# Title: Lesson 4 Molecular Shapes

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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 Only: Model, draw, and explain the shapes of molecules with 5 and 6 charge centres around a central atom

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 180o bond angle

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

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

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

Boardworks AS Chemistry Structure and Shape
Shapes of molecules Teacher notes Students could be reminded that the double bond in carbon dioxide means that, although the shape of the molecule is linear, it contains four bonding pairs, not two.

Working out basic shapes of molecules
Boardworks AS Chemistry Structure and Shape Working out basic shapes of molecules

Molecular shape calculations
Boardworks AS Chemistry Structure and Shape Molecular shape calculations

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

BeF2 2 . Be F F Be 2 Beryllium Fluoride MOLECULE AND FORMULA
DOT-AND-CROSS DIAGRAM SHAPE and BOND ANGLE(S) BP LP CN BeF2 2 . F Be 180º F Be NB Be does NOT have an octet of e- LINEAR 2

TRIGONAL (or TRIANGULAR) PLANAR
MOLECULE AND FORMULA Boron Trifluoride DOT-AND-CROSS DIAGRAM SHAPE and BOND ANGLE(S) BP LP CN BF3 F F B 3 . . . B 120º TRIGONAL (or TRIANGULAR) PLANAR NB B does NOT have an octet of e- 3

CF4 4 C . C F . . F 4 Tetrafluoromethane MOLECULE AND FORMULA
DOT-AND-CROSS DIAGRAM SHAPE and BOND ANGLE(S) BP LP CN CF4 4 F F C 109.5º . . . C 4 TETRAHEDRAL

PF5 5 . . . F P P F 5 Phosphorus Pentafluoride MOLECULE AND FORMULA
DOT-AND-CROSS DIAGRAM SHAPE and BOND ANGLE(S) BP LP CN PF5 F F 90º . . . 5 P P 120º TRIGONAL BIPYRAMID NB P does NOT have an octet of e- 5

SF6 6 F S S F 6 Sulphur Hexafluoride MOLECULE AND FORMULA
DOT-AND-CROSS DIAGRAM SHAPE and BOND ANGLE(S) BP LP CN SF6 F F 6 90º S S NB S does NOT have an octet of e- OCTAHEDRAL 6

SUMMARY : Symmetrical molecules  NO LP on central atom,
only BP belonging to single bonds. BP are identical  REPEL each other equally.  SYMMETRICAL shape  180o, 120o, 109.5o or 90o

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 A covalent bond will repel another covalent bond BOND BOND PAIRS SHAPE ANGLE(S) EXAMPLE 2 LINEAR º BeCl2 3 TRIGONAL PLANAR 120º AlCl3 4 TETRAHEDRAL º CH4 5 TRIGONAL BIPYRAMIDAL 90º & 120º PCl5 6 OCTAHEDRAL 90º SF6

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

Effect of lone pairs on shape
Boardworks AS Chemistry Structure and Shape 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. tetrahedral pyramidal V-shaped Teacher notes Students could be taught the ‘square planar’ molecular shape, as exemplified by xenon tetrafluoride (XeF4). In this structure, four bonding pairs are arranged in the same plane around the central atom with a bond angle of 90°, with a lone pair both above and below the central atom. V-shaped molecules may also be described as “non-linear”. replacing one bonding pair with a lone pair replacing another bonding pair with a lone pair

Effect of lone pairs on bond angles
Boardworks AS Chemistry Structure and Shape Effect of lone pairs on bond angles Teacher notes It may be worth pointing out to students that the dotted line represents a bond extending behind the plane of the screen, and the wedge-shaped bond represents a bond extending in front of the plane of the screen.

NH3 3 .. N . H N 1 . . H 3 Ammonia MOLECULE AND FORMULA
DOT-AND-CROSS DIAGRAM SHAPE and BOND ANGLE(S) BP LP CN NH3 H N .. modified tetrahedron 3 H . . . N 1 107º TRIGONAL PYRAMID 3

H2O 2 O H . O H 2 . . 2 Water MOLECULE AND FORMULA
DOT-AND-CROSS DIAGRAM SHAPE and BOND ANGLE(S) BP LP CN H2O H O .. modified tetrahedron 2 H . . . O 2 105º V-SHAPE (BENT) 2

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

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

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

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

Shape of CF3+ ? C F C Gp 4  4 e- 3F 3 bonds  3 e- Charge 1+  - 1 e-
Total number e- around C = 6 F C +  3 e- pairs  3 BP + 0 LP  symmetrical 120º  trigonal planar

Shape of SCl4 ? Cl S S Gp 6  6 e- 4Cl 4 bonds  4 e- Charge  0 e-
Charge  0 e- Total number e- around S = 10  5 e- pairs  4 BP + 1 LP Cl S : 88º  modified trigonal bipyramid 118º  see-saw shape

Shape of SeF42- ? F S Se Gp 6  6 e- 4F 4 bonds  4 e- 2- Charge
Total number e- around Se = 12 S F 2- ..  6 e- pairs  4 BP + 2 LP  modified octahedron 90º  square shape

Draw the dot-and-cross diagram for each of the following molecules
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 (b) H2S (c) XeF (d) PCl3 (e) CH (f) XeF2 (g) PCl (h) ICl5 (i) PCl4+ (j) AlCl3

SiCl4 H2S XeF4 PCl3 CH3+ XeF2 PCl6- ICl5 (i) PCl4+ (j) AlCl3
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°

H C SHAPES OF ALKENE MOLECULES eg Ethene 121º 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 118º Because of mutual repulsions, the three bonds are arranged as far apart as possible. 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

H CH3 C H H3C CH3 C 121º Similarly, for : 118º Propene 121º But-2-ene
NB Other atoms (i.e. H3) are NOT in same plane

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

Shapes of molecules activity
Boardworks AS Chemistry Structure and Shape Shapes of molecules activity