CAN DO MOLECULAR POLARITY WE HAVE TO BE ABLE TO DO MOLECULAR STRUCTURE, MOLECULAR SHAPES, AND THE METHOD THAT WE RE GOING TO USE FOR THIS IS CALLED THE VALENCE SHELL ELECTRON PAIR REPULSION MODEL. IT S IN THE TEXT, YOU DON T NEED TO WORRY ABOUT INDIVIDUAL WORDS, BUT IT S THE VALENCE, MEANS THOSE EIGHT ELECTRONS THAT WE RE TALKING ABOUT, THE S AND P. VALENCE SHELL, THAT JUST MEANS THAT SUBLEVEL, VALENCE SUBLEVEL ELECTRON PAIR. EVERY CHEMICAL BOND MAKING UP THEN A COVALENTLY BONDED SPECIES IS A PAIR OF ELECTRONS, AND THESE THEN REPEL EACH OTHER. IF WE RE GOING TO MANY PAIRS OF ELECTRONS AROUND ONE NUCLEUS THEY RE GOING TO REPEL EACH OTHER. SO VALENCE SHELL ELECTRON PAIR REPULSION. BUT BEFORE WE CAN USE VALENCE SHELL ELECTRON PAIR REPULSION TO PREDICT THE GEOMETRIC SHAPE WE HAVE TO KNOW HOW MANY ELECTRON PAIRS AROUND THE CENTRAL ATOM AND TO DOT HAT THEN WE MUST DO A LEWIS STRUCTURE AND WE HAVE A SET OF RULES THEN THAT WE FOLLOW TO DO THIS. WE LL SCAN THROUGH THESE VERY QUICKLY THEN WE WILL APPLY IT AND I THINK THEY LL MAKE MORE SENSE AS WE GO ALONG. FIRST OF ALL WE DETERMINE THE TOTAL NUMBER OF VALENCE ELECTRONS IN THE MOLECULE, OR IT CAN BE AN ION ALSO THAT HAS COVALENT BONDING IN IT. ALRIGHT, TOTAL NUMBER OF VALENCE ELECTRONS. WE ATTACH EACH ATOM TOGETHER WITH A SINGLE BOND. THAT MEANS THAT WE RE USING UP ONE PAIR OF ELECTRONS THEN TO FORM THAT CHEMICAL BOND. DETERMINE THE NUMBER OF ELECTRONS NEEDED TO GIVE EVERY ATOM AN OCTET. REMEMBER THAT MOLECULAR OR THE BONDING STABILITY OCCURS WHEN THINGS ACHIEVE A NOBLE GAS-LIKE STRUCTURE. SO

5 CHM 105 & 106 MO1 UNIT THREE, LECTURE ELEVEN 5 WE RE GOING TO FIND OUT HOW MANY DO WE NEED TO MAKE SURE THAT EVERYTHING ENDS UP WITH AN OCTET AND OF COURSE FOR HYDROGEN ONLY TWO. THEN IT SAYS, IF WE DON T HAVE ENOUGH ELECTRONS TO GIVE EVERYTHING AN OCTET, RULE 4, WE MAY NEED TO PUT IN A MULTIPLE BOND. WE TALKED ABOUT MULTIPLE BONDS BETWEEN TWO OXYGENS, DOUBLE BOND, OR A TRIPLE BOND THREE PAIRS OF ELECTRONS IN NITROGEN. OKAY, ALRIGHT, SO A MULTIPLE BOND NEEDS TO BE PUT IN, AND IT SAYS ONE MULTIPLE BOND IS NEEDED FOR EACH PAIR OF ELECTRONS THAT WE ARE SHORT. SO IN OTHER WORDS IF WE NEEDED EIGHT ELECTRONS WE HAD ONLY SIX ELECTRONS WE HAVE TO PUT IN ONE DOUBLE BOND TO GIVE EVERYTHING AN OCTET. BUT ALSO WE MUST REMEMBER THERE ARE ONLY A FEW SELECT ELEMENTS THAT CAN FORM MULTIPLE BINDS. WE MENTIONED THE OTHER DAY CARBON AND NITROGEN CAN FORM DOUBLE OR TRIPLE BONDS AND THEN WE SEE OXYGEN, SILICON, PHOSPHORUS AND SULFUR AND TO SOME EXTENT SELENIUM CAN FORM DOUBLE BONDS. SO IF WE RE BONDING SOMETHING ELSE, IF WE RE BONDING TE OR I OR BORON WE CAN T PUT A DOUBLE BOND IN THERE BECAUSE THEY CANNOT FORM MULTIPLE BONDS. SO ONLY THAT VERY SMALL GROUP RIGHT UP HERE THEN CAN WE ESSENTIALLY USE AND THE SELENIUM. THOSE SEVEN ELEMENTS THAT CAN HAVE A MULTIPLE BOND. SAYS SOMETIMES WE RE NOT GOING TO BE ABLE TO GIVE EVERYTHING AN OCTET, AND THAT S ESPECIALLY TRUE FOR THE METALS IN GROUP 2A AND 3A. OCCASIONALLY THESE METALS WILL FOR COVALENT BONDS. NOW YESTERDAY WE MENTIONED THAT NORMALLY THESE METALS OVER HERE, WHEN ATTACHED TO THE NON-METALS, ARE IONIC BONDS. THAT S BECAUSE THE ELECTRONEGATIVITY DIFFERENCE