Chapter 2. Section 2.1 The Formation of Ionic and Covalent Bonds Solutions for Selected Review Questions Student Edition page 63

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1 Chapter 2 Chemical Bonding Section 2.1 The Formation of Ionic and Covalent Bonds Solutions for Selected Review Questions Student Edition page Review Question (page 63) Predict whether the bond between each pair of atoms will be non-polar covalent, slightly polar covalent, polar covalent, or mostly ionic. a. carbon and fluorine e. silicon and hydrogen b. oxygen and nitrogen f. sodium and fluorine c. chlorine and chlorine g. iron and oxygen d. copper and oxygen h. manganese and oxygen You need to classify the nature of a bond between two elements. You know the atoms that are bonded. You can find the electronegativity of the atoms in the periodic table. Find the electronegativity of each given element in the periodic table. Subtract the electronegativity of the two elements that are bonded. Classify the nature of the bond based upon the electronegativity difference, EN. Electronegativity Difference, EN greater than 1.7 between less than 0.4 zero Classification of Bond mostly ionic polar covalent slightly polar covalent non-polar covalent Chapter 2 Chemical Bonding MHR 1

2 a. carbon and fluorine EN = EN F EN C = = 1.4 The bond is classed as polar covalent. b. oxygen and nitrogen EN = EN O EN N = = 0.4 The bond is classed as polar covalent. c. chlorine and chlorine EN = EN Cl EN Cl = = 0.0 The bond is classed as non-polar covalent. d. copper and oxygen EN = EN O EN Cu = = 1.5 The bond is polar covalent. e. silicon and hydrogen EN = EN H EN Si = = 0.3 The bond is slightly polar covalent. f. sodium and fluorine EN = EN F EN Na = = 3.1 The bond is mostly ionic. g. iron and oxygen EN = EN O EN Fe = = 1.6 The bond is polar covalent Chapter 2 Chemical Bonding MHR 2

3 h. manganese and oxygen EN = EN O EN Mn = = 1.8 The bond is mostly ionic. Check to be certain that the electronegativity values have been recorded and subtracted correctly. The EN are consistent with the type of bond. 12. Review Question (page 63) Arrange the bonds in each group below in order of increasing polarity. a. hydrogen bonded to chlorine, oxygen bonded to nitrogen, carbon bonded to sulfur, sodium bonded to chlorine b. carbon bonded to chlorine, magnesium bonded to chlorine, phosphorus bonded to oxygen, nitrogen bonded to nitrogen You need to arrange bonds in order of increasing polarity. You know the atoms that are bonded. You can find the electronegativity of the atoms in the periodic table. Find the electronegativity of each given element in the periodic table. Subtract the electronegativity of the two elements that are bonded, EN The bond of highest EN is the most polar. List the pairs of bonded elements from lowest to highest EN. a. hydrogen and chlorine EN = EN Cl EN H = = 1.0 oxygen and nitrogen EN = EN O EN N = = Chapter 2 Chemical Bonding MHR 3

4 carbon and sulfur EN = EN C EN S = = 0.0 sodium and chlorine EN = EN Cl EN Na = = 2.3 In order from lowest to highest polarity: C S < N O < H Cl < Na Cl b. carbon and chlorine EN = EN Cl EN C = = 0.6 magnesium and chlorine EN = EN Cl EN Mg = = 1.9 phosphorus and oxygen EN = EN O EN P = = 1.2 nitrogen and nitrogen EN = EN N EN N = = 0.0 In order from lowest to highest polarity: N N < C Cl < P O < Mg Cl Section 2.2 Writing Names and Formulas for Ionic and Molecular Compounds Solutions for Practice Problems Student Edition page Practice Problem (page 73) Write the name of P 4 S 7. You need to determine the name of P 4 S Chapter 2 Chemical Bonding MHR 4

