1 / 77

Types of Chemical Bonds

Types of Chemical Bonds. Covalent bond Electrons shared Both atoms have high affinity Ionic bond Atoms have very different affinities Atom with greater affinity takes electron Forms ions which are held together because opposite charges attract. Types of Chemical Bonds. Metallic bonds

hkinney
Télécharger la présentation

Types of Chemical Bonds

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Types of Chemical Bonds • Covalent bond • Electrons shared • Both atoms have high affinity • Ionic bond • Atoms have very different affinities • Atom with greater affinity takes electron • Forms ions which are held together because opposite charges attract

  2. Types of Chemical Bonds • Metallic bonds • Atoms have weak affinities & few valence electrons • Easily lost electrons are shared among many

  3. Covalent Bonds • Atoms • Have similar electronegativities • Are only a few missing valence electrons • Are usually nonmetals • Share valence electrons

  4. Covalent Bonds • Bonding pair • Electrons shared in the bond • Located between nuclei of atoms • Form area of negative charge, hold atoms together  electrostatic force

  5. Diatomic Elements • Strong affinities, bond quickly • Exist in diatomic, pure form • Ex: H2, N2, O2, F2, Cl2, Br2, I2

  6. Lewis Dot Structure • Shows bond between different atoms • - = 2 electrons • Nonbonding electrons shown as dots

  7. Ionic Bonding • Large difference in electronegativities • Form ionic compounds • Cations and anions held by electrostatic attraction • No distinct molecules • Sum of charges is 0 • Formula unit: ratio of cations to anions  neutral • Forms crystal lattice (orderly arrangement of ions)

  8. Metallic Bonding • Electron sea theory • Metals are a crystal lattice of metal cations in a “sea” of mobile electrons • Electrons shared among many • Delocalized Electrons • Mobile • Hold cations together

  9. 6C Properties of Compounds • Covalent Compounds • Form molecules • Held by weak forces, separated easily • Poor conductors, fairly low melting points • Wide array (range) of colors • EXCEPTION: Network covalent substances • Hardest known substances • Ex: diamonds, silicates, quartz • Hard, brittle crystals

  10. Properties of Compounds • Ionic Compounds • Dense, brittle, hard solids because held by strong electrostatic forces • High melting points • Split or cleave along flat surfaces • Poor conductors because valence electrons are held tightly • Conduct electricity if soluble in water

  11. Properties of Metals & Alloys • Due to delocalized electrons, metals are… • Conduct heat and electricity • Have luster (e- absorb & reemit light) • Malleable (e- act as glue for metal cations)

  12. Orbitals & Valence Bond Theory • Orbitals: • Each holds 2 electrons • Areas where electrons may be found • In different sublevels (s,p,d,f) of energy levels • S – 1 orbital • P – 3 orbitals • D – 5 orbitals • F – 7 orbitals

  13. Orbitals with unpaired electrons overlap  forms a bond • Called valence bond (a model that explains how covalent bonds form)

  14. Sigma bonds (σ) • End to end overlap of spherical s orbitals • Region of high electron density forms on the “line” that connects the two nuclei • Strongest type of covalent bond • Why? Electrostatic forces only pull nuclei together • Ex: H2

  15. Double bonds • End-to-end overlap (sigma bond) and side-to-side overlap (pi bond) • Pi bond: two lobes overlap • Bond is parallel to bond axis (above & below) • 2 regions but 1 orbital! • Double bond = 1 sigma & 1 pi bond

  16. 7B Molecular Geometry • VSEPR: valence shell electron pair repulsion theory • Focuses on location of highest electron densities around central atom • 1 area = Single bond, double bond, triple bond, or unbonded pair of electrons • Charge repulsion determines shape • Areas of negative charge repel to be as far apart as possible

  17. # of regions & shapes • 4 regions • Tetrahedral, 109.5 degrees apart • Ex: CCl4 • Pyramidal also has 4 regions • Ex: NH3 • 3 regions • Trigonal planar • Y shaped on single plane • Ex: CH2O • 2 regions • Linear, 180 degrees apart • Ex: diatomic molecules, CO2

  18. Other (4 Region) Shapes • Bent • Typically 109.5 degree angle but 104.5 degrees in water • Nonbonding electrons repel more than bonding pairs • Linear • Ex: HF