IS GREATER THAN 1.6. BUT THERE ARE CASES WHERE SOME OF THESE METALS, LIKE BARIUM WITH SOMETHING LIKE IODINE, MIGHT ACTUALLY FORM A SHARING OR PARTIAL SHARING BOND AND THEREFORE THEN WE RE NOT GOING TO BE ABLE TO GIVE IT AN OCTET. SO METALS IN THE 2A AND METALS IN THE 3A IT SAYS WE RE PROBABLY GOING TO FIND OUT THAT WE CAN T SATISFY THE OCTET. THEN WHEN WE GET THE LEWIS STRUCTURE BUILT IT SAYS WE NOW CAN LOOK AT THE NUMBER OF ELECTRON PAIRS AROUND THE SIMPLE ATOM AND THAT THEN TELLS US THE GEOMETRY. AGAIN, WE RE GOING TO DETERMINE THE ELECTRON PAIRS AROUND THE CENTRAL ATOM AND ONCE WE VE

6 CHM 105 & 106 MO1 UNIT THREE, LECTURE ELEVEN 6 DETERMINED THE ELECTRON PAIRS AROUND THE CENTRAL ATOM WE CAN DETERMINE GEOMETRIC SHAPE. NOW IF THERE ARE TWO, WE RE LINEAR. THREE, PLANAR TRIANGLE. FOUR, TETRAHEDRAL. IF WE HAVE FOUR BUT WE HAVE ONE NON-BONDING HERE WE RE PYRAMIDAL. IF WE HAVE FOUR AROUND THE CENTER AND TWO NON-BONDING IT S GOING TO BE BENT OR ANGULAR IN SHAPE. ALSO ONE OTHER THING THAT WE LL HAVE TO REMEMBER IF A DOUBLE OR A TRIPLE BOND IN A MOLECULE, AS FAR AS GEOMETRY, COUNTS THE SAME AS JUST ONE BOND. ALRIGHT, SO WE RE ADDING THEM UP IT COUNTS AS JUST ONE. ALRIGHT, NOW WE VE COVERED A LOT THERE IN THOSE RULES, BUT I THINK IT WILL, HOPEFULLY AT LEAST, IT WILL BEGIN TO MAKE SENSE AS WE PUT A FEW OF THESE TOGETHER. WE RE GOING TO TAKE THE SERIES BERYLLIUM, BORON, CARBON, NITROGEN, OXYGEN IN THAT ROW AND LOOK AT THEM RELATIVE TO EACH OTHER SO WE CAN SEE, REMEMBER THE NICE THING ABOUT THE PERIODIC TABLE, IF WE SAY SOMETHING ABOUT BERYLLIUM IT TELLS THE SAME THING ABOUT MAGNESIUM, CALCIUM, STRONTIUM, ET CETERA. OR, IF WE TALK ABOUT NITROGEN IT S ALSO TRUE OF PHOSPHORUS AND ARSENIC, ANTIMONY. ALRIGHT, SO WE ONLY NEED ONE REPRESENTATIVE ELEMENT TO THEN SHOW THE OVERALL PROCESS. BEH2. BERYLLIUM, SECOND COLUMN - TWO VALENCE ELECTRONS. HYDROGEN, FIRST COLUMN ONE VALENCE ELECTRON BUT THERE ARE TWO OF THEM SO IT S TWO, WHICH GIVES US THEN FOUR VALENCE ELECTRONS TO WORK WITH. NOW SEEING THAT CHEMICAL BONDS ARE ALWAYS PAIRS OF ELECTRONS, I USUALLY WORK IT IN PAIRS RATHER THAN SINGLES SO I WOULD SAY I HAVE TWO ELECTRON PAIRS. RULE NUMBER ONE FIND OUT HOW MANY VALENCE ELECTRONS WE HAVE. RULE NUMBER TWO HOOK EACH ATOM TO THE CENTRAL ATOM WITH A SINGLE BOND, A PAIR OF ELECTRONS. SO WE START WITH BE, WE USE A DASH TO SHOW A PAIR OF SHARED ELECTRONS, AND H. WE THEN SEE THAT HYDROGEN IS SATISFIED BECAUSE IT NOW HAS A PAIR OF ELECTRONS SO IT HAS BECOME LIKE HELIUM. BERYLLIUM ISN T. IT HASN T REACHED AN OCTET, AND SO IT IS NOT THEN SATISFIED, AND WE WOULD ASK THEN QUESTION HOW MANY MORE ELECTRON PAIRS WOULD WE NEED TO GIVE BERYLLIUM AN OCTET. WELL WE WOULD NEED ONE MORE HERE AND ONE MORE HERE. BUT WE DON T HAVE ANY. WE VE ALREADY USED UP THE TWO THAT WE HAD. SO WE WOULD NEED FOUR ELECTRON PAIR TO GIVE