5 You are given the formula P 4 S 7. Phosphorus and sulfur are both non-metals. The compound is molecular and prefixes are needed. Sulfur is the second element. Change the ending to -ide. There are four phosphorus atoms. The prefix for four is tetra-. There are seven sulfur atoms. The prefix is hepta-. The name is tetraphosphorus heptasulfide. The name describes the correct number of atoms of the corresponding elements. 2. Practice Problem (page 73) Write the name of Pb(NO 3 ) 2. You need to determine the name of Pb(NO 3 ) 2. You are given the formula Pb(NO 3 ) 2. Pb is the metal lead. NO 3 is a polyatomic ion and the name is nitrate. Since the compound is ionic, prefixes are not needed. A lead ion can have a charge of 2+ or 4+. Nitrate has a charge of 1 and there are two of these ions. The lead ion must have a charge of 2+ to make the compound neutral. Add (II) to the name of lead. The name is lead(ii) nitrate. The charges on the ions add up to zero. 3. Practice Problem (page 73) Write the formula for manganese(iv) chloride. You need to write the formula for manganese(iv) chloride Chapter 2 Chemical Bonding MHR 5

6 You know the name is manganese(iv) chloride. Manganese (Mn) is a metal. You know its charge must be 4+ because (IV) is part of the name. Chlorine (Cl) is a non-metal. Its charge is 1. The ending ide indicates that the compound is binary. It is an ionic compound. The charges are not the same, so subscripts are needed. The ratio of the charges on the two ions is 1:4. This does not reduce to a simpler whole number ratio. The ratio of the number of atoms is also 1:4. The subscript (number of atoms) of each element is the same as the magnitude of the charge of the other ion. 1 Mn 4+ ion balances the charge on 4 Cl ions. Therefore, the formula is MnCl 4. The charges on the ions add up to zero. 4. Practice Problem (page 73) Write the formula for nitrogen triiodide. You need to write the formula for nitrogen triiodide. You know the name is nitrogen triiodide. Nitrogen and iodine are both non-metals. The ending is ide. The compound is binary and molecular. Nitrogen has no prefix so there will be one atom of nitrogen and no subscript in the formula. Iodine has the prefix tri- so there must be three iodine atoms. Its subscript is 3. The formula is NI 3. The formula describes the correct number of atoms of the corresponding elements Chapter 2 Chemical Bonding MHR 6

7 5. Practice Problem (page 73) Write the name of CuBr. You need to determine the name of CuBr. You are given the formula CuBr. Cu is copper and it is a metal. Br is bromine and it is a non-metal. The compound is ionic. Prefixes are not needed. A copper ion can have a charge of 1+ or 2+. Bromine has a charge of 1 and there is one of these ions. The copper ion must have a charge of 1+ to make the compound neutral. Add (I) to the name of copper. The name is copper(i) bromide. The charges on the ions add up to zero. 6. Practice Problem (page 73) Write the formula for iron(iii) oxide. You need to write the formula for iron(iii) oxide. You know the name is iron(iii) oxide. Iron (Fe) is a metal. Its charge must be 3+ because (III) is part of the name. Oxygen(O) is a non-metal. Its charge is 2. The ending ide indicates that the compound is binary. It is an ionic compound. The charges are not the same so subscripts are needed. The ratio of the charges on the two ions is 2 : 3. This does not reduce to a simpler whole number ratio. The ratio of the number of atoms is also 2:3. The subscript (number of atoms) of each element is the same as the magnitude of the charge of the other ion. 2 Fe 3+ ions balance the charge on 3 O 2 ions. Therefore, the formula is Fe 2 O Chapter 2 Chemical Bonding MHR 7

8 The charges on the ions add up to zero. 7. Practice Problem (page 73) Write the formula for silicon dioxide. You need to write the formula for silicon dioxide. You know the name is silicon dioxide. Silicon and oxygen are both non-metals. The ending is ide. The compound is binary and molecular. Silicon has no prefix so there will be one atom of silicon and no subscript in the formula. Oxygen has the prefix di- so there must be two oxygen atoms. Its subscript is 2. The formula is SiO 2. The formula describes the correct number of atoms of the corresponding elements. 8. Practice Problem (page 73) Write the name of SeF 6. You need to determine the name of SeF 6. You are given the formula SeF 6. Selenium and fluorine are both non-metals so the compound is molecular and you need prefixes Fluorine is the second element. Change its ending to -ide. There is one selenium atom. It is the first element in the name and no prefix is needed. There are six fluorine atoms. Its prefix is hexa Chapter 2 Chemical Bonding MHR 8

