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BONDING TOPIC 4. Terms Covalent Bonding Bonds –Breaking them takes energy –Making them gives off energy.

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Presentation on theme: "BONDING TOPIC 4. Terms Covalent Bonding Bonds –Breaking them takes energy –Making them gives off energy."— Presentation transcript:

1 BONDING TOPIC 4

2 Terms Covalent Bonding Bonds –Breaking them takes energy –Making them gives off energy

3 Exothermic –More energy is given off than put in Endothermic –More energy is absorbed than given off Intramolecular Forces –Forces within molecules (ionic, covalent and metallic) Intermolecular Forces (IMF) –Forces between particles

4

5 Metal: K Non-Metal: Cl Ionic Bonding If the electronegative difference between the atoms involved is =>1.8 –There are always exceptions to this rule! Will conduct electricity in its molten or aqueous state (This test proves ionic) Less e - = Less e - repulsionMore e - = e - more repulsion. +-

6 Intramolecular Forces Ionic Bonding Just use the valence shell Be sure to include square brackets and charge after electron exchange. Drawing Ionic Bonding Na Cl X Electrons are in pairs + Special Note: The ionic bond is the electrostatic attraction between oppositely charged ions! - Lewis Dot Diagram

7 Lewis Diagrams Lewis Dot diagrams us the atoms valance shell electrons Combine C Cl Be Br Al F Fe Cl Mg O

8 Intramolecular Forces NaCl When in molten or aqueous state, ionic substances WILL conduct electricity, by the movement of (+) and (-) ions. This is different from how METALS conduct electricity! Decomposition 2Na + (aq) + 2Cl - (aq)  2Na (s) + Cl 2(g) CATHODE (-) ANODE (+) Conductivity is FINITE

9 Metal: K Non-Metal: Cl Ionic Compounds No bonds are made!!! Static attractions holds them together. (opposites attract) When a force is applied, ionic compounds will make a clean break. Force Like charges repel Giant Ionic Lattices Physical characteristics Hard and brittle Solid doesn’t conduct Electricity More soluble in water than other solvents High MP and BP Cation Anion

10 Table Salt NaCl Cubic or Isometric Giant Ionic Lattices

11 Cassiterite SnO 2 Tetragonal Giant Ionic Lattices

12 Agagonite CaCO 3 Orthorhombic Also found in mollusk shells and coral Giant Ionic Lattices

13 Beryl Be 3 Al 2 (SiO 3 ) 6 Hexagonal Giant Ionic Lattices

14 Quartz SiO 2 Trigonal Giant Ionic Lattices

15 Beryl Be 3 Al 2 (SiO 3 ) 6 Ionic Bonding Giant Ionic Lattices

16 Copper(II) Sulfate CuSO 4 Triclinic Giant Ionic Lattices

17 Intramolecular Forces Multiple Ions Transition metals can have multiple ions. Ones you should know. Transition Metals Cu 2+ Cu + Fe 2+ Fe 3+ Copper(I) Oxide Iron (III) Oxide Iron(II) Oxide Copper(II) Oxide

18 Ions Polyatomic Ions Be sure to review your polyatomic ions!!! Reminder NO 3 - SO 4 -2 PO 4 -3 OH - CO 3 -2 NH 4 + HCO 3 -

19 Covalent Bonding Topic 4

20 Intramolecular Forces COVALENT BONDING If the electronegative difference between the atoms involved is <1.8 Will NOT conduct electricity Electrons are shared Covalent Bonding X H H Cl Differences |3-2.1| =0.9 Special Note: The covalent bond is the electrostatic attraction between pairs of e - and positively charged nuclei!

21 Questions For ionic compounds to form the valance shells of both metal and non-metal must be full!! Na Ca Li Na Cl O CO 3 SO 3 What is the chemical formula? What is the names for each? K+ NO 3 Review

22 Intramolecular Forces COVALENT BONDING Structural formula Lewis structure Covalent Bonding H H C H H H H H H Cl X X X X H H H H H H H H

23 Intramolecular Forces COVALENT BONDING 1) Sum all valence e - 2) Subtract 2e - for every bond 3) Place e - around periphery atoms to form octets. The remaining around central atom 4) All atoms MUST be paired!!!!!! Lewis Structures H2OH2O H H O - 4 = 4 Hydrogen can only hold 2e - remaining must be paired on Oxygen

24 Intramolecular Forces Covalent Bonding Draw the following Lewis structures H 2 Cl 2 O 2 N 2 HCNC 2 H 6 C 2 H 4 C 2 H 2 Lewis Structures HL: PCl 5, PCl 4 +, PCl 6 - and XeF 4

25 Intramolecular Forces Covalent Bonding Coordinate or dative covalent bonds When both e - are shared from the same atom. (Not one from each as before) Occurs when a non bonding e - pair donates an e - to an e - deficient atom. Special Lewis Structures H H N H + Electrophile H Lone pair of e - +

26 Intramolecular Forces Covalent Bonding Draw the following Lewis structures CO H 3 O + Special Lewis Structures

27 Intramolecular Forces CO 3 2- More bonds = more strength & shorter bonds Resonance structures –Bond length is longer than a double bond but shorter than a single bond Length, Strength & Hybrid Resonance C OO O 2- C OO O C OO O Don’t forget to show the e - pairs!!