7 CHM 105 & 106 MO1 UNIT THREE, LECTURE ELEVEN 7 BERYLLIUM AN OCTET. NOW THAT SAYS THEN IF WE ARE ELECTRON PAIRS SHORT OKAY, WE NEED FOUR, WE HAVE TWO, THAT WOULD SAY THAT WE NEED TO PUT IN TWO MULTIPLE BONDS IF WE CAN. CAN BERYLLIUM FORMA MULTIPLE BOND? NO. CAN HYDROGEN FORM A MULTIPLE BOND? NO. WE CAN T PUT IN ANY MULTIPLE BONDS, SO THIS IS A CASE WHERE MOLECULE IS GOING TO BE ELECTRON DEFICIENT AS FAR AS THE CENTRAL ATOM. SO OUR FINAL THEN PICTURE OF THE MOLECULE, LEWIS STRUCTURE, LOOKS LIKE THAT. THAT S ALL WE HAVE. THOSE ARE ALL THE ELECTRONS INVOLVED. NOW, VALENCE SHELL ELECTRON THEORY SAYS PUT THE PAIR OF BONDING ELECTRONS AS FAR FROM EACH OTHER AS POSSIBLE. WELL OBVIOUSLY TO PUT THEM AS FAR FROM EACH OTHER AS POSSIBLE YOU D PUT THEM WHERE RELATIVE TO THE BERYLLIUM NUCLEUS? 180 DEGREES AWAY FROM EACH OTHER. WHAT IS THE SHAPE OF THIS MOLECULE? IT IS CALLED A LINEAR MOLECULE. ALRIGHT, TWO ELECTRON PAIRS AROUND THE CENTRAL ATOM WILL ALWAYS GIVE US A LINEAR MOLECULE. THAT S TRUE OF MAGNESIUM HYDROGEN, BARIUM HYDROGEN, ET CETERA, IT S TRUE OF ALL OF THOSE. ALRIGHT, NOW THIS IS ACTUALLY A COVALENT MOLECULE BECAUSE BERYLLIUM HAS ELECTRONEGATIVITY OF 1.1 AND HYDROGEN 2.1. SO IT IS THEN A COVALENT, SHARED MOLECULE, BUT IT DOES HAVE A POLAR BOND. IN OTHER WORDS, WE WOULD HAVE PARTIALLY POSITIVE ON THAT HYDROGEN. SO IT IS A POLAR BOND, POLAR COVALENT MOLECULE. ALRIGHT, LET S LOOK AT THE NEXT ONE IN THE SERIES, BH3. BORON IS IN THE THIRD COLUMN, SO IT HAS THREE VALENCE ELECTRONS. THERE ARE THREE HYDROGENS. SO WE WOULD ADD THREE. WE HAVE SIX. WE THEREFORE HAVE THREE ELECTRON PAIRS THAT WE CAN WORK WITH. THE FIRST THING THAT WE DO THEN IS TO ATTACH EACH ATOM TOGETHER USING ONE PAIR OR A SINGLE BOND. NOW EVERY HYDROGEN IN HERE IS SATISFIED, THEY EACH HAVE TWO, BUT AGAIN, THE BORON ONLY HAS SIX. SO THEREFORE WE WOULD NEED ANOTHER PAIR. WE NEED FOUR, WE HAVE THREE. WHAT THAT SAYS IS THAT IF WE CAN WE SHOULD FORM A MULTIPLE BOND. HOWEVER, HYDROGEN CAN T FORM A MULTIPLE BOND. BORON CAN T FORMA MULTIPLE BOND, SO WE RE STUCK. WE CAN T DO ANYTHING ABOUT IT, AND THAT IS THEN THE FINAL LEWIS STRUCTURE RIGHT THERE. NOW WHAT ABOUT THE GEOMETRIC SHAPE? THERE ARE THREE ELECTRON PAIRS AROUND THE CENTRAL ATOM. WHERE WOULD WE PUT

8 CHM 105 & 106 MO1 UNIT THREE, LECTURE ELEVEN 8 THOSE SO THAT THEY WOULD BE AS FAR FROM EACH OTHER AND EQUIVALENT FROM EACH OTHER? WELL IF WE PUT THE BORON HERE, PUT ONE HERE, 120 DEGREES DOWN THIS WAY. WE WOULD PUT THE OTHER ONE 120 DEGREES THIS WAY. WE WOULD PUT THE OTHER ONE, EACH OF THESE WOULD BE 120 DEGREES APART, AND WE WOULD CALL THIS THEN, ITS SHAPE IS REFERRED TO AS EQUILATERAL TRIANGULAR, ACTUALLY SOMETIMES IT S JUST CALLED TRIANGULAR SO EQUILATERAL IS NOT TOTALLY NECESSARY, EQUILATERAL TRIANGULAR PLANAR, WHICH MERELY MEANS THAT ALL OF THE ATOMS ARE IN AN X AND A Y PLANE. WE CAN SHOW THIS PRETTY QUICKLY HERE, IF THIS WAS THE BORON HERE AND THESE REPRESENT THE THREE BONDS ON THAT BORON AND WE PUT THE THREE HYDROGENS ON NOW AND IF WE DID THAT WE NOTE THAT THE ENTIRE MOLECULE STAYS FLAT. IT S PLANAR, THAT S WHAT WE MEAN BY PLANAR. IT DOES NOT, HAS ONLY TWO DIMENSIONS, AN X AND A Y, AND WE NOTICE THE BOND ANGLES IN THERE ARE THE 120 DEGREE BOND ANGLES AND IT DOESN T MAKE ANY DIFFERENCE WHICH WAY WE TURN IT, IT S THE SAME SHAPE NO MATTER WHICH WAY. DOES THIS MOLECULE HAVE A POLAR BOND? WELL AGAIN WE WOULD HAVE TO LOOK AT ELECTRONEGATIVITY. WE KNOW HYDROGEN IS 2.1, AND LOOKING AT THE ELECTRONEGATIVITY TA BLE WE SEE THAT BORON IS 2.0. SO THEREFORE