Chm 116 (Sp 2004) - Review of Chm 115

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1 Chm 116 (Sp 2004) Review of Chm 115 Conversions within the SI system Examples (with Answers): 1. Convert 1.29 x mg into the following units: kilograms, grams, micrograms, and nanograms. In order from the smallest unit to the largest unit, the answers are: 1.29 x ng = 1.29 x microg = 1.29 x mg = 1.29 x g = 1.29 x 10 1 kg 2. Convert 7.44 x 10 6 m into the following units: angstroms (Å), nanometers, millimeters, and centimeters. In order from the smallest unit to the largest unit, the answers are: 7.44 x Å = 7.44 x nm = 7.44 x 10 3 mm = 7.44 x 10 4 cm = 7.44 x 10 6 m Note: An Angstrom (Å) is 1 x meter. Angstrom is not a prefix for other units (like grams or liters). It refers only to length and is not an SI unit (However, it is a convenient and common unit of length for the atomic scale). Solubility Rules and list of strong acids and strong bases Review from Chapter 4 and know the following: Solubility Rules NO 3 C 2 H 3 O 2 Cl Br 2 2 CO 3 OH S 2 NO 3 I SO 4 Na +, K +, Rb +, Cs + +, NH 4 s S s s s s Mg 2+ s S s i i d Ca 2+ s S sl s i i d Ba 2+ s S i i sl s d Al 3+ s S s * i d Pb 2+ s I i i i i Ag + s I sl s i i i 2+ Hg 2 s I sl s i * * Hg 2+ s sl s sl s i i i s = soluble, sl s = slightly soluble, i = insoluble, d = decomposes in or reacts with water, * = compound does not exist

2 Classification of Acids and Bases Strong Acids and Strong Bases an acid or base that is virtually 100 percent dissociated into ions in aqueous solution (note that this means that strong acids and bases are also strong electrolytes) Weak Acids and Weak Bases an acid or base that partly dissociates into ions in aqueous solution (note that this means that weak acids and bases are also weak electrolytes) Common Strong Acids and Bases Acid Name of Acid Base Name of Base HCl hydrochloric acid LiOH lithium hydroxide HBr hydrobromic acid NaOH sodium hydroxide HI hydroiodic acid KOH potassium hydroxide HNO 3 nitric acid RbOH rubidium hydroxide H 2 SO 4 sulfuric acid CsOH cesium hydroxide HClO 4 perchloric acid Ca(OH) 2 calcium hydroxide Sr(OH) 2 Ba(OH) 2 strontium hydroxide barium hydroxide Reactions: writing and balancing Examples (with Answers): 1. Write the balanced chemical equation for the combustion of C 4 H 6 O 4 (assume the only products are carbon dioxide and water). 2 C 4 H 6 O O 2 > 8 CO H 2 O 2. Write balanced full molecular, full ionic, and net ionic equations for the following aqueous reactions: o phosphoric acid reacts with potassium hydroxide (Weak Acid Strong Base) H 3 PO 4 (aq) + 3 KOH (aq) ===> K 3 PO 4 (aq) + 3 H 2 O (l) Full Ionic Representation (only strong electrolytes are shown dissociated into ions) H 3 PO 4 (aq) + 3 K + (aq) + 3 OH (aq) ===> 3 K + (aq) + PO 4 3 (aq) + 3 H 2 O (l)