28 Intramolecular Forces CO 3 2- Compare the two molecules Ethyne has stronger and shorter bonds C=O bond is stronger and shorter due to Oxygen being more electronegative Length & Strength Ethene R = Functional Group C ROH O C H H C H H Carboxylic Acid C H C H Ethyne

29 Intramolecular Forces Covalent Bonding Non-Metals are fighting for e - Atom with larger electronegativity will hold the e - closer to itself. Atoms become slightly charged. Bond Polarity Cl X H H δ-δ- δ+δ+ Dipole Moment

30 Intramolecular Forces Covalent Bonding BF 3 Actual structure: Boron is e - deficient This is known because of its reactivity towards electron rich molecules such as NH 3 CNOF all obey the octet rule. Exceptions to the Octet Rule FF B F

31 SO 4 2- Single bonds (8 e - around S) Double bonds (12 e - around S) Formal Charge = (# valence e - on free atom) – (# valence e - assigned to the atom in the molecule) (Valence e - ) assigned = (# lone pair e - ) + ½ (# of shared e - ) 1) Molecules attempt to achieve Formal Charge as close to 0 as possible. 2) Any negative Formal charge will reside on most electronegative atom. Formal Charge Covalent Bonding Intramolecular Forces

32 Covalent Bonding VSEPR ( Valence Shell Electron Pair Repulsion ) Paired e - attempt to get as far away from each other as possible. Multiple bonds still count only as 1 pair!! VSEPR (shape) OO C FF B F 3 Pairs of e o 2 Pairs of e o O O C O 2-

33 Intramolecular Forces Covalent Bonding Tetrahedral Lone pair e - have increased charge density and require more room More repulsion from lone pair will decrease bond angle. VSEPR H H C H H 4 Pairs of e o H H N H Lone pair 107 o H O H Lone pair o

34 Intramolecular Forces Covalent Bonding Predict the shape AND bond angles H 2 SPbCl 4 H 2 COSO 2 NO 3 - PH 3 NO 2 - NH 2 - POCl 3 CO 2 Home Work

35 HL VSEPR MoleculeShapeTotal valance electrons Bond PairsNon Bonding Electron pair Angle BeF 2 Linear180 BeF 3 Triangular Planar 120 SO 2 Bent117 CH 4 Tetrahedral109.5 NH 3 Trigonal pyramidal 107 H2OH2OBent104

36 HL VSEPR MoleculeShapeTotal Valance electrons Bond PairsNon Bonding Electron pair Angle PCl 5 Triangular Bipyramidal 90 & 120 SF 4 Seesaw90 & ≈117 T-Shape90 CF 6 Octahedral180 IF 5 Square Pyramidal 90 XeF 4 Square Planar ≈88

37 Expanded Valance Shell (14.1) Molecules with more than 8 electrons Electron promotion:

38 Dipole Moment Covalent Bonding Polarity effects state change (physical change) Unequal sharing causes a dipole moment to form Q: Why is BF 3 non-polar whereas PF 3 is polar? Molecule Polarity (4.2.6) δ+δ+ 2δ-2δ- δ+δ+ H O H Cl H δ-δ- δ+δ+ Non Polar H H C H δ-δ- δ+δ+ H H C Cl

39 Hybridization (14.2.2) Sigma bond: σ (single bond) – Axial overlap of orbital’s Cl 1s 1 H H 2p x 2 p y 2 pz2pz2

40 Hybridization (14.2) Sigma bond: σ (single bond) – Axial overlap of orbital’s Cl

41 Hybridization (14.2) Pi bond: π(Double bond, one σ bond) – Parallel overlap of orbital’s N N N N O O O O

42 Hybridization (14.2.3) Hybridization electron promotion – New Orbital sp 3 2p x 2 p y 2 pz2pz2 2s 2 Ground State C C Excited State 4 Equal orbital`s capable of holding a maximum of 2 electrons each

43 Hybridization (14.2) How to determine Hybridized orbital`s – Look at the shape ShapeHigh Electron dense regions Hybridized Orbital sp2Tetrahedral Sp 2 3Trigonal planar sp 3 4Linear sp 3 d5Trigonal bi-pyramidal sp 3 d 2 6Octahedral (Square bi-pyramidal)

44 Allotropes Giant Covalent 1) Diamond (Tetrahedron, localized e - ) –Very hard and does not conduct electricity 2) Fullerenes (C 60 ) Hexagonal and pentagonal rings –Nanotubes Carbon C CC C C

45 Allotropes Giant Covalent 3) Graphite (Planar, delocalized e - ) –Weak pi bonding between sheets cause it to conduct electricity and be slippery. –Bonds are shorter than a tetrahedral due to the pi bonding Carbon C C C C C C HL: sp hybrid Delocalized electrons able to move Weak Pi Bonds

46 C6H6C6H6 C6H6C6H6 Benzene (14.3) C C C C C C Pi bonds overlap allowing for electrons to be delocalized over the entire molecule.

