6/5/2015. Transferring of electrons (ionic) Sharing of electrons (molecular)

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1 Chem 20 Final Review Chemical Bonding Transferring of electrons (ionic) Sharing of electrons (molecular) 1

2 Neil Bohr s model of the atom: Valence Electrons: These are the electrons in the highest energy level around the nucleus. They are found in the outer most energy level of an atom. 2, 8, 8, 18 The number of energy levels equals the period number The number of valence electrons is equal to the last digit of the group number valence electrons are involved in the chemical reaction. They are either going to be exchanged, so as to form an ionic bond Orbitals may be occupied by 0, 1 or 2 electrons, but no more than 2 electrons Or they are going to be shared, so as to form a covalent/molecular bond 2

3 lone pair: contains 2 electrons bonding electrons: electrons that are not paired Rules for drawing Lewis Dot Diagrams: Write out the atomic symbol place a dot on the left side of the atomic symbol Continue placing dots around the symbol in a clockwise order 3e 8e 2e 13 p + Al as three Valence Electrons These are bonding electrons Al Electronegativity gradually increases as we move from left to right in a period And the Electronegativity decreases as we move down the group 3

4 Ionic Bonds difference between the electronegativities are quite large magnesium oxide Mg O [Mg] 2+ [ O ] 2 calcium fluoride Ca F F [Ca] 2+ 2x[ F ] 1 Ionic bonds are very strong bonds and thus require a large amount of energy to break them apart. This explains why ionic compounds: Are solid at room temperature ave high melting and boiling points 4

5 A Covalent Bond is formed when both electronegativities are high. The two atoms end up sharing the bonding electrons The atom with the most possible bonding electrons becomes the Central Atom Let s look at Cl 2 Let s look at O 2 (O has 2 bonding e ) sharing of electrons forms a single bond Can also be written as: Cl Cl sharing of electrons forms a double bond Can also be written as: O=O 5

6 Three hydrogen atoms are bonded to one nitrogen atom. This compound is ammonia. N 3x( ) We should have 8 electrons in total VSEPR stands for: Valence Shell Electron Pair Repulsion N We have 8 electrons in total According to VSEPR theory Only the valence electrons of the central atom are important in determining the molecule s shape Double and triple bonds can be treated just as single bonds for predicting shapes using VSEPR theory Bonded pairs and lone pairs of electrons are treated approximately equally 6

of the plane of the page AX 2 AX3 AX 4 AX 3 E AX 2 E 2 If there is unequal sharing

7 A solid line represents a bond in the plane of the page A dashed line represents a bond behind the plane of the page A wedged or triangle represents a bond in front of the plane of the page AX 2 AX3 AX 4 AX 3 E AX 2 E 2 If there is unequal sharing of the electrons the bond is considered to be a polar covalent bond Cl δ δ

Non-Polar Covalent Molecule: dipoles cancel each other out (net dipole of zero) carbon dioxide CO (g) 2 O C O C methane C (g) 4 2.2 2.

8 Non-Polar Covalent Molecule: dipoles cancel each other out (net dipole of zero) carbon dioxide CO (g) 2 O C O C methane C (g) Molecules that are symmetrical are non-polar because their dipole moments balance. 2.2 ere we can see that the dipoles cancel each other out, so C 4 is considered a non-polar compound 8

9 Polar molecules: The bond dipoles do not cancel out leaving a non-zero net dipole O water 2 O(l) O ere we can see that the dipoles do not cancel each other out, so water is considered a polar compound N 2.2 ammonia N 3 (g) N 2.2 ere the dipoles do not cancel each other out, so ammonia is considered a polar compound δ + δ N δ + δ + When in comes to solubility remember that like dissolves like 9

10 London Dispersion forces Dipole-Dipole forces The strength of the London force is: directly related to the number of electrons in the molecule ydrogen bonding Dipole Dipole Forces These forces are only found in polar compounds The strength of the dipole dipole force is dependent on the overall polarity of the molecule ydrogen Bond: For hydrogen bonding to occur, it must be bonded to a highly electronegative atom: nitrogen, oxygen or fluorine. 10

covalent bond (intramolecular) hydrogen bond (intermolecular) + covalent bond (intramolecular) Ranking the intermolecular bonds from strongest to weakest we get: 1. ydrogen bonding 2.