THERE IS A SLIGHT POLARITY TO THE BOND. VERY SMALL, ONLY A.1 DIFFERENCE. SO THIS WOULD BE THEN YES, JUST SLIGHTLY POLAR, PARTIALLY POSITIVE AND VERY PARTIALLY, WHOOPS, THE OTHER WAY. PARTIALLY POSITIVE IN THE HYDROGEN, PARTIALLY NEGATIVE IN THIS CASE. OKAY, ALRIGHT, THE NEXT ONE IN OUR SERIES WOULD BE THE CARBON WITH THE HYDROGEN, AND SO LET S TAKE A LOOK AT IT. CARBONS IN THE FOURTH COLUMN, FOUR VALENCE ELECTRONS. EACH HYDROGEN HAS ONE. WE HAVE A TOTAL OF EIGHT. WE HAVE FOUR ELECTRON PAIR. WE ATTACH THEN THE HYDROGENS TO THE CARBON, EACH WITH A SINGLE BOND. THIS IS WHAT WE MEAN BY THE CENTRAL ATOM. CENTRAL ATOM IS USUALLY THE ONE IN THE LEAST AMOUNT IN THE CHEMICAL FORMULA. WE NOTICE THAT THE HYDROGEN HAS TWO EACH SO IT S LIKE HELIUM. THE CARBON NOW SEES 2,4,6,8. IT HAS AN OCTET SO WE VE SATISFIED EVERYTHING IN THIS MOLECULE. EVERYTHING HAS NOBLE GAS-LIKE STRUCTURE. THE ONLY THING LEFT THEN TO DETERMINE THEN IS WHAT IS THE GEOMETRIC SHAPE. SO WHERE ARE WE

9 CHM 105 & 106 MO1 UNIT THREE, LECTURE ELEVEN 9 GOING TO PUT THOSE FOUR PAIRS OF ELECTRONS TO PUT THEM AS POSSIBLE BUT EQUAL DISTANCE FROM EACH OTHER? WELL THE FIRST TENDENCY MIGHT BE TO PUT HEM JUST AS WE HAVE HERE, TO MAKE IT LOOK LIKE A SQUARE. EACH WOULD BE 90 DEGREES FROM EACH OTHER, EXCEPT THIS ONE WOULD BE 90 DEGREES FROM THAT ONE BUT THIS ONE WOULD BE 180 DEGREES FROM THE ONE BELOW IT. AND SO THEY RE NOT ALL EQUAL DISTANT FROM EACH OTHER AND THEREFORE WE CANNOT FORM A PLANAR MOLECULE. SQUARE PLANAR WOULD NOT BE THE WAY THAT IT WOULD BE. SO WE HAVE TO GO TO A THREE DIMENSIONAL SHAPE, AND IT LOOKS SOMETHING LIKE THIS THEN. CARBON AND SOLID LINE HERE REPRESENTS THAT THAT BOND IS IN THE PLANE OF THE PAPER. A BOND THAT WOULD STICK DOWN AND BACK. SO THESE ARE STICKING OUT BEHIND THE SCREEN, BEHIND THE BOARD. WE WOULD HAVE OUR HYDROGENS ON THERE AND THEN THE OTHER ONE WOULD BE STICKING OUT AT US. WE USE A LITTLE TRIANGLE LIKE THIS TO SHOW THE ONE STICKING OUT AT US. SO IT WOULD BE A THREE- DIMENSIONAL MOLECULE AND THE NAME OF THIS IS TET RAHEDRAL. TETRAHEDRAL GEOMETRY. HEDRAL MEANS FACE, AND TETRA OF COURSE IF FOUR. SO FOUR FACES, IF WE LOOK AT IT WE WOULD HAVE A FACE HERE AND HERE, ONE IN THE BOTTOM AND ONE BEHIND. SO FOUR-FACED ITEM. AND IF WE WERE TO PUT TOGETHER A LITTLE MODEL OF IT THEN FIRST OF ALL WE WOULD START OUT HERE WITH THE BONDS. THE BLACK SPHERE HERE, WHICH I KNOW IS GOING TO BE A LITTLE HARD TO SEE FROM WAY BACK THERE, BUT THE BLACK SPHERE IS THE CARBON. WE CAN SEE THAT IT HAS THESE FOUR AND IT DOESN T REALLY MAKE ANY DIFFERENCE WHICH WAY WE PUT IT, THEY RE ALWAYS THE SAME. SO THEY RE EQUAL DISTANT FROM EACH OTHER. SO THEY RE IN WHAT WE CALL THEN A TETRAHEDRAL GEOMETRIC SHAPE AT THIS POINT. NOW WHEN WE PUT ON THE FOUR HYDROGENS THEN WE CAN SEE THAT THE MOLECULE SHAPE IS ALSO THE SAME. IT IS THREE DIMENSIONAL THEN WE SEE HOW IT STANDS THERE SO THAT WE DO HAVE, IT S NOT PLANAR, I CAN PUT MY FINGER UNDERNEATH, SO IT S A THREE-DIMENSIONAL MOLECULE CALLED A TETRAHEDRAL MOLECULAR SHAPE. A FACE, A FACE, A FACE, A FACE, AND THAT S WHERE THE NAME TETRAHEDRAL COMES FROM. ALRIGHT, THE NEXT ONE IN OUR FAMILY THAT WE RE LOOKING AT HERE IS NH3. NITROGEN IS IN THE FIFTH COLUMN. IT HAS FIVE VALENCE ELECTRONS. HYDROGEN IS IN THE FIRST COLUMN SO WE