47 Intramolecular Forces Silicon Tetrahedron Configuration Similar to diamond Silicon Si

48 Intramolecular Forces Quartz Single bonds formed between Oxygen to satisfy the octet. HL: Less overlap in the P-sub orbital due to atomic size difference therefore Pi bonds do not form. Silicon & Silicon dioxide Si OO O O SiO 2 but based on a network of SiO 4

49 Metallic Bonding Topic 4

50 Intramolecular Forces Metallic Bonding In solid state Outer e - are delocalized and free to move about Bond is a result of electrostatic attraction between Fixed positive metal ions and delocalized e - Metallic Bonding Conductivity is INFINITE - - Sea of electrons

51 Metallic Bonding The ability for a material to be pounded into thin sheets. Aluminum Foil Swords and Folding Malleability Physical Properties

52 Ductility Metallic Bonding The ability for a material to be pulled into wire Or in this case extruded into a wire Physical Properties Electrons have been excluded

53 Metallic Bonding Because e - can move easily it can conduct energy. (Heat or electricity) MP related to attractive force (between atoms) 1) Size of Cation(+) 2) # of valence e - 3) Atom packing Size increases MP decreases: Giant Covalent substances have very high mp Physical Properties

54 Metallic Bonding Same element but different structure Carbon Diamond Graphite Fulluron Allotrops

55 INTERMOLECULAR FORCES Topic 4

56 Intermolecular Forces IMF Van der Waals Forces van der Waals’ ForcesIntermolecular Forces (4.3.1) Charge Induction      

57 Intermolecular Forces IMF Polar molecules (polar covalent) have slightly charged ends Opposites attract. Large electronegative difference = stronger attraction. Dipole-Dipole (4.3.1) H H C Cl   H H C  

58 Intermolecular Forces IMF Hydrogen Bonding (F, O or N bonded to H) Due to small size and high electronegativity of non metals Creates a large charge difference Basically a super strong dipole-dipole bond van der Waals’ ForcesHydrogen Bonding (4.3.1)   H O H    H O H 

59 Get a picture of group 4,5,6,7 boiling points for hydrides Key question is why does water have an abnormally high BP? H bonding with O, F and N IMF Phase change when IMF are overcome Be sure to explain using the words IMF and how they affect the bonds BETWEEN particles. Van der Waals’ Forces are ALWAYS present!!! Boiling Point Trends (4.3.2) Intermolecular Forces

60 Van der Waal’s: Lowest MP, Non polar Butane (C 4 H 10 ) Dipole-dipole: Slightly miscible Propanone C 3 H 6 O Hydrogen Bonding: Miscible with polar substances H 2 O Ionic Bonding: Only conducts electricity when liquid or aqueous. (Decomposition when it does) NaCl Metallic Bonding: Conducts electricity, not water soluble, MP regulated by, valance, size and packing. Fe Giant Covalent: Highest MP, Insoluble in both non- polar and polar solvents. Does not conduct electricity except for graphite. Diamond and Graphite (Allotropes) Physical Properties Increasing Melting Point

61 Bonding Questions Compare the following for B.P HF and HCl H 2 O and H 2 S NH 3 and PH 3 CH 3 OCH 3 and CH 3 CH 2 OH CH 3 CH 2 CH 3, CH 3 CHO and CH 3 CH 2 OH

62 HL Material

63 Hybridization (14.2) Sigma bond: σ (single bond) – Axial overlap of orbital’s

64 Hybridization (14.2) Sigma bond: σ (single bond) – Axial overlap of orbital’s

65 Hybridization (14.2) Sigma bond: σ (single bond) – Axial overlap of orbital’s

66 Hybridization (14.2) Sigma bond: σ (single bond) – Axial overlap of orbital’s

67

68

69 Lattice Formation Where the heat comes from Route 1: A + B + C + E Route 2: F Hess’s law: A + B +C + E = F (-349) + E = -411 E = -787 kJ mol -1

70 Intramolecular Forces NaCl 1) Production of Gaseous atoms 2) Formation of Gaseous ions 3) Production of solid ionic lattice Lattice Enthalpy Na (s) + ½ Cl 2(g)  Na + Cl - or NaCl 1) Na (s)  Na (g) ½ Cl 2(g)  Cl (g) 3) Na + (g) + Cl - (g)  NaCl (s) 2) Na (g)  Na + (g) + e - Cl (g) + e -  Cl - (g)

71 Born-Haber Cycle Na + (g) + Cl - (g) Na (s) + Cl 2(g) Na (g) Cl (g) NaCl (s) ΔH θ I.E. 1 st Ionization of Na +496 kJ mol -1 ΔH θ at Atomization of Na -107 kJ mol -1 ΔH θ at Atomization of Cl +122kJ mol -1 ΔH θ E.A. 1 st electron affinity of Cl -349 kJ mol -1 ΔH θ f Formation of NaCl -411 kJ mol -1 Lattice Enthalpy BA D C E F Exothermic Endothermic

72 OO C Spare Parts O C Cl X H H δ-δ- δ+δ+ H H C H N H H N H + H - - H O H H H C H H C C C C C C


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