11 covalent bond (intramolecular) hydrogen bond (intermolecular) + covalent bond (intramolecular) Ranking the intermolecular bonds from strongest to weakest we get: 1. ydrogen bonding 2. Dipole-dipole bonding 3. London dispersion bonding Remember if it is an ionic compound it will have a higher boiling point Gases Comparing Kelvin and Celsius Scales To convert degrees Celsius to Kelvin, you add 273. K = C To convert Kelvin to degrees Celsius, you subtract 273. C = K

12 Atmospheric Pressure Standard Temperature and Pressure (STP) = kpa and 0 C Gas Laws They are based on the temperature, pressure and volume relationships that all gases have in common 1. Boyle s Law P 1 V 1 = P 2 V 2 Standard Ambient Temperature and Pressure (SATP) = 100 kpa and 25 C 2. Charles Law V 1 = V 2 T 1 T 2 3. Combined Gas Law P 1 V 1 =P 2 V 2 T 1 T 2 Example: Law of Combining Volumes Use the law of combining volumes to predict the volume of oxygen required for the complete combustion of 120 ml of butane gas from a lighter. 2C410(g) + 13O2(g) 8CO2(g) + 102O(g) 120 ml V =? Molar Volume At STP, molar volume = 22.4 L/mol At SATP, molar volume = 24.8 L/mol VO2: 120 ml C410 x ( 13 ml O2) 2 ml C410 = 780 ml 12

13 Calculate the volume occupied by mol of carbon dioxide at SATP IDEAL GAS LAW IDEAL GAS does not really exist Follows all gas laws perfectly under all conditions Does not condense when cooled Assumes that the particles have no volume and are not attracted to each other REAL GAS does not follow gas laws exactly, it deviates at low temperatures and high pressures Condenses to liquid or sometimes solid when cooled or under pressure Particles are attracted to each other and have volume Behaves like an ideal gas at higher temperatures and lower pressures Using the Ideal Gas Law Example One: What mass of neon gas should be introduced into an evacuated 0.88L tube to produce a pressure of 90 kpa at 30 C? PV = nrt Solutions n = PV RT n = (90kPa)(0.88L) (8.314 L kpa/mol K)(303K) n = mol x g 1 mol = 0.63 g 13

14 Dissociation vs. Ionization What is the difference between dissociation and ionization? Both produce (aq) ions... Dissociation, however, is the separation of ions that already exist before dissolving in water M + X - (s) M + (aq) + X - (aq) Ionization involves the production of new ions, specifically hydrogen ions Molecular Ionic SUMMARY Substance Process General Equation Base (ionic hydroxide) Acid Reference pg. 201 Disperse as individual, neutral molecules Dissociate into individual ions Dissociate into positive ions and hydroxide ions Ionize to form hydrogen ions and anions XY (s/l/g) XY (aq) MX (s) M + (aq) + X - (aq) MO (s) M + (aq) + O - (aq) X (s/l/g) + (aq) + X - (aq) X 0 (aq) + (aq) + X - (aq) 14

15 Amount Concentration (aka Molar Concentration) (mol/l) C = n solute (mol) V solution (L) Al 2 (SO 4 ) 3 (aq) 2 Al 3+ (aq) + 3 SO 4 2- (aq) c = 0.30 mol/l [Al 3+ (aq)] = 0.30 mol/l x (2) = 0.60 mol/l 1 [SO 4 2- (aq) ] = 0.30 mol/l x (3) = 0.90 mol/l 1 Determining the volume of Stock Solution for a Standard Solution dilution formula: C 1 V 1 = C 2 V 2 Example: ow would you prepare 100 ml of 0.40 mol/l MgSO 4(aq) from a solution of 2.0 mol/l MgSO 4(aq) C 1 V 1 = C 2 V 2 (2.0mol/L) (V 1 ) = (0.40mol/L) (100mL) V 1 = 20 ml 15