  19. Orbital Hybridization • Valence bond theory fails to explain how C bonds (should form 2 bonds since 1s2s 2p) • Suggested explanation: • C forms 4 hybrid sp orbitals • Hybridization • Process by which new orbitals with equal energies are formed by combining orbitals with different energies

  20. Why hybridize? • Forms larger lobes, which are able to overlap other orbitals more effectively & form stronger bonds (more stable)

  21. Polar Covalent Bonds • Electrons unevenly shared • Creates partial charges • Affects chemical properties • Polarity shown by Lewis dot structure symbol (+ ) • Arrow points to more electroneg. atom • Remember: greater ΔEN = more polar bond 

  22. Dipole Moment • Molecules A through D are shown below. Determine whether each one is polar or nonpolar. • A – Nonpolar • B – Polar • C – Polar • D – Nonpolar D A B C

  23. If a molecule has polar bonds, will it be polar? • For asymetrical molecules: YES • One end is + and other is – • Shapes: bent & pyramidal (always polar) • For symetrical molecules: Maybe • Tetrahedral, trigonal planar, & linear shapes are… • Not polar if outermost atoms are the same • Polar if outermost atoms are different

  24. Oxidation Numbers & Formulas • Oxidation Numbers (ON) • Represent number of electrons element must gain/lose to return to neutral • Negative ON = bonded atom gained e- • Positive ON = bonded atom lost e- • May refer to charge of ionic compound/ion

  25. Oxidation Number Rules • #1: Free-element Rule • ON of pure elements (& covalently bonded elements) is zero • #2: Ion Rule • ON of monoatomic ion is equal to charge of ion

  26. Oxidation Number Rules • #3: Zero-sum Rule • Sum of ON of all atoms in compound must add up to zero • Element with highest electronegativity often determines ON of other atoms

  27. #4: Specific ON Rule • Alkali metals: +1 • Alkali earth metals: +2 • Hydrogen: usually +1 if bonded to nonmetal, -1 if bonded to metal • Forms metallic hydrides • Oxygen: -2 unless bonded to F (-1 in rare cases) • Halogens: -1 when bonded to metals • F always has -1 ON • *If one rule contradicts another, rule listed first should be followed

  28. Finding Oxidation Numbers • Algebra can be used (because sum of neutral compound equals zero) • ON numbers used to write chemical formulas • For binary (2 element) ionic compounds, use criss-cross method to find chemical formula

  29. List the elements (positive ON always first) Write the ON underneath Draw the criss-cross arrows Rewrite the formula showing subscripts Formulas should show simplest ratio Mg F +2 -1 Mg F +2 -1 MgF2 1:2 is simplest ratio Criss-Cross Method

  30. Binary Covalent Compounds • Use Greek prefixes • Write least electronegative element first • Change ending of 2nd element to -ide • NOT acids • Mono only for 2nd element • Omit double vowels (mono-oxide becomes monoxide)

  31. Binary Covalent Compounds • Application: CO2 P2S CO • Use Greek prefixes • Write least electronegative element first • Change ending of 2nd element to -ide • NOT acids • Mono only for 2nd element • Omit double vowels (mono-oxide becomes monoxide)

  32. Binary Ionic Compounds • Greek prefixes not used • Positive ions keep original (parent) name • Negative ions have –ide ending • Name cation, then anion • If metal hydride (Metal + H), name metal 1st

  33. Binary Ionic Compounds • Application: MgO Al2O3 MgH2 • Greek prefixes not used • Positive ions keep original (parent) name • Negative ions have –ide ending • Name cation, then anion • If metal hydride (Metal + H), name metal 1st

  34. Ionic Compounds with Multiple Oxidation States • Use Stock System • Place Roman numeral after the metal to show its oxidation number • For Fe, Co, Cu, Mg, Pb, Sn • Ex: FeCl2 iron (II) oxide

  35. Binary Acids • H as 1st element • Ex: hydrogen chloride HCl • When dissolved in water, add prefix “hydro-” & suffix “-ic acid” • Ex: HCl in water becomes hydrochloric acid

  36. Chemical Equations • Show composition of substances (what & how much) Ca(HCO3)2 + Ca(OH)2  H2O + CaCO3 • Account for all atoms involved (law of mass conservation) • Balanced using coefficients (called balancing by introspection) • Is the formula balanced?