3 H 3 PO 4 (aq) + 3 OH (aq) ===> 3 H 2 O (l) PO 4 3 (aq) o hydrochloric acid reacts with ammonia (Strong Acid Weak Base) HCl (aq) + NH 3 (aq) ===> NH 4 Cl (aq) Full Ionic Representation (only strong electrolytes are shown dissociated into ions) H + (aq) + Cl (aq) + NH 3 (aq) ===> NH 4 + (aq) + Cl (aq) H + (aq) + NH 3 (aq) ===> NH 4 + (aq) o perchloric acid reacts with sodium hydroxide (Strong Acid Strong Base) HClO 4 (aq) + NaOH (aq) ===> NaClO 4 (aq) + H 2 O (l) Full Ionic Representation (only strong electrolytes are shown dissociated into ions) H + (aq) + ClO 4 (aq) + Na + (aq) + OH (aq) ===> Na + (aq) + ClO 4 (aq) + H 2 O (l) H + (aq) + OH (aq) ===> H 2 O (l) o copper (II) chloride reacts with sodium carbonate CuCl 2 (aq) + Na 2 CO 3 (aq) ===> CuCO 3 (s) + 2 NaCl (aq) Full Ionic Representation Cu 2+ (aq) + 2 Cl (aq) + 2 Na + (aq) + CO 3 2 (aq) ===> CuCO 3 (s) + 2 Na + (aq) + 2 Cl (aq) Cu 2+ (aq) + CO 3 2 (aq) ===> CuCO 3 (s)

4 o barium nitrate reacts with copper (II) sulfate Ba(NO 3 ) 2 (aq) + CuSO 4 (aq) ===> BaSO 4 (s) + Cu(NO 3 ) 2 (aq) Full Ionic Representation Ba 2+ (aq) + 2 NO 3 (aq) + Cu 2+ (aq) + SO 4 2 (aq) ===> BaSO 4 (s) + Cu 2+ (aq) + 2 NO 3 (aq) Ba 2+ (aq) + SO 4 2 (aq) ===> BaSO 4 (s) o potassium hydroxide reacts with iron (III) chloride 3 KOH (aq) + FeCl 3 (aq) ===> Fe(OH) 3 (s) + 3 KCl (aq) Full Ionic Representation 3 K + (aq) + 3 OH (aq) + Fe 3+ (aq) + 3 Cl (aq) ===> Fe(OH) 3 (s) + 3 K + (aq) + 3 Cl (aq) Fe 3+ (aq) + 3 OH (aq) ===> Fe(OH) 3 (s) Write balanced the following redox reactions (for the aqueous reactions you may need to add components, H +, H 2 O, or OH, to balance the reactions): o magnetite reacts with hydrogen (balanced easily by inspection) Fe 3 O 4(s) + 4 H 2(g) 4 H 2 O (g) + 3 Fe (l) Fe 3 O 4(s) is the oxidizing agent and H 2(g) is the reducing agent o zinc reacts with dichromate (balanced using the halfreaction method for acidic conditions) 3 * ( Zn (s) Zn 2+ (aq) + 2 e ) 3 * (oxidation halfreaction) 14 H + (aq) + 6 e 2 + Cr 2 O 7 (aq) 2 Cr 3+ (aq) + 7 H 2 O (l) + the reduction halfreaction 3 Zn (s) + 14 H + 2 (aq) + Cr 2 O 7 (aq) 3 Zn 2+ (aq) + 2 Cr 3+ (aq) + 7 H 2 O (l) = complete redox rxn Cr 2 O 7 2 (aq) is the oxidizing agent and Zn (s) is the reducing agent