Empirical Definitions Acids and Bases Acid a substance which dissolves in water to produce a solution that: Tastes sour Turns blue litmus red Conducts electricity Reacts with active metals to produce

16 Empirical Definitions Acids and Bases Acid a substance which dissolves in water to produce a solution that: Tastes sour Turns blue litmus red Conducts electricity Reacts with active metals to produce 2 (g) Neutralizes Bases Base a substance which dissolves in water to produce a solution that: Tastes bitter; feels slippery Turns red litmus blue Conducts electricity Neutralizes acids a) Arrhenius: Theoretical Definitions Acid a substance that forms an acidic solution by dissolving in water to produce free hydrogen ions ( + (aq)) in solution Example: Cl (aq) + (aq) + Cl - (aq) b) Modified Definition: Theoretical Definitions Acid a species that forms an acidic solution by reacting with water to produce hydronium ions ( 3 O + (aq)) Example: Cl (aq) + 2 O (aq) 3 O + (aq) + Cl - (aq) Base a substance that forms a basic solution by dissolving in water to produce free hydroxide ions (O - (aq)) in solution Example: NaO (aq) Na + (aq) + O (aq) Base a species that forms a basic solution by reacting with water to produce hydroxide ions (O - (aq)) Example: N 3 (aq) + 2 O (aq) N 4 + (aq) + O (aq) 16

Summary Acid Base Indicators p = -log [ 3 O + (aq)] po = -log [O - (aq)] [ 3 O + (aq)] =10 p [O - (aq)] =10 po Common name Color of In(aq) p range of colour change Color of In - (aq) Bromothymol blue

0 Pink The number of digits following the decimal point in a p or po value is equal to the number of significant digits in the corresponding hydronium or hydroxide concentration.

17 Summary Acid Base Indicators p = -log [ 3 O + (aq)] po = -log [O - (aq)] [ 3 O + (aq)] =10 p [O - (aq)] =10 po Common name Color of In(aq) p range of colour change Color of In - (aq) Bromothymol blue Yellow Blue Phenolphthalein Colourless Pink The number of digits following the decimal point in a p or po value is equal to the number of significant digits in the corresponding hydronium or hydroxide concentration. For both p and po, an inverse relationship exist between the ion concentration and the p or po. The greater the hydronium ion concentration, the lower the p is. More Practice... I (aq) explain acidic properties I (aq) + 2 O (l) 3 O + (aq) + I - (aq) Try Three More... 3 PO 4(aq) explain acidic properties 3 PO 4(aq) + 2 O (l) 3 O + (aq) + 2 PO 4 - (aq) NaC 3 COO (aq) explain basic properties NaC 3 COO (aq) Na + (aq) + C 3 COO - (aq) Simple Dissociation C 3 COO - (aq) + 2 O (l) C 3 COO (aq) + O - (aq) Na 2 SO 4 (aq) explain basic properties Na 2 SO 4 (aq) 2 Na + (aq) + SO 4 2- (aq) Simple Dissociation SO 4 2- (aq) + 2 O (l) SO 4 - (aq) + O - (aq) OCl (aq) explain acidic properties OCl + 2 O (l) 3 O + (aq) + OCl - (aq) Sr(O) 2(aq) explain basic properties Sr(O) 2(aq) Sr 2+ (aq) + 2 O - (aq) 17

The Difference: Using the Modified Arrhenius Theory Strong Acids: have high conductivity, high rate of reaction w/ metals and carbonates and a

conductivity, a lower rate of reaction w/ active metals and carbonates and a relatively high p Based on this evidence, a weak acid reacts

Strong Bases all soluble ionic hydroxides that dissociate completely (>99%) to release hydroxide ions NaO (s) Na + (aq) + O - (aq) Weak Bases an