  37. Balancing by Introspection • 1. Write the formula for reactants & products • 2. Check to see if equation is balanced • 3. Adjust coefficients to balance • 4. Make sure coefficients are in simplest whole-number ratio • Practice: Work through example problem (p. 200-201), then do SRQ 8C 2a,b,c • Take notes on “Special symbols & equations” section

  38. eggs shoes Molar mass is the mass of 1 mole of in grams marbles atoms For any element atomic mass (amu) = molar mass (grams) 1 12C atom = 12.00 amu 1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g 1 mole 12C atoms = 12.00 g 12C 1 mole lithium atoms = 6.941 g of Li 3.2

  39. 1 mol K 6.022 x 1023 atoms K x x = 1 mol K 39.10 g K 6.022 x 10^23 6.022 x 10^23 6.022 x 10^23 amu amu amu How many atoms are in 0.551 g of potassium (K) ? 1 mol K = 39.10 g K 1 mol K = 6.022 x 1023 atoms K 0.551 g K 8.49 x 1021 atoms K 3.2

  40. 1S 32.07 amu 2O + 2 x 16.00 amu SO2 SO2 64.07 amu Molecular mass of a compound can be found by Adding together the molar masses of the elements (think of it as the total amu) *units: amu For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO2 = 64.07 amu molar mass = 1 mole SO2 = 64.07 g SO2  molecular mass 64.07 g SO2 3.3

  41. Types of Formulas • Structural Formula • Shows the types of atoms • Exact composition • Arrangement of chemical bonds • Molecular Formula • Shows the types and numbers of atoms • More convenient • Does not show the shape, location or types of bonds

  42. Types of Formulas • Empirical Formula • Tells what elements are present using the simplest whole number ratio • Formula units [ NaCl, CA(OH)2 ] are always empirical formulas • Example - Ethene • Molecular – C2H4 • Empirical – CH2

  43. Percent composition • Shows what percentage of a compound’s total mass comes from each element • Based on the model, estimate what percent of the mass comes from H & what percent comes from O 3.5

  44. Percent composition • Based on the formula, what percent of the mass comes from H and what percent comes from O? Total = O + H + H = 16.00 + 1.008 + 1.008 = 18.016 amu % Oxygen = 16.00 / 18.016 = 88.8% % Hydrogen = (2 x 1.008) / 18.016 = 11.2% 3.5

  45. 2 x (12.01 g) 6 x (1.008 g) 1 x (16.00 g) n x molar mass of element %C = %H = %O = x 100% = 52.14% x 100% = 13.13% x 100% = 34.73% x 100% 46.07 g 46.07 g 46.07 g molar mass of compound C2H6O Percent composition • Shows what percentage of a compound’s total mass comes from each element nis the number of moles of the element in 1 mole of the compound 52.14% + 13.13% + 34.73% = 100.0% 3.5

  46. 53.28 g 6.72 g %H = %O = x 100% = 11.2% x 100% = 88.80% 60.00 g 60.00 g Percent composition of an element in a compound A 60.0 g sample of water is decomposed into its elements of 53.28 g of O and 6.72 g of H. Find the percent composition. H2O 11.2 + 88.80% = 100.0% 3.5

  47. 128 g 91.96 g 64.14 g %O = %Na= %S = x 100% = 45.05% x 100% = 32.37% x 100% = 22.58% 284.1 g 284.1 g 284.1 g Percent composition of an element in a compound Find the percent composition of a substance with 91.96 g Na, 64.14 g S, and 128 g O. 32.37% + 22.58% + 45.05% = 100.0% 3.5

  48. 4 x 16.00 g 32.07 g 2 x 22.99 g %Na= %O = %S = x 100% = 22.58% x 100% = 45.05% x 100% = 32.37% 142.05 g 142.05 g 142.05 g Percent composition of an element in a compound Find the percent composition of Na2SO4. Na2SO4 32.37% + 22.58% + 45.05% = 100.0% 3.5

  49. Percent Composition Elemental (sometimes called “laboratory”) Analysis • Listing the mass of each element present in 100 g of a compound • Can be used to find the empirical formula What is the chemical formula for a compound with The following elemental analysis? 49.5% C 5.2% H 28.8%N 16.5% O Steps: 1) convert % to g 2) convert to mol 3) divide by smallest number 4) round

  50. C = H = N = O = 49.5% 5.2% 28.8% 16.5% Percent composition of an element in a compound In 100 g of a sample you would have 49.5 g of C In 100 g of a sample you would have 5.2 g of H In 100 g of a sample you would have 28.8 g of N In 100 g of a sample you would have 16.5 g of O

More Related