Presentation on theme: "Chapter #10 - The Shapes of Molecules"— Presentation transcript:
1 Chapter #10 - The Shapes of Molecules Depicting Molecules and Ions with Lewis StructuresUsing Lewis Structures and Bond Energies to CalculateHeats of ReactionValence-Shell Electron-Pair Repulsion (VSEPR)Theory( VB ) Theory and Molecular ShapeMolecular Shape and Molecular Polarity
2 Lewis Structures1) Only the valence electrons appear in a Lewis structure.2) The line joining two atoms represents a pair of electrons shared between two atoms.single bond - two shared electrons, one linedouble bond - four shared electrons, two linestriple bond - six shared electrons, three lines3) Dots placed next to an atom represent nonbonding electrons.
3 Lewis Structures of the Elements by Group in the Periodic Table IIIIIIVVVI..... ...H...B. .. .C..OBe.N. ...Li. ..Mg... ...NaAl..... ..SiP. .S... ..CaVIIIVII. .. .. .. .... .. .. .. .. .. .. .FClHeNeAr. .. .. .. .
5 Writing Lewis Structures for molecules with one central atom. Problem: Write a Lewis structure for the molecule CHCl3, Chloroform,a molecules that has been used to put people to sleep!Solution: Step 1: Place the atoms next to each other with carbon in thecenter, since it is the lowest element in a group with more that oneelectron. Place the others around the carbon in the four locations.Step 2: Count valence electrons.Step 3: Draw single bonds between the atoms, and subtract 2 electronsper bond. 26 electrons - 8 electrons = 18 electrons.Step 4: Distribute the remaining electron in pairs beginning with thesurrounding atoms.Check:[1xC(4e-)] [1xH(1e-)] [3xCl(7e-)] = 26 electrons1.CClHCClHCClH3.4...................
6 Lewis Structures of Simple Molecules - I . ... .. ... .H Cl.. .. .. .. ..F FH H. .. .Hydrogen ChlorideMolecular Fluorine. .H H. .. .C CH.. .. .F F. .. .H H. .H F. .H HHydrogen FluorideEthaneMolecularHydrogen--. .. .-. .. .. .. .. ..ClMg+2Cl. .. ... .Na+Cl. .MagnesiumChlorideSodium Chloride
7 .. .. .. .. Writing lewis Structures for Molecules with more than one Central atom!Problem: Write the Lewis structure for Hydrogen peroxide(molecularformul, H2O2 ) an important household bleech.Solution: Step 1. Place the atoms in the best geometry, with thehydrogen atoms having only one bond, they are on the ends or outside,and oxygen can have up to two bonds so put them in the middle.H O O HStep 2. Find the sum of electrons: [2 x H(1e-)] + [2 x O(6e-)] = 14e-Step 3. Add single bonds and subtract 2e- for each bond:H - O - O - H14e- - 6e- = 8e-Step 4. Add the remaning electron in pairs around the oxygen atoms asHydrogen can only have two!....H - O - O - H....Check: oxygen has an octet of 8e-and hydrogen ha sit’s two electrons.
8 Writing Lewis Structures for Molecules with Multiple Bonds. Problem: Write lewis structures for Oxygen and Acetylene(C2H2):Plan: We begin with the first 4 steps we have done: placing atoms,counting electrons, placing single bonds, and completing octets, and ifneeded we finishas follows by placing multiple bonds in the molecules.Solution:......a) For oxygen: O O - O....Change one of the lone pairs to a bonding pair. The oxygen on theright has an octet of electrons, while the oxygen aton on the leftonly has six electrons, so we convert the lone pair to another bondingpair between the two oxygen atoms.......O O......b) For Acetylene: C2H2H - C - C - HNeither of the carbon atoms has an octet, or if they are placed aroundone atom, the other has only 4! so no octet! Therefore place both pairsinto forming multiple bonds, a tripple bond!H - C C - H
9 Lewis Structures of Simple Molecules - II ....HCHO......HCH..Cl ClHCl2 ChlorineC2H4O2 Acetic AcidCH4 Methane..........OCO.........Cl..CO2 Carbon Dioxide....CO.........ClCCl......CO Carbon MonoxideOClHHH2O WaterHydrogen OxideCCl4 Carbon Tetrachloride
10 Lewis Structures of Simple Molecules - III Multiple Bonds. .N N. .NitrogenN2. .H C NH C C HHydrocyanic acidHydrogen Cyanide : HCNAcetylene : C2H2Molecular Oxygen : O2Ethylene : C2H4. .O OHHC C. .O O. .HH. .