18 The Difference: Using the Modified Arrhenius Theory Strong Acids: have high conductivity, high rate of reaction w/ metals and carbonates and a relatively low p >99% Cl (aq) + 2 O (l) 3 O + (aq) + Cl - (aq) The Difference: Using the Modified Arrhenius Theory Weak Acids: have low conductivity, a lower rate of reaction w/ active metals and carbonates and a relatively high p Based on this evidence, a weak acid reacts incompletely (<50%) with water to form relatively few hydronium ions <50% C 3 COO (aq) + 2 O (l) 3 O + (aq) + C 3 COO - (aq) Strong and Weak Bases Strong Bases all soluble ionic hydroxides that dissociate completely (>99%) to release hydroxide ions NaO (s) Na + (aq) + O - (aq) Weak Bases an ionic or molecular substance that reacts partially (<50%) with water to produce relatively few hydroxide ions Stoichiometry <50% N 3 (aq) + 2 O (aq) O (aq) + N 4 + (aq) 18

19 What do you remember? Using the solubility table: In Science 10 you learned about five reaction types, can you match them up Composition (Formation) Decomposition Combustion Single Replacement Double Replacement C 4(g) + O 2(g) CO 2(g) + 2 O (g) Mg (s) + O 2(g) MgO (s) Cu (s) + AgNO 3(aq) Ag (s) + Cu(NO 3 ) 2(g) CaCl 2(aq) + Na 2 CO 3(aq) CaCO 3(s) + NaCl (aq) 2 O (l) O 2(g) + 2(g) Practice Write the net ionic equation for the reaction of aqueous barium chloride and aqueous sodium sulfate. (Refer to the solubility table) 1) BaCl 2(aq) + Na 2 SO 4(aq) BaSO 4(s) + 2NaCl (aq) 2) Ba 2+ (aq) + 2Cl - (aq) +2Na + (aq) + SO 4 2- (aq) BaSO 4(s) + 2Na + (aq) + 2Cl - (aq) (Complete ionic equation) 3) Ba 2+ (aq) + 2Cl - (aq) +2Na + (aq) + SO 4 2- (aq) BaSO 4(s) + 2Na + (aq) + 2Cl - (aq) Limiting and Excess Reagents When no further changes appear to be occurring, we assume that all of the AgNO 3(aq) that was initially present has now been completely reacted. A limiting reagent is the reactant whose entities are completely consumed in a reaction, meaning the reaction stops. In order to make sure this happens, more of the other reactant must be present than is required An excess reagent is the reactant whose entities are present in surplus amounts 4) Ba 2+ (aq)) + SO 4 2- (aq) BaSO 4(s) (Net ionic equation) 19

20 Stoichiometry Calculations (Measured quantity) solids/liquids m n solids/liquids m n (Required quantity) mole ratio Practice #3 (Mass Stoichiometry) What mass of iron (III) oxide is required to produce g of iron? Fe 2 O 3(s) + 3 CO (g) 2 Fe (s) + 3 CO 2(g) m =? m = 100.0g M = g/mol M = g/mol m Fe 2 O 3(s) : g x 1 mol x 1 mol x g = g Fe 2 O g 2 mol 1 mol Percent Yield for Reactions Percent Yield Example #1 Percent yield = actual yield x 100 predicted yield Example: In a chemical analysis, 3.00 g of silver nitrate in solution was reacted with excess sodium chromate to produced 2.81 g of filtered, dried precipitate. Predicted value: 2AgNO 3(aq) + Na 2 CrO 7(aq) Ag 2 CrO 7(s) +2NaNO 3(aq) m = 3.00 g m = M = g/mol M = g/mol 3.00g x 1 mol x 1 mol x g = g g 2 mol 1 mol Percent yield = actual yield x 100% = 2.81g x 100% = 95.9% predicted yield 2.93g 20