9 The name is selenium hexafluoride. The name describes the correct number of atoms of the corresponding elements. 9. Practice Problem (page 19) Write the name of CaO. You need to determine the name of CaO. You are given the formula CaO. Ca is calcium and it is a metal. O is oxygen and it is a non-metal. The compound is ionic. Prefixes are not needed. Calcium has a charge of 2+. Oxygen has a charge of 2. In this example, the charges on each element are in the ratio of 2:2. This ratio reduces to a simpler whole number ratio of 1:1for the number of atoms. The name is calcium oxide. The charges on the ions add up to zero. 10. Practice Problem (page 73) Write the formula for cobalt(iii) nitrate. You know the name is cobalt(iii) nitrate. Cobalt (Co) is a metal. You know its charge must be 3+ because (III) is part of the name. Nitrate is a polyatomic ion that has the formula and charge NO 3. The compound is ionic. The charges are not the same so subscripts are needed. The ratio between the cobalt and nitrate ions is 1:3. This ratio does not reduce to a simpler whole number ratio. The subscript (number of atoms) of each element is the same as the magnitude of the charge of the other ion Chapter 2 Chemical Bonding MHR 9

10 1 Co 3+ ion balances the charge on 3 NO 3 ions. Therefore, the chemical formula is Co(NO 3 ) 3. The charges on the ions add up to zero. Section 2.2 Writing Names and Formulas for Ionic and Molecular Compounds Solutions for Selected Review Questions Student Edition page Review Question (page 75) Write the name of each compound. a. Al 2 O 3 e. NH 4 Cl b. HgI 2 f. LiClO 4 c. Na 3 P g. HNO 3 (aq) d. K 3 PO 4 h. LiOH (aq) a. aluminum oxide e. ammonium chloride b. mercury(ii) iodide f. lithium perchlorate c. sodium phosphide g. aqueous hydrogen nitrate or nitric acid d. potassium phosphate h. lithium hydroxide 6. Review Question (page 75) Write the formula for each compound. a. zinc oxide d. magnesium iodide b. iron(ii) sulfide e. cobalt(iii) chloride c. potassium hypochlorite f. sodium cyanide a. ZnO d. MgI 2 b. FeS e. CoCl 3 c. KClO f. NaCN 8. Review Question (page 75) The following six compounds contain nitrogen and oxygen: NO, NO 2, N 2 O, N 2 O 3, N 2 O 4, and N 2 O 5 Write the names of these compounds. NO, nitrogen monoxide NO 2, nitrogen dioxide N 2 O, dinitrogen monoxide N 2 O 3, dinitrogen trioxide N 2 O 4, dinitrogen tetroxide N 2 O 5, dinitrogen pentoxide Chapter 2 Chemical Bonding MHR 10

11 9. Review Question (page 75) Write the formula for each compound. a. phosphorus pentachloride d. silicon tetrabromide b. difluorine monoxide e. cobalt(ii) hydroxide c. sulfur trioxide f. sulfur hexafluoride a. PCl 5 d. SiBr 4 b. F 2 O e. Co(OH) 2 c. SO 3 f. SF Review Question (page 75) Write the name of each compound. a. CO e. SiO 2 b. BCl 3 f. PI 3 c. CS 2 g. Ba(OH) 2 d. CCl 4 h. H 3 BO 3 a. carbon monoxide e. silicon dioxide b. boron trichloride f. phosphorus triiodide c. carbon disulfide g. barium hydroxide d. carbon tetrachloride h. trihydrogen borate Section 2.3 Comparing the Properties of Ionic and Molecular Compounds Solutions for Selected Review Questions Student Edition page Review Question (page 82) If a compound has very high melting and boiling points, is the compound likely to be soluble in water? Explain the relationship between these two properties of a compound. Compounds having a very high melting and boiling point are likely to be soluble in water. The high melting and boiling points are due to strong forces of attraction between charged particles called ions. The ionic solids in which these particles are held together can be pulled apart from their crystal lattice structure when surrounded by polar water molecules. This is the dissolving process. 11. Review Question (page 82) Glycerol is a compound that dissolves readily in water. The water solution of glycerol, however, will not conduct an electric current. What would you predict about the properties of glycerol? Since the glycerol dissolves, it is likely made up of polar molecules. Glycerol dissolves as polar water molecules surround the polar glycerol molecules Chapter 2 Chemical Bonding MHR 11