10 CHM 105 & 106 MO1 UNIT THREE, LECTURE ELEVEN 10 HAVE THREE, SO WE HAVE 7, 8, SO WE HAVE FOUR ELECTRON PAIRS, AND THE FIRST THING I DO THEN IS I ATTACH EACH OF THE HYDROGENS TO THE NITROGEN. WE VE USED UP THREE OF THE ELECTRON PAIR. THE RULE IS, CHECK TO SEE THAT WE CAN GIVE EVERYTHING A NOBLE GAS- LIKE STRUCTURE. HOW MANY MORE ELECTRONS WOULD NITROGEN NEED TO HAVE AN OCTET? TWO. AND WE HAVE TWO, BECAUSE WE HAD FOUR ELECTRON PAIRS, WE HAVE ONLY USED THREE FOR THE BONDING SO WE DO HAVE ONE MORE ELECTRON PAIR, VALENCE ELECTRON PAIR LEFT AND WE WILL GIVE THAT THEN TO THE NITROGEN AND SO NOW NITROGEN SEES ONE, TWO, THREE, FOUR ELECTRON PAIR AROUND IT. IT HAS AN OCTET AND OF COURSE WE HAVE THE FOUR ELECTRON PAIR SATISFIED. WE HAD THAT MANY, WE USED THAT MANY. EVERYTHING IS A NOBLE GAS-LIKE STRUCTURE. NOW, LOOKING AT THIS, THIS IS A NON- BONDING ELECTRON PAIR. WE MENTIONED THIS IN OUR PREVIOUS LECTURE WHEN WE TALKED ABOUT TWO FLUORINES COMING TOGETHER OR TWO OXYGENS COMING OR TWO NITROGENS, THERE WERE SOME OF THESE NON-BONDING ELECTRONS. ELECTRONS THAT ARE STRICTLY WITH ONE ATOM. THEY RE NOT BEING SHARED ARE NON-BONDING. SO AT THIS PARTICULAR CASE WE HAVE FOUR ELECTRON PAIRS AROUND THE CENTRAL ATOM. OKAY, SO WE HAVE FOUR ELECTRON PAIRS, UT WE HAVE ONE NON-BONDING PAIR. IF WE WERE TO LOOK AT THE TABLE IT WOULD TELL US THAT THE SHAPE OF THIS MOLECULE SHOULD BE PYRAMIDAL, WHOOPS, PYRAMIDAL OKAY. SHOULD BE A PYRAMIDAL MOLECULE. NOW HOW DOES THAT OCCUR? HOW DO WE GET A PYRAMID OUT OF THIS PARTICULAR STRUCTURE? WELL FIRST OF ALL, FOUR ELECTRON PAIRS AROUND THE CENTRAL ATOM SHOULD BE WHERE? WHERE SHOULD THOSE FOUR ELECTRON PAIRS I HAVE FOUR, WHERE WOULD I PUT FOUR AROUND THE CENTRAL ATOM? TETRAHEDRALLY, JUST LIKE WE DID WITH THE CARBON. OKAY, NOW THE ONLY DIFFERENCE BETWEEN THIS AND THE PREVIOUS ONE IS THIS TIME I M GOING TO PUT ON THREE HYDROGENS FOR AMMONIA WITH THE ONE HYDROGEN AND THERE S THE NON-BONDING ELECTRON PAIR. NOW WHEN WE LOOK AT MOLECULES, WE CAN LOOK AT THEM BY USING X- RAY DIFFRACTION, WHEN WE LOOK AT MOLECULES THE ONLY THING WE CAN SEE ARE THE NUCLEI. WE CAN NEVER SEE THE ELECTRONS. WELL IF WE CAN T SEE THE ELECTRONS THEN I CAN T SEE THAT SO THE SHAPE OF THE MOLECULE LOOKS LIKE THAT. WELL WHAT SHAPE

11 CHM 105 & 106 MO1 UNIT THREE, LECTURE ELEVEN 11 WOULD WE GIVE FOR THAT? A PYRAMID, A THREE SIDED PYRAMID THEN, A TRIANGULAR PYRAMIDAL STRUCTURE, AND SO THAT S WHERE THE SHAPE PYRAMID COMES FROM. ANY TIME WE HAVE FOUR ELECTRON PAIR AROUND A CENTRAL ATOM AND ONE OF THOSE ELECTRON PAIRS IS NON-BONDING IT LL BE THREE DIMENSIONAL, PYRAMIDAL. HOW CAN WE DRAW THAT PARTICULAR STRUCTURE IF WE RE GOING TO DRAW IT? WELL WE CAN SHOW NITROGEN DASH HYDROGEN DASH HYDROGEN, MEANING THOSE HYDROGENS ARE GOING BACK INTO THE BOARD AND DOWN AND THEN A LITTLE TRIANGLE AND THE HYDROGEN, AND THAT WOULD BE OUR PYRAMIDAL SHAPE. IF WE WANTED TO WE COULD SHOW THE NON-BONDING ELECTRON PAIR THERE, BUT THAT IS NOT PART OF MOLECULAR GEOMETRY. ALRIGHT, THE LAST ONE IN THE FAMILY THAT WE RE GOING, I SHOULDN T SAY FAMILY SERIES THAT WE RE GOING ACROSS, WOULD BE