21 Gas Stoichiometry If 300g of propane burns in a gas barbecue, what volume of oxygen measured at SATP is required for the reaction? Remember: 24.8L/mol for SATP C 3 8(g) + 5O 2(g) 3CO 2(g) O (g) m = 300g V =? 44.11g/mol 24.8L/mol 300 g x 1 mol x 5 mol x 24.8 L = 843 L O 2(g) g 1 mol 1 mol Gas Stoichiometry What volume of ammonia at 450kPa and 80 o C can be obtained from the complete reaction of 7.5kg of hydrogen with nitrogen? 2N 2(g) + 3 2(g) 2N 3(g) m = 7500g V =? M = 2.02 g/mol P = 450kPA T = K 7500 g x 1 mol x 2 = mol N 3(g) 2.02 g 3 PV = nrt V = nrt = ( mol)(8.314 kpa L / mol K )(353.15K) P (450kPa) = L 1.6 x 10 4 L of N 3(g) Solution Stoichiometry In an experiment, a ml sample of sulfuric acid solution reacts completely with 15.9 ml of mol/l potassium hydroxide. Calculate the amount concentration of the sulfuric acid. 2 SO 4(aq) + 2KO (aq) 2 2 O (l) + K 2 SO 4(aq) V = 10.00mL V = 15.9 ml c =? mol/l Chemical Analysis 15.9mL x mol x 1mol x 1 = mol/l 1 L 2 mol 10.0 ml 21

22 2) If 10.0g of copper is placed in solution of 20.0g of silver nitrate, which reagent will be the limiting reagent? All reactants must be converted to moles, then using the mole ratio, determine which reactant will run out first. Cu (s) + 2 AgNO 3(aq) 2 Ag (s) + Cu(NO 3 ) 2(aq) 10.0g 20.0g given: mol mol n Cu(s) : 10.0g x 1 mol = mol x 2 = mol g 1 n AgNO3 : 20.0g x 1 mol = mol x 1 = mol g 2 Chemical Analysis by Titration Titration is a common experimental design used to determine the amount concentration of substances in solution. The solution of known concentration may be either the titrant or the sample; it makes no difference to the analysis Titration breakdown: Carefully adding a solution (titrant) from a burette into a measured, fixed volume of another solution (sample) in an Erlenmeyer flask until the reaction is judged to be complete Chemical Analysis by Titration Burette precisely marked glass cylinder with a stopcock at one end. Allows precise, accurate measurement and control of the volume of reacting solution. When doing a titration, there will be a point at which the reaction is complete; when chemically equivalent amounts of reactants have combined. This is called the equivalence point: Equivalence point the point during a titration at which the exact theoretical chemical amount of titrant has been added to the sample. (QUANTITATIVE) To measure this equivalence point experimentally, we look for a sudden change in an observable property, such as color, p, or conductivity. This is called the endpoint. (QUALITATIVE) Chemical Analysis by Titration An initial reading of the burette is made before any titrant is added to the sample. Then the titrant is added until the reaction is complete; when a final drop of titrant permanently changes the colour of the sample. The final burette reading is then taken. The difference between the readings is the volume of titrant added. Near the endpoint, continuous gentle swirling of the solution is important 22

Sample Problem Determine the concentration of hydrochloric acid in a commercial solution. A 1.59g mass of sodium carbonate, Na 2 CO 3(s), was dissolved to make 100.0mL of solution. Samples (10.

23 Sample Problem Determine the concentration of hydrochloric acid in a commercial solution. A 1.59g mass of sodium carbonate, Na 2 CO 3(s), was dissolved to make 100.0mL of solution. Samples (10.00mL) of this standard solution were then taken and titrated with the hydrochloric acid solution. The titration evidence collected is below. Methyl orange indicator was used. Trial Final burette reading (ml) Initial burette reading (ml) Volume of Cl (aq) added Indicator colour Red Orange Orange Orange TIP: In titration analysis, the first trial is typically done very quickly. It is just for practice, to learn what the endpoint looks like and to learn the approximate volume of titrant needed to get to the endpoint. Greater care is taken with subsequent trials General Rule Strong Acid to Weak Base: p at equivalence point is always lower than 7 Strong Base to Weak Acid: p at equivalence point is always higher than 7 23

Chemistry 20 Lesson 36 The Whole Enchilada

Unit I: Science 10 Review Chemistry 20 Lesson 36 The Whole Enchilada 1. Classify the substances as ionic (i), molecular (m), or acid (a) and provide the IUPAC name and the state of matter at SATP where