12 However, an aqueous solution of glycerol will not conduct an electric current. When electrodes are place in a solution containing polar molecules, the molecules orient themselves so that their positive ends point to the negative electrode and their negative ends to the positive electrode. There is no movement of electrons. Since the glycerol molecules are polar, the melting point and boiling point would be expected to be intermediate in value, lower than that of ionic solids, but higher than that of molecular, non-polar solids. 12. Review Question (page 82) Under what two conditions can an ionic compound conduct an electric current? Ionic compounds can conduct in the liquid state and in aqueous solution. In both these instances, the ions are free to move. 13. Review Question (page 82) Can polar molecular compounds conduct electric current under either of the conditions that you described in question 12? Explain why or why not. Polar molecules cannot conduct electricity in the liquid state or in aqueous solution. In aqueous solution, these molecules orient themselves so that their positive end points to the negative electrode and their negative ends to the positive electrode. There is no movement of electrons. Acids are an exception to this since they react with water to form ions. In the liquid state, the polar molecules are not made up of separate positively and negatively charged particles. This is necessary if the solution is to conduct an electric charge. Each molecule has a positive and a negative end; overall the molecule is neutral. 14. Review Question (page 82) To be transported throughout the body in the bloodstream, fat molecules must be bound to protein molecules, as shown in the following figure. Explain why you think this is necessary Chapter 2 Chemical Bonding MHR 12

13 Fat molecules are not soluble because they are only slightly polar. Protein molecules are polymers, very large molecules, made up of amino acids with many sites that are polar, thus increasing the solubility. Molecules that make up fats can readily adhere to these protein molecules and are carried through the blood. 15. Review Question (page 82) You might have heard the saying, Like dissolves like. From what you have learned about solubility, comment on the validity of this statement. This is a general statement that is often valid. Like refers to similarity in structure: ionic compounds with polar solvents; polar molecules with polar solvents; non-polar solids with non-polar solvents. In nature, many factors affect the events we see and often explanations are not clear cut. There are many exceptions to Like dissolves like For example, the molecules of all alcohols have a polar end but not all alcohols are soluble in water. At some point, the non-polar chain in the alcohol molecule dominates the solubility process. 16. Review Question (page 82) Two molecular compounds, X and Y, have similar masses. Compound X is solid at room temperature, has a melting point of 146 C, and is soluble in water. Compound Y is liquid at room temperature, has a melting point of 10 C, and is not soluble in water. a. What would you predict about the polarities of compound X and compound Y? b. Based on your predictions, explain the differences in their melting points and solubilities. a. The compound having the higher melting point, X, would be predicted to be more polar. Melting point depends upon the electrical attractions between molecules. The more polar the molecule, the greater the attractions and the higher the melting point. Compounds X is polar; compound Y is less polar or possibly non- polar. b. Compound X has a melting point of intermediate value. It is lower than that of ionic solids, but higher than that of molecular, non-polar solids. Compound X would be expected to be made up of polar molecules. This compound will dissolve in water as its polar molecules become surrounded by polar water molecules. Since it has a low melting point, compound Y is likely made up of non-polar (or only slightly polar) molecules with low attractions between molecules. Polar water molecules will not be attracted to these molecules and the solubility would be low Chapter 2 Chemical Bonding MHR 13