IB Chemistry Power Points Topic 4 Bondingwww.pedagogics.ca LECTURE Shapes of Molecules
Much taken from AN INTRODUCTION TO BONDING andSHAPES OF MOLECULES Great thanks to JONATHAN HOPTON & KNOCKHARDY PUBLISHING www.knockhardy.org.uk/sci.htm
VALENCE SHELL ELECTRON PAIRREPULSION (VSPER) THEORY “THE SHAPE ADOPTED BY A SIMPLE MOLECULE OR ION IS THAT WHICH KEEPS REPULSIVE FORCES TO A MINIMUM” Bonds are closerMolecules contain covalent together so repulsivebonds. As covalent bonds forces are greaterconsist of a pair of Alelectrons, each bond will Bonds are furtherrepel other bonds. apart so repulsive forces are lessBonds will therefore push eachother as far apart as possible All bonds areto reduce the repulsive forces. equally spaced out as far apartBecause the repulsions are Al as possibleequal, the bonds will also be Note – you mustequally spaced. think of spacing in 3D not just 2DA lone pair of electrons willexert even greater repulsiveforces on bonds.
2 Bonding Pairs- BERYLLIUM CHLORIDE Be Cl Cl Be ClBeryllium - has two electrons to pair up Two covalent bonds are formedChlorine - needs 1 electron for „octet‟ Beryllium still has an incomplete shell BOND PAIRS 2 180° LONE PAIRS 0 Cl Be Cl BOND ANGLE... 180° Geometry ... LINEAR
MOLECULES WITH DOUBLE BONDS Treat as a single region of electron density The shape of a compound with a double bond is calculated in the same way. A double bond repels other bonds as if it was single e.g. carbon dioxide C O O C OCarbon - needs four electrons to complete its shell The atoms share two electronsOxygen - needs two electron to complete its shell each to form two double bonds DOUBLE BOND PAIRS 2 180° LONE PAIRS 0 O C O Double bonds behave exactly as single BOND ANGLE... 180° bonds for repulsion purposes so the shape will be the same as a molecule with Geometry ... LINEAR two single bonds and no lone pairs.
3 Bonding Pairs - ALUMINIUM CHLORIDE Al Cl Cl Cl AlAluminium - has three electrons to pair up ClChlorine - needs 1 electron to complete „octet‟Three covalent bonds are formed; aluminiumstill has an incomplete outer shell. BOND PAIRS 3 LONE PAIRS 0 Cl 120° Cl Al BOND ANGLE... 120° Cl Geometry ... TRIGONAL PLANAR
4 Bonding Pairs- METHANE H H C H C H HCarbon - has four electrons to pair up Four covalent bonds are formedHydrogen - 1 electron to complete shell C and H now have complete shells BOND PAIRS 4 H LONE PAIRS 0 109.5° C H BOND ANGLE... 109.5° H H Geometry ... TETRAHEDRAL
4 regions of electron density - AMMONIA H BOND PAIRS 3 N H H N H LONE PAIRS 1 TOTAL PAIRS 4• The shape is based on a tetrahedron but not all the repulsions are the same• LP-BP REPULSIONS > BP-BP REPULSIONS• The N-H bonds are pushed closer together• Lone pairs are not included in the shape! N H 107° H H ANGLE... 107° SHAPE... PYRAMIDAL
4 regions of electron density - WATER H BOND PAIRS 2 O H H O LONE PAIRS 2 TOTAL PAIRS 4• The shape is based on a tetrahedron but not all the repulsions are the same• LP-LP REPULSIONS > LP-BP REPULSIONS > BP-BP REPULSIONS• The O-H bonds are pushed even closer together• Lone pairs are not included in the shape O 104.5° H O O HH H ANGLE... 104.5° H H SHAPE... BENT
HOW TO DETERMINESHAPES OF IONS H BOND PAIRS 3 PYRAMIDALNH3 N H N LONE PAIRS 1 H-N-H 107° H H +NH4 N + H N + H BOND PAIRS 4 TETRAHEDRAL LONE PAIRS 0 H-N-H 109.5° H H - BOND PAIRS 2 BENTNH2 N H N LONE PAIRS 2 H-N-H 104.5°
5 Bonding Pairs (HL only) - PHOSPHORUS(V) FLUORIDE F P F F F PPhosphorus - has five electrons to pair up FFluorine - needs one electron to complete „octet‟ FFive covalent bonds are formed; phosphorus canmake use of d orbitals to expand its „octet‟ F BOND PAIRS 5 90° LONE PAIRS 0 F 120° P F F BOND ANGLE... 120° & 90° F Geometry ... TRIGONAL BIPYRAMIDAL
