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Title: Lesson 4 Molecular Shapes Learning Objectives: Model, draw, and explain the shapes of molecules with 2-4 charge centres around a central atom HL.

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Presentation on theme: "Title: Lesson 4 Molecular Shapes Learning Objectives: Model, draw, and explain the shapes of molecules with 2-4 charge centres around a central atom HL."— Presentation transcript:

1 Title: Lesson 4 Molecular Shapes Learning Objectives: Model, draw, and explain the shapes of molecules with 2-4 charge centres around a central atom HL Only: Model, draw, and explain the shapes of molecules with 5 and 6 charge centres around a central atom

2 Main Menu VSEPR – a brief introduction  Valence Shell Electron Pair Repulsion (aka ‘vsepr’)  Pairs of electrons around an atom repel each other – this determines a molecule’s shape  Pairs of electrons are known as ‘charge centres’ and include both:  The electrons in a covalent bond  a double/triple bond only counts as one charge centre!  Lone pairs / non-bonding pairs  Example: ethyne  The carbon has two charge centres (the C-H bond and the C  C bond)  They push as far away from each other as possible making a 180 o bond angle

3 Main Menu Drawing shapes in three dimensions:  Draw as many atoms as you can in the same plane ‘flat’ on the paper  Use solid wedges to show atoms coming out of the plane of the paper towards you  Use dashed wedges to show atoms going back into the plane of the paper away from you

4 3D-SHAPES OF MOLECULES Explained and predicted using V.S.E.P.R. theory. Electron pairs repel each other and GET AS FAR APART AS POSSIBLE to minimise repulsions. (VALENCE SHELL ELECTRON PAIR REPULSION) If bond pairs (BP) ONLY present,  totally symmetrical shape If LP also present  slight distortion of symmetrical shape [no lone pairs, (LP)]

5 VSEPR Summary Repulsion applies to electron domains Number of domains around central atom determines the geometrical arrangement. Shape determined by angles between bonded atoms Lone pairs have a higher concentration of charge (not shared between 2 atoms) so have more repulsion. lone pair – lone pair > lone pair- bonded pair > bonded par – bonded pair

6 6 of 33© Boardworks Ltd 2009 Shapes of molecules

7 7 of 33© Boardworks Ltd 2009 Working out basic shapes of molecules

8 8 of 33© Boardworks Ltd 2009 Molecular shape calculations

9 CN = COORDINATION NUMBER of central atom SYMMETRICAL MOLECULES Have at least THREE planes of symmetry at 90 . BP = NUMBER OF BOND PAIRS ON CENTRAL ATOM LP = NUMBER OF LONE PAIRS ON CENTRAL ATOM = number of atoms bonded to central atom In the following slides:

10 MOLECULE AND FORMULA Beryllium Fluoride Beryllium Fluoride DOT-AND-CROSS DIAGRAM SHAPE and BOND ANGLE(S) BP LP CN F F BeF 2 Be º LINEAR NB Be does NOT have an octet of e - F F Be

11 StopwatchStopwatch Graph HomeGraphHome ADDING ANOTHER ATOM - ANIMATION

12 MOLECULE AND FORMULA Boron Trifluoride Boron Trifluoride DOT-AND-CROSS DIAGRAM SHAPE and BOND ANGLE(S) BP LP CN..... F F F BF 3 B º TRIGONAL (or TRIANGULAR) PLANAR NB B does NOT have an octet of e - F F F B

13 StopwatchStopwatch Graph HomeGraphHome ADDING ANOTHER ATOM - ANIMATION

14 MOLECULE AND FORMULA Tetrafluoromethane Tetrafluoromethane DOT-AND-CROSS DIAGRAM SHAPE and BOND ANGLE(S) BP LP CN F F F F CF 4 C F F F F C 109.5º TETRAHEDRAL

15 MOLECULE AND FORMULA PhosphorusPentafluoride DOT-AND-CROSS DIAGRAM SHAPE and BOND ANGLE(S) BP LP CN PF º TRIGONAL BIPYRAMID P F F F F F NB P does NOT have an octet of e - F F F F F P 120º

16 MOLECULE AND FORMULA Sulphur Hexafluoride Sulphur Hexafluoride DOT-AND-CROSS DIAGRAM SHAPE and BOND ANGLE(S) BP LP CN SF OCTAHEDRAL F F F F F F S NB S does NOT have an octet of e - 90º S F F F F F F

17  180 o, 120 o, o or 90 o Symmetrical molecules  NO LP on central atom, only BP belonging to single bonds. BP are identical  REPEL each other equally.  SYMMETRICAL shape SUMMARY :

18 REGULAR SHAPES Molecules, or ions, possessing ONLY BOND PAIRS of electrons fit into a set of standard shapes. All the bond pair-bond pair repulsions are equal. All you need to do is to count up the number of bond pairs and chose one of the following examples... C 2LINEAR 180ºBeCl 2 3TRIGONAL PLANAR 120ºAlCl 3 4TETRAHEDRAL 109.5ºCH 4 5TRIGONAL BIPYRAMIDAL 90º & 120º PCl 5 6OCTAHEDRAL 90ºSF 6 BOND PAIRS SHAPE ANGLE(S)EXAMPLE A covalent bond will repel another covalent bond

19  “normal” (symmetrical) angles reduced by about 2  per LP. LP are NEARER to the nucleus than BP  BP pushed closer to each other by stronger LP repulsions  modifications of trigonal, tetrahedral, trigonal bipyramid and octahedral shapes What happens if LP are present? LP by BP repulsions LP by LP repulsions BP by BP repulsions  LP are closer to BP than BP’s are to eachother and LP are even closer to each other INCREASING strength of repulsions