H2O. TWO HYDROGENS, SO WE HAVE TWO VALENCE ELECTRONS. OXYGEN IS IN THE SIXTH COLUMN SO WE HAVE SIX ELECTRONS, AND SO WE HAVE EIGHT ELECTRONS OR FOUR ELECTRON PAIRS. WE RE GOING TO TAKE THEN THE OXYGEN, HOOK THE HYDROGENS TOGETHER. WE VE USED UP TWO ELECTRON PAIRS. HYDROGEN IS ALREADY SATISFIED. HOW MANY MORE ELECTRON PAIRS WOULD WE NEED TO GIVE OXYGEN AN OCTET? WELL WE D NEED ONE THERE AND WE D NEED ANOTHER ONE. DO WE HAVE THAT MANY? YES WE DO. WE HAVE FOUR, WE NEED FOUR. WE VE NOW GIVEN OXYGEN AN OCTET AND NO MULTIPLE BONDS YET. SO WE RE DONE WITH THE LEWIS STRUCTURE. NOW WE LOOK FOR THE GEOMETRY. WE SAY WELL WE HAVE FOUR ELECTRON PAIRS AROUND THE CENTRAL ATOM, BUT THIS TIME WE NOTICE THAT TWO ELECTRON PAIRS ARE NON-BONDING ELECTRON PAIRS. WELL WHAT DOES THAT DO TO THE GEOMETRIC SHAPE THEN? WELL, LET S GO BACK TO OUR BASIC STARTING BLOCK ONCE AGAIN, THIS TIME WITH THE FOUR ELECTRON PAIRS. WHERE ARE THEY GOING TO BE? THEY RE GOING TO BE TETRAHEDRAL AREN T THEY? FOUR ELECTRON PAIRS WILL ALWAYS TAKE THIS TETRAHEDRAL GEOMETRY. NOW THIS TIME HOWEVER, IN THE CASE OF OXYGEN BEING THE CENTRAL ATOM, I M GOING TO PUT TWO HYDROGENS. SO I HAVE TWO BONDED AND I HAVE TWO NON-BONDING ELECTRON PAIRS. BUT I CAN T SEE THE NON- BONDING ELECTRON PAIRS WHEN I LOOK AT GEOMETRY. SO WHAT S THE SHAPE OF A WATER MOLECULE? WE SAY THAT IT IS BENT OR ANGULAR IS THE TERM THAT WE USE FOR THIS, BENT

12 CHM 105 & 106 MO1 UNIT THREE, LECTURE ELEVEN 12 OR ANGULAR. NOT ONLY IS THAT TRUE OF H2O, THAT S TRUE FOR H2S, H2TE, H2SE. IT S TRUE FOR ALL OF THOSE ATOMS WITH THAT OXYGEN. OXYGEN, IF IT HAS ONLY SINGLE BONDS IS ALWAYS GOING TO HAVE THIS BENT OR ANGULAR SHAPE. ALRIGHT, LET S GO O N AND LOOK AT ANOTHER ONE NOW THAT S NOT IN THAT SERIES. LET S LOOK AT THIS ONE HERE, NBR3. NITROGEN HAS FIVE, BROMINE HAS HOW MANY? SEVEN. RIGHT, IT S IN THE SEVENTH COLUMN. A HALOGEN OVER THERE, THREE OF THOSE, THAT S 21. SO WE HAVE A TOTAL OF 26 ELECTRONS. WE HAVE 13 ELECTRON PAIR THAT WE CAN WORK WITH. SO WE START WITH THE NITROGEN AND WE ATTACH THE BROMINES. SO WE VE USED UP THREE ELECTRON PAIRS. SO LET S SUBTRACT THREE ELECTRON PAIR HERE FOR BONDING. SO THAT LEAVES US WITH 10 ELECTRON PAIR LEFT. NOW HOW MANY ELECTRON PAIRS DO WE NEED TO GIVE EVERYTHING AN OCTET IN THIS MOLECULE? WELL THIS BROMINE WOULD NEED ONE I M GONNA USE A DASH INSTEAD OF X S THIS TIME ONE, TWO, THREE, FOUR, FIVE, SIX, SEVEN, EIGHT, NINE, TEN. HOW MANY ELECTRON PAIRS DID I HAVE AVAILABLE? TEN. I AHD ENOUGH ELECTRON PAIRS TO GIVE EVERYTHING AN OCTET AND ONLY USE SINGLE BONDS. SO THEREFORE WE HAVE NOW WRITTEN THE CORRECT LEWIS STRUCTURE, NOW WE NEED TO DETERMINE GEOMETRY. WELL, HOW MANY ELECTRON PAIRS DO WE HAVE AROUND THE CENTRAL ATOM? FOUR ELECTRON PAIRS, BUT WE NOTICE THAT ONE IS A NON-BONDING ELECTRON PAIR, AND WHAT S THE GEOMETRY WHEN WE HAVE FOUR WITH ONE NON-BONDING? PYRAMIDAL RIGHT? AGAIN, THEN IT WOULD BE LIKE HERE S NITROGEN AND DOWN HERE WOULD BE A BROMINE AND A BROMINE AND A BROMINE STICKING OUT AT US AND THE PAIR OF NON-BONDING ELECTRONS. SO A PYRAMIDAL MOLECULE. QUESTION? (STUDENT NOT AUDIBLE) I HAVEN T HAD TO USE ANY MULTIPLE BONDS YET. THE QUESTION WAS WHY AM I NOT USING ANY DOUBLE OR TRIPLE BONDS? WE HAVEN T BEEN ABLE TO IN THOSE CASES WHERE WE COULDN T GET AN OCTET BECAUSE THERE WASN T AN ATOM THAT COULD FORM MULTIPLE BONDS AND WE HAVEN T NEEDED TO IN THOSE CASES WHERE WE DID HAVE THINGS LIKE CARBON AND OXYGEN AND NITROGEN BUT WE LL SEE ONE IF YOU LL JUST HOLD ON A MINUTE AND WE LL LOOK AT ONE WHERE THAT IS INVOLVED. ALRIGHT, SOME IONS WHICH WE REFER TO AS POLYATOMIC, MEANING THAT IT S AN ION THAT CONTAINS TWO OR MORE DIFFERENT ELEMENTS, OKAY,