11 Writing Lewis Structures - IV Step 1) Place the atoms relative to each other: For compounds offormula ABn , place the atom with the lower group number inthe center, the one that needs more electrons to attain an octet.In NF3 (nitrogen trifluoride), the N (Group 5A) has fiveelectrons so it needs three, whereas F (Group 7A) has seven soit needs only one; thus, N goes in the center with the threeF atoms around it.Step 2) Determine the total number of valence electrons available:For molecules, add up the valence electrons of all atoms (thenumber of valence electrons equals the A-group number). InNF3, N has five valence electrons, and each F has seven. Forpolyatomic ions, add one e - for for each negative charge, orsubtract one e - for each positive charge.
12 Writing Lewis Structures - V Step 3) Draw a single bond from each surrounding atom to thecentral atom, and subtract two valence electrons for eachbond. There must be at least a single bond between bondedatoms.Step 4) Distribute the remaining electrons in pairs so that eachatom obtains eight electrons (or two for H). Place lone pairson the surrounding (more electronegative) atoms first to giveeach an octet. If electrons remain, place them around the centralatom. Then check that each atom has 8e -.
13 Lewis Structures of Simple Molecules - VI HHH..CH4HCC..OHMethaneHHEthyl Alcohol (Ethanol)........CF..OK+....Cl..........O..OCF4..KClO3Potassium ChlorateCarbon Tetrafluoride
14 Lewis Structures of Simple Molecules - VII HHHN. .HH. .. .NNAmmoniaHHCH+. .O. .HNHUreaHAmmonium Ion
18 Lewis Structures of Simple Molecules Resonance Structures -IIINitrate......O..N....O....O......O....ON....N......O....O........OO
19 Lewis Structures of Simple Molecules -VIII Determine the Lewis structure of molecular Nitrogen, N2N2 is a covalent compound.There are 10 valence electrons.N-N use 2 e-, leaving 8 around the 2 atoms.Three pairs are placed around one atom, leaving 1 pair.Provisional structure:N N Calculate FCFormal ChargeN = 5 valence -(1 bonding+ 2 nonbonding) = +2+ 6 nonbonding) = -2Move electrons in to make a triple bond.N N. .. .. .. .. .. .
20 Lewis Structures for Octet Rule Exceptions ......BCl........F........FCl....FEach Chlorine atom has8 electrons associated.Boron has only 6!Each Fluorine atom has8 electrons associated.Chlorine has 10 electrons!.............N....ClBeCl......OOEach Chlorine atom has8 electrons associated. TheBeryllium has only 4 electrons.NO2 is an odd electron atom.The nitrogen has 7 electrons.
25 Calculating H from Bond Energies - I Problem: Using the bond energies in Table 9.2, calculate the H ofthe reaction between methane and chlorine and fluorine togive freon-12 (CCl2F2).Plan: Look up the bond energies of the reactants and products intable 9.2, and subtract the product bands from the reactant bonds.Solution:CH4 (g) +2 Cl2 (g) + 2 F2 (g) CF2Cl2 (g) + 2 HF(g) + 2 HCl (g)Reactant bonds broken:For Methane : mol C - H bondsFor Molecular Chlorine : 2 mol Cl - Cl bondsFor Molecular Fluorine : mol F - F bondsProduct bonds formed:For Freon - 12 : mol C - F bonds , 2 mol C - Cl bondsFor HF : mol H - F bondsFor HCl : mol H - Cl bonds
26 Calculating H from Bond Energies - II Solution cont.Reactant bonds broken:4 mol C - H bonds = 4 mol x 413 kJ/mol = 1652 kJ2 mol Cl - Cl bonds = 2 mol x 243 kJ/mol = kJ2 mol F - F bonds = 2 mol x 159 kJ/mol = kJH0bonds broken = _________Product bonds formed:2 mol C - F bonds = 2 mol x 453 kJ/mol = kJ2 mol C - Cl bonds = 2 mol x 339 kJ/mol = 678 kJ2 molH - F bonds = 2 mol x 565 kJ/mol = kJ2 mol H - Cl bonds = 2 mol x 427 kJ/mol = 854 kJH0bonds formed = ________H0rxn = H0bonds broken H0bonds formedH0rxn = _______ - _________ = _________
27 Fig 10.4 (P 375)Two Three Four Five SixNumber of Electron Groups
30 AX2 Geometry - Linear............Molecular Geometry =Linear ArrangementClBeClBeCl21800GaseousBeryllium Chloride is an example of a molecule in which thecentral atom - Be does not have an octet of electrons, and is electrondeficient.Other alkaline earth elements also have the same valence electronconfiguration, and the same geometry for molecules of this type.Therefore this geometry is common to group II elements.........OCOCO21800Carbon Dioxide also has the same geometry, and is a linear molecule,but in this case, the bonds between the carbon and oxygens are doublebonds.