20 20 of 33© Boardworks Ltd 2009 Effect of lone pairs on shape The number of lone pairs in a molecule is calculated by subtracting the number of bonding pairs from the total number of electron pairs in the outer principal energy level. The shape of a molecule with lone pairs is based on the basic shape for the total number of outer electron pairs, but with a lone pair replacing one of the bonds. tetrahedralpyramidalV-shaped replacing one bonding pair with a lone pair replacing another bonding pair with a lone pair

21 21 of 33© Boardworks Ltd 2009 Effect of lone pairs on bond angles

22 MOLECULE AND FORMULA Ammonia Ammonia DOT-AND-CROSS DIAGRAM SHAPE and BOND ANGLE(S) BP LP CN H H H NH 3 N º TRIGONAL PYRAMID H H H N.. modified tetrahedron

23 MOLECULE AND FORMULA Water Water DOT-AND-CROSS DIAGRAM SHAPE and BOND ANGLE(S) BP LP CN 105º H H O H H H2OH2O O V-SHAPE (BENT)

24 Modifications of TRIGONAL PLANAR shape 3 BP ; 0 LP 120º 2 BP ; 1 LP 118º V-shape LP pushes BP closer

25 Modifications of TETRAHEDRAL shape 4 BP ; 0 LP 109.5º 3 BP ; 1 LP 107º 2 BP ; 2 LP 105º like AMMONIA like WATER Trigonal pyramid V - shape LP pushes BP closer 2LP push BP even closer

26 Modifications of TRIGONAL BIPYRAMID shape 5 BP ; 0 LP 90º and 120º 4 BP ; 1 LP 88º and 118º See-saw 120º Trigonal 180º Linear 2 BP ; 3 LP3 BP ; 2 LP

27 Modifications of OCTAHEDRAL shape 6 BP ; 0 LP 90º 5 BP ; 1 LP 88º Umbrella 90º Square Planar 4 BP ; 2 LP 88º T-shape 180º Linear 3 BP ; 3 LP2 BP ; 4 LP

28 Shape of CF 3 + ? CGp 4  4 e - 3F3 bonds  3 e - Charge  - 1 e - Total number e- around C =6  3 e- pairs  3 BP + 0 LP  symmetrical  trigonal planar 120º F F F C + 1+

29 Shape of SCl 4 ? SGp 6  6 e - 4Cl4 bonds  4 e - Charge  0 e - Total number e- around S =10  5 e- pairs  4 BP + 1 LP  modified trigonal bipyramid 0 88º 118º Cl S :  see-saw shape

30 Shape of SeF 4 2- ? SeGp 6  6 e - 4F4 bonds  4 e - Charge  2 e - Total number e- around Se =12  6 e- pairs  4 BP + 2 LP  modified octahedron 2- 90º S F F F F 2-..  square shape

31 Draw the dot-and-cross diagram for each of the following molecules. Then predict and DRAW the 3D shapes. State the approximate bond angles in each case. (a) SiCl 4 (b) H 2 S (c) XeF 4 (d) PCl 3 (e) CH 3 + (f) XeF 2 (g) PCl 6 - (h)ICl 5 (i) PCl 4 + (j) AlCl 3

32 (a)SiCl 4 (b)H 2 S (c)XeF 4 (d)PCl 3 (e)CH 3 + (f)XeF 2 (g)PCl 6 - (h)ICl 5 (i) PCl 4 + (j) AlCl 3 Si has 4BP and 0 LP  TETRAHEDRAL  109.5° S has 2BP and 2 LP  V-SHAPED  105° Xe has 4BP and 2 LP  SQUARE-PLANAR  90° P has 3BP and 1 LP  TRIGONAL PYRAMID  107° C has 3BP and 0 LP  TRIGONAL PLANAR  120° Xe has 2BP and 3 LP  LINEAR  180° P has 6BP and 0 LP  OCTAHEDRAL  90° I has 5BP and 1 LP  UMBRELLA  88° P has 4BP and 0 LP  TETRAHEDRAL  109.5° Al has 3BP and 0 LP  TRIGONAL PLANAR  120°

33 H H H H CC Because of mutual repulsions, the three bonds are arranged as far apart as possible. 118º 121º This creates a TRIGONAL shape with the bonds about 120º apart. However, the 4 electrons of the C=C bond (compared to the 2 electrons of the C-H bonds) cause stronger repulsions. C-H bonds “pushed” slightly closer  118º Because of the structure of the C=C bond, the six atoms involved with, or bonded to, a C=C bond ALL LIE IN THE SAME PLANE SHAPES OF ALKENE MOLECULES eg Ethene

34 121º H H CH 3 H CC H H3CH3C H CC 118º 121º 118º Similarly, for : Propene But-2-ene NB Other atoms (i.e. H 3 ) are NOT in same plane

35 THE SHAPES OF CARBON DIOXIDE and SULPHUR DIOXIDE C O O Two equivalent double bonds (2 e- bond pairs) to C and no lone pairs 2 double bonds get as far apart as possible  linear molecule O=C=O 180º S O O Two equivalent double bonds (2 e- bond pairs) to S and one lone pair 2 double bonds and lone pair get as far apart as possible  V-shaped molecule S SS 120º

36 36 of 33© Boardworks Ltd 2009 Shapes of molecules activity

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