6 Bonding Pairs (HL only) - SULPHUR(VI) FLUORIDE F S F F F SSulphur - has six electrons to pair up F FFluorine - needs one electron to complete „octet‟ FSix covalent bonds are formed; sulphur canmake use of d orbitals to expand its „octet‟ F BOND PAIRS 6 90° LONE PAIRS 0 F F S BOND ANGLE... 90° F F Geometry ... OCTAHEDRAL F
HL only - XENON TETRAFLUORIDE F F BOND PAIRS 4 F Xe LONE PAIRS 2 Xe TOTAL PAIRS 6 F F• As the total number of electron pairs is 6, the shape is BASED on an octahedron• There are two possible spatial arrangements for the lone pairs• The preferred shape has the two lone pairs opposite each other F F Xe F F F Xe F F F ANGLE... 90° SHAPE ... SQUARE PLANAR
SUMMARY FUNDAMENTAL SHAPES – no lone pairsMolecules, or ions, possessing ONLY BONDINGPAIRS of electrons fit into a set of standardshapes. All the bond pair-bond pair Crepulsions are equal.All you need to do is to count up the number A covalent bond will repelof bond pairs and chose one of the following another covalent bondexamples... BOND ELECTRON BOND PAIRS GEOMETRY ANGLE(S) EXAMPLE 2 LINEAR 180º BeCl2 3 TRIGONAL PLANAR 120º AlCl3 4 TETRAHEDRAL 109.5º CH4 HL ONLY 5 TRIGONAL BIPYRAMIDAL 90º & 120º PCl5 6 OCTAHEDRAL 90º SF6
Effect of Lone Pairs on Molecular ShapeIf a molecule, or ion, has lone pairs on the central atom, the shapes are slightlydistorted away from the regular shapes. This is because of the extra repulsioncaused by the lone pairs. BOND PAIR - BOND PAIR < LONE PAIR - BOND PAIR < LONE PAIR - LONE PAIR O O O As a result of the extra repulsion, bond angles tend to be slightly less as the bonds are squeezed together.
SUMMARY - CALCULATING THE SHAPE OF IONSThe shape of an ion or molecule is determined by...• calculating the number of electrons in the outer shell of the central species *• pairing up electrons, making sure the outer shell maximum is not exceeded -• calculating the number of bond pairs and lone pairs (regions of e density)• using ELECTRON PAIR REPULSION THEORY to calculate shape and bondangle(s)Note for ions* the number of electrons depends on the charge on the ion* if the ion is positive you remove as many electrons as there are positivecharges* if the ion is negative you add as many electrons as there are negative charges - e..g. for PF6 add one electron to the outer shell of P + for PCl4 remove one electron from the outer shell of P
OTHER EXAMPLES TO TRY 2-SO4 O BOND PAIRS LONE PAIRS O S O- SHAPE ANGLE O-BrF3 F BOND PAIRS LONE PAIRS F Br SHAPE ANGLE FBrF5 BOND PAIRS F F LONE PAIRS Br SHAPE F F ANGLE F
OTHER EXAMPLES TO TRY 2- OSO4 O BOND PAIRS 4 LONE PAIRS 0 O S O- TETRAHEDRAL S O- ANGLE 109.5° O O- O-BrF3 F F BOND PAIRS 3 LONE PAIRS 2 F Br ‟T‟ SHAPED F Br ANGLE <90° F FBrF5 BOND PAIRS 5 F F Br LONE PAIRS 1 F F Br SQUARE PYRAMID F F F F ANGLES 90° <90° F F
TEST QUESTIONSFor each of the following ions/molecules, state the number of bond pairs state the number of lone pairs state the bond angle(s) state, or draw, the shapeBF3SiCl4 +PCl4 -PCl6 2-SiCl6H2S
TEST QUESTIONSFor each of the following ions/molecules, state the number of bond pairs state the number of lone pairs state the bond angle(s) state, or draw, the shapeBF3 3 bp 0 lp 120º trigonal planar boron pairs up all 3 electrons in its outer shellSiCl4 4 bp 0 lp 109.5º tetrahedral silicon pairs up all 4 electrons in its outer shell +PCl4 4 bp 0 lp 109.5º tetrahedral as ion is +, remove an electron in the outer shell then pair up -PCl6 6 bp 0 lp 90º octahedral as the ion is - , add one electron to the 5 in the outer shell then pair up 2-SiCl6 6 bp 0 lp 90º octahedral as the ion is 2-, add two electrons to the outer shell then pair upH2S 2 bp 2 lp 92º bent planar sulphur pairs up 2 of its 6 electrons in its outer shell - 2 lone pairs are left