13 CHM 105 & 106 MO1 UNIT THREE, LECTURE ELEVEN 13 MONOATOMIC IONS OBVIOUSLY THEY RE JUST IONS, BUT WE DO HAVE GROUPS OF ATOMS THAT HAVE CHARGES WHICH ARE HELD TOGETHER BY COVALENT BONDS. IN OTHER WORDS, THE SULFUR AND HYDROGEN HERE ARE HELD TOGETHER BY COVALENT BONDS. NOW LET S TAKE A LOOK HERE. WE HAVE SULFUR, SIXTH COLUMN, ONE HYDROGEN. BUT NOTICE THE MINUS ONE CHARGE THIS TIME. HOW DOES A SPECIES GET TO BE A MINUS ONE? IT HAS TO GAIN A NEGATIVE PARTICLE OR IN OTHER WORDS IT HAS TO GAIN AN ELECTRON. SO IN OTHER WORDS, IT HAS ONE ADDITIONAL WITH THAT MINUS ONE CHARGE. SO WE HAVE EIGHT, BUT WE HAVE FOUR ELECTRON PAIR. SO WE TAKE THE SULFUR AND WE HOOK IT TO THE HYDROGEN. WE VE USED UP ONE ELECTRON PAIR, WE HAVE THREE LEFT. HOW MANY ELECTRON PAIRS DO WE NEED TO GIVE EVERYTHING A NOBLE GAS-LIKE STRUCTURE.? WELL THE SULFUR WOULD NEED THREE MORE WOULDN T IT? HOW MANY DID WE HAVE? WE HAVE THREE. WE DON T NEED ANY MULTIPLE BONDS, SO THEREFORE THAT IS THE CORRECT LEWIS STRUCTURE AND SO WE WOULD SHOW IT THEN AS S DASH H, NON-BONDING PAIRS, BUT WE DO HAVE TO PUT A PARENTHESIS AND A MINUS SO THAT WE KNOW THAT WE RE TALKING ABOUT THEN A CHARGED SPECIES, NOT A NEUTRAL MOLECULE. LET S LOOK AT ONE, WELL YOU MIGHT SAY WHAT ABOUT THE SHAPE? WELL ANY TIME YOU HAVE TWO THINGS HOOKED TOGETHER WHAT S THE SHAPE GONNA BE? LINEAR. I MEAN, IT CAN T BE ANYTHING BUT HOOKED TOGETHER IN A LINE. SO ALL TWO- ATOM SPECIES ARE LINEAR. LET S LOOK AT THIS ONE THEN. PHOSPHORIC ACID. WE MENTIONED THIS IN AN EARLIER CHAPTER. NUMBER SEVEN, INDUSTRIAL CHEMICAL, PHOSPHORIC ACID. WHAT DOES IT LOOK LIKE AS FAR AS A MOLECULE? THREE ELECTRONS IN THE HYDROGEN. PHOSPHORUS IS IN THE FIFTH ROW. FIVE ELECTRONS FROM THAT. OXYGEN IS IN THE SIXTH ROW. SIX TIMES FOUR WOULD BE TWENTY-FOUR. WE HAVE THEN A TOTAL OF 32 ELECTRONS OR 16 ELECTRON PAIR TO WORK WITH. NOW, WE WOULDN T NECESSARILY KNOW IN WHAT ORDER THESE THINGS WE RE HOOKED TOGETHER. HOWEVER, THIS IS ONE OF THE QUESTIONS OUT OF THE TEXT AND IT IDENTIFIES WHAT THINGS SHOULD BE HOOKED TOGETHER. THE OXYGENS ARE HOOKED TO THE PHOSPHORUS SO WE START BY DOING THAT AND THE HYDROGENS ARE HOOKED TO THE OXYGENS AND WE LL DO THAT. IT DOESN T MAKE ANY DIFFERENCE WHICH ONES WE HOOK THEM TO. SO WE HOOK IT LIKE THAT ALRIGHT, NOW