32 AX3 Geometry - Trigonal Planar ............All of the Boron Family(IIIA)elements have the samegeometry. Trigonal Planar !FFBF3BBoron Trifluoride1200......FAX2E SO2..-........OSO....NO3-..1200..N........OOThe AX2E molecules have a pair ofElectrons where the third atomwould appear in the space around thecentral atom, in the trigonal planargeometry.1200Nitrate Anion
52 Using VSEPR Theory to Determine Molecular Shape 1) Write the Lewis structure from the molecular formula to see therelative placement of atoms and the number of electron groups.2) Assign an electron-group arrangement by counting all electrongroups around the central atom, bonding plus nonbonding.3) Predict the ideal bond angle from the electron-group arrangementand the direction of any deviation caused by the lone pairs or doublebonds.4) Draw and name the molecular shape by counting bonding groupsand non-bonding groups separately.
54 Predicting Molecular Shapes Problem: Determine the molecular shape and ideal bond angles for:a) NCl3 b) COCl2Solution:
55 Predicting Molecular Shapes Problem: Determine the molecular shape and ideal bond angles for:a) NCl3 b) COCl2Solution: a) for NCl31) Write the Lewis structure:........NCl..2) Assign the electron arrangement:Four electron groups around N,( three bonding, and one non-bonding), so we have the tetrahedralarrangement.3) For the tetrahedral arrangement, the ideal angle is Sincethere is one lone pair, the actual bond angle should be less than4) Draw and name the molecular shape:....NNCl3 Has a trigonal pyramidal shape......Cl....Cl......Cl
56 ....Solution:b) for COCl21) Write the Lewis structure:O........Cl..C..Cl2) Assign the electron-group arrangement: three electron groups aroundcarbon ( two single, and one double bond ) which gives the trigonalplanar arrangement.3) Predict the bond angles: the ideal angle is 1200, but the double bondbetween the Carbon and the oxygen should compress the Cl - C - Clbond angle by repelling the chlorine atoms, and the bonds betweenthem and the carbon atom.4) Draw and name the molecular shape:....O124.50....C......Cl..Cl1110
57 Predicting Molecular Shapes with five or six Electron GroupsProblem: Determine the molecular shape and predict the bond angles( relative to the ideal angles ) of (a) AsI5 (b) BrF5Solution:..........I(a) 1)Lewis structure for AsI5:........II2) Electron grouparrangement with fivegroups, this is thetrigonal bipyramidalarrangement...As................I..II....I900......1200As..I..3) Bond angles: since all the groups andsurrounding atoms are indentical, thebond angles are ideal: 1200 between equatorialgroups and 900 between axial and equatorial groups...I......I4) Molecular arrangement: Trigonal bipyramidal
58 ..........b) BrF51) Lewis structure for BrF5:2) Electron Group arrangement6 electron groups- octhedral!3) Bond angles: Lone pair shouldmake all angles less than 900.4) Molecular shape: one lone pair,and five bonding pairs give square pyrimidal:F......F....F..........FBr..F..........F......F..F............FBr..F
67 .. .. .. Predicting The Polarity of Molecules -I Problem: From electronegativity, and their periodic trends, predictwhether each of the following molecules is polar and show the directionof bond dipoles and the overall molecular dipole.(a) Phosphine, PH3(b) Carbon Disulfide, CS2 (atom sequence SCS)(c) Auminum Chloride, AlCl3Plan: First we draw and name the molecular shape. Then using relativeEN values, we decide on the direction of each bond polarity. Finally wedecide upon the polarity of the molecule based upon the geometry.Solution:......(a)PPPHHHHHHHHHMolecular DipoleBond dipolesMolecular shape