5 o zinc reacts with nitrate (balanced using the halfreaction method for basic conditions) 4 * ( Zn (s) Zn 2+ (aq) + 2 e ) 4 * (oxidation halfreaction) NO 3 (aq) + 8 e + 6 H 2 O (l) NH 3 (aq) + 9 OH (aq) + the reduction halfreaction 4 Zn (s) + NO 3 (aq) + 6 H 2 O (l) 4 Zn 2+ (aq) + 9 OH (aq) + NH 3 (aq) = complete redox reaction NO 3 (aq) is the oxidizing agent and Zn (s) is the reducing agent Identify the oxidizing and reducing reagents in each of the redox reactions you just balanced. See above (under each reaction) Mass, Moles, Molarity, and Stoichiometric Calculations Examples (with Answers): 1. Convert grams of MgCl 2 into moles of MgCl 2. How many moles of Mg +2 and Cl does this correspond to? grams of MgCl 2 = mole of MgCl 2 = mole of Mg +2 = mole of Cl 2. What mass of MgCl 2 is required to prepare 2.00 L of M MgCl 2? 28.6 grams of MgCl 2 is required to prepare 2.00 L of M MgCl 2 3. What volume of 2.00 M MgCl 2 (aq) would be required to prepare 2.00 L of M MgCl 2 (aq)? L (or 150 ml) of 2.00 M MgCl 2 (aq) is needed to prepare 2.00 L of M MgCl 2 (aq) 4. Refer to the balanced chemical equation you wrote for the reaction of perchloric acid with sodium hydroxide. If ml of perchloric acid is titrated to neutralization by ml of M NaOH, what is the molarity of the perchloric acid in the initial ml sample? M ( mole / L ) HClO 4 5. Refer to the balanced chemical equation you wrote for the reaction of potassium hydroxide with iron (III) chloride. When 50.0 ml of M KOH reacts with 25.0 ml of M iron (III) chloride, what is the maximum amount of precipitate that can form? (give your answer in grams) g of Fe(OH) 3 (s) is the maximum amount of ppt.

6 Gas Laws Know and understand how to use simple gas laws (ex. Boyles), the ideal gas law (PV = nrt), Datlon s law of partial pressures, and Grahams law of effusion. There are example problems available in the online text at Dr. Mencer s web site. Electronic structure and periodic trends Review Chapter 7 and 8 of your text and answer the following: 1. The arrangement of the elements in the periodic table are in order of increasing atomic number. 2. Elements in the same group of the periodic table have similar chemistry. Why? Same valence electron configuration leads to similar chemistry... For example: Li has a single 2s electron and Na has a single 3s electron and they both typically form +1 ions. 3. How does atomic size vary as you proceed down a group? Why? Atomic size increases down a group because the outermost electrons are stored in increasingly large orbitals.. For example: 3s orbital is larger than 2s orbital, therefore, Na is larger than Li 4. How does atomic size vary from left to right across a period? Why? [think about the effective nuclear charge that is experienced by the outermost electrons] The effective nuclear charge experienced by the valence electrons increases from left to right across a period because electrons in the same principle energy level (and particularly in the same subshell) do not screen each other well from the nucleus positive charge. 5. The equation for the first ionization step for Al can be written as: Al > Al +1 + e write equations for the second and third ionization steps. Explain why Al is not likely to undergo a fourth ionization process. Second step: Al +1 > Al +2 + e Third step: Al +2 > Al +3 + e Fourth step: Al +3 > Al +4 + e Al has a valence shell electron configuration of 3s 2 3p 1... losing three electrons to form Al 3+ only requires the removal of electrons from the valence shell. In order to form Al +4 an electron would have to be lost from the 2p energy level. This requires too much energy.

7 6. Describe and explain the variation in ionization energy down a group and across a period. Ionization Energy (IE) decreases as one proceeds down a group because the outermost valence electron is further from the nucleus and more loosely held (i.e. the valence electron is in a well shielded large orbital). IE increases from left to right across a period because the effective nuclear charge experienced by the valence electrons increases from left to right (see #4 above). 7. The equation for the first electron addition step for O can be written as: O + e > O 1 write equations for the second electron addition step. Explain why O is not likely to undergo a third electron addition process. Second step: O 1 + e > O 2 Third step: O 2 + e > O 3 In order for oxygen to become O 2 a neutral oxygen atom must gain two electrons. Since oxygen has a valence electron configuration of 2s 2 2p 4 gaining two electrons completes the 2p energy level (where the electrons should feel a reasonably large effective nuclear charge). Therefore, the Electron Affinity (EA) for this process is favorable. In order to form O 3 an additional electron would have to be gained and would occupy the 3s orbital (where it would feel a very weak effective nuclear charge). 8. Describe and explain the variation in electron affinity (the relative ability for a gas phase atom to gain an electron) down a group and across a period. Elements on the righthand side of the periodic table have a greater ability to gain an electron. This is because the effective nuclear charge is higher on the righthand side of the Periodic Table. Elements toward the bottom of the groups have lower ability to gain electrons. The electron is gained into a larger orbital that is better shielded and the energetic favorability of the process is not as great. 9. When metals react with nonmetals, ionic compounds are formed. Use the concepts of ionization energy and electron affinity to explain: (a) which gains and which loses electrons [metals or nonmetals] and (b) the relative charges acquired in these ionic compounds. Ex. Why is NaCl a correct formula but MgCl is incorrect? (a) Metals lose electrons because they have low IE and unfavorable EA (see # 4 7 above for the reasons). Nonmetals gain electrons because they have favorable EA and do not lose electrons easily due to their high IE. (see # 4 7 for the reasons). (b) Metals (representative) tend to lose all their valence electrons (but many transition metals do not and some representative metals do not). Nonmetals tend to gain enough electrons to complete their valence shell.