14 CHM 105 & 106 MO1 UNIT THREE, LECTURE ELEVEN 14 WE HAVE ALL OF THE ATOMS NOW, EVERYTHING WITH SINGLE BONDS HOOKED TOGETHER AND WE WOULD THEN SAY ALRIGHT WE VE USED 1, 2, 3, 4, 5, 6, WE VE USED 7 ELECTRON PAIR FOR BONDING. SO WE HAVE 9 ELECTRON PAIR LEFT. QUESTION IS HOW MANY ELECTRON PAIRS DO WE NEED TO GIVE EVERYTHING AN OCTET? WELL HYDROGEN DOESN T NEED ANY BUT HIS OXYGEN NEEDS TWO. THIS OXYGEN WOULD NEED THREE. THIS OXYGEN NEEDS TWO AND THIS OXYGEN NEEDS TWO. ONE, TWO, THREE, FOUR, FIVE, SIX, SEVEN, EIGHT, NINE. WE HAD NINE, SO THAT IS THE CORRECT LEWIS STRUCTURE FOR PHOSPHORIC ACID. WHAT ABOUT THE GEOMETRIC SHAPE? WELL WE WOULD ACTUALLY HAVE TO TAKE THE MOLECULE APART AND LOOK AT IT IN DIFFERENT PARTS. IF WE LOOKED AT IT RELATIVE TO THE CENTER PART THEN PHOSPHORUS HAS FOUR ELECTRON PAIRS AROUND IT AND FOUR ATTACHED GROUPS OF THE GEOMETRY RELATIVE TO THE PHOSPHORUS CENTER WOULD BE TETRAHEDRAL. TETRAHEDRAL. IF WE LOOK AT THIS PART OF THE MOLECULE THE OXYGEN HAS FOUR ELECTRON PAIRS AROUND IT BUT IT HAS TWO NON-BONDING PAIRS AND THEREFORE IT WOULD BE BENT. SO THIS WOULDN T BE IN A STRAIGHT LINE IT WOULD BE BENT AND EVERYTHING AROUND THIS PHOSPHORUS WOULD BE IN A TETRAHEDRAL GEOMETRIC SHAPE. ALRIGHT, WE DON T USUALLY WORRY ABOUT THE GEOMETRY OF THE MORE COMPLEX ONES, SO WE LL PASS ON THAT FOR A SECOND. NO2, MINUS ONE ION. FIVE PLUS TWO TIMES SIX, TWELVE PLUS ONE FOR THE MINUS ONE CHARGE EQUALS 18, NINE ELECTRON PAIR. WE HOOK THE TWO OXYGENS TO THE CENTRAL ATOM, THE NITROGEN. SO WE HAVE USED UP TWO, MINUS TWO ELECTRON PAIRS. WE HAVE SEVEN ELECTRON PAIR REMAINING. HOW MANY DO WE NEED TO GIVE EVERYTHING AN OCTET? ONE, TWO, THREE, FOUR, FIVE, SIX, SEVEN, EIGHT, OH-OH. WE NEED EIGHT ELECTRON PAIR. WE HAVE ONLY SEVEN ELECTRON PAIR. IF WE NEED MORE THAN WE HAVE WE HAVE TO LOOK FOR A MULTIPLE BOND. WE RE SHORT ONE ELECTRON PAIR, WE NEED ONE MULTIPLE BOND. DOES THIS SPECIES INVOLVE THINGS THAT CAN FORM MULTIPLE BONDS? YES. NITROGEN, OXYGEN, CARBON, SILICON, PHOSPHORUS, SULFUR CAN FORM MULTIPLE BONDS SO I CAN PUT IN A MULTIPLE BOND. SO I M GOING TO CANCEL THAT BECAUSE I CAN T HAVE IT THAT WAY AND I M GOING TO RE-HOOK THIS TOGETHER NOW AND PUT IN A DOUBLE BOND. WHERE I PUT IT DOESN T MAKE ANY DIFFERENCE. WHICH SIDE YOU GIVE IT TO DOESN T MAKE ANY

15 CHM 105 & 106 MO1 UNIT THREE, LECTURE ELEVEN 15 DIFFERENCE. OKAY, NOW WHEN I PUT IN THE DOUBLE BOND AND THEN I GIVE EVERYTHING AN OCTET, ONE, TWO, THREE, THAT MEANS THIS NITROGEN STILL NEEDS A PAIR. THIS OXYGEN WOULD NEED TWO PAIR. LET S ADD UP THE TOTAL NUMBER OF ELECTRON PAIRS. WE NOW HAVE ONE, TWO, THREE, FOUR, FIVE, SIX, SEVEN, EIGHT, NINE. HOW MANY DID WE HAVE? WE HAD NINE TOTAL SO WE RE ALRIGHT. WE HAVE THE NUMBER THAT WE NEED. WE WOULD NEED TO PUT A PARENTHESIS HERE WITH A MINUS BECAUSE IT S NOT A NEUTRAL MOLECULE. THE VERY LAST THING IN THE LAST MINUTE HERE QUICKLY, WHAT IS THE SHAPE OF THE NO2 ION? WELL WE LOOK AT THE NITROGEN. NOW, HOW MANY EFFECTIVE ELECT RON PAIRS SURROUND THE NITROGEN? REMEMBER THE ONE RULE, THE DOUBLE OR TRIPLE BOND COUNTS AS ONE. ALRIGHT, SO THE QUESTION IS HOW MANY ELECTRON PAIR AROUND THE CENTRAL NITROGEN FOR SHAPE? THREE. WHAT S THE SHAPE OF SOMETHING THAT HS THREE ELECTRON PAIRS AROUND IT? EQUILATERAL TRIANGULAR, AND SO THIS THEN SHOULD HAVE A SHAPE THAT WOULD BE LIKE THIS: NITROGEN-DOUBLE BOND-OXYGEN-BOND-OXYGEN, AND OF COURSE THIS NON-BONDING PAIR WHICH WE CAN T SEE AND SO THE SHAPE OF THIS ION WOULD BE BENT. OKAY, WOULD BE A BENT ION. ALRIGHT, WELL, IN THE NEXT LECTURE WE LL DO SOME MORE OF THESE AND WE LL THEN ALSO TIE THIS IN WITH NOW PREDICTING MOLECULAR POLARITY AS WELL.