5 Names and Formulas of Compounds

Transcription

1 47374_05_p33-42.qxd 2/9/07 7:52 AM Page 33 5 Names and Formulas of Compounds 5.1 a. When a sodium atom loses its valence electron, its second energy level has a complete octet. b. Group 1A (1) and 2A (2) elements can lose 1 or 2 electrons to attain a noble gas arrangement. Group 8A (18) elements already have an octet of valence electrons, so they do not lose or gain electrons and are not normally found in compounds. 5.2 a. When a chlorine atom gains a valence electron, its third energy level has a complete octet. b. Group 7A (17) elements can gain 1 electron to attain a noble gas arrangement. Group 8A (18) elements already have an octet of valence electrons, so they do not lose or gain electrons and are not normally found in compounds. 5.3 Atoms with 1, 2, or 3 valence electrons lose those electrons to form ions. a. 1 b. 2 c. 3 d. 1 e Atoms with 5, 6, or 7 valence electrons gain electrons when they form ions. a. 1 b. 2 c. 3 d. 1 e Atoms form ions by losing or gaining electrons to achieve the same electron arrangement of their nearest noble gas. a. Na has an electron arrangement 2, 8, which is the same