8 For example: Cl (3s 2 3p 5 ) + e > Cl (3s 2 3p 6 ) Mg (3s 2 ) > Mg 2+ (3s 0 ) + 2 e Then, of course, when the ions combine to form compounds, they do so in ratios determined by their relative charges. For example: Mg Cl > MgCl Write e configurations (spdf format) for: Fe, Cu, Mo, Pb, Fe 2+, Fe 3+, Cu 1+, Cu 2+, Pb 2+, and Pb 4+ Fe, 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 4s 2... or [Ar]3d 6 4s 2 Fe 2+, 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6... or [Ar]3d 6 Fe 3+, 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5... or [Ar]3d 5 Cu, 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 1... or [Ar]3d 10 4s 1 Cu 1+, 1s 2 2s 2 2p 6 3s 2 3p 6 3d or [Ar]3d 10 Cu 2+, 1s 2 2s 2 2p 6 3s 2 3p 6 3d 9... or [Ar]3d 9 Mo, 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 5 5s 1... or [Kr]4d 5 5s 1 Pb, 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 10 5s 2 5p 6 4f 14 5d 10 6s 2 6p 2... or [Xe]4f 14 5d 10 6s 2 6p 2 Pb 2+, 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 10 5s 2 5p 6 4f 14 5d 10 6s 2... or [Xe]4f 14 5d 10 6s 2 Pb 4+, 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 10 5s 2 5p 6 4f 14 5d or [Xe]4f 14 5d 10 Lewis structures and hybridization Know and be able to apply all relevant rules for Lewis structures (octet rule w/exceptions, formal charge, resonance, etc.) and understand how to describe the bonding using hybridized orbitals. Additional Points of Importance 1. The chemistry of the first member of each family (group) often differs in several important ways from that of subsequent members. One cause for the differences is atomic size. What are two other causes? Hint: compare OF 2 with F compounds of S AND compare CO 2 with SiO 2 and consider the kinds of bonds that can be formed. 2. Period 2 elements cannot form sp 3 d and sp 3 d 2 hybrid orbitals. Thus, oxygen can only form OF 2 using sp 3 orbitals, but sulfur can form SF 2 (sp 3 ),SF 4 (sp 3 d), and SF 6 (sp 3 d 2 ). Ability to form pi bonds is greater for the first member of each group. Thus, carbon forms CO 2 (g) (the Lewis structure contains two double bonds) while silicon forms SiO 2 (s) which has no double bonds. The carbon dioxide is a nonpolar gaseous species whereas the silicon dioxide is a network covalent solid.

9 Thermochemistry Understanding of thermochemical principles and calculations was the topic of chapter 6. This material will be relevant for a complete understanding of chapter 11. An additional Point of Importance Whether a reaction is endothermic or exothermic depends upon the types of bonds broken and formed (if the reaction is a gas phase reaction nothing else matters, when other phases are present then other energetic terms also should be considered). Solutions Review and be familiar with the following concentration units: molarity, % by mass, % by volume, and mole fractions... we'll cover these quickly when we do Chapter 12 (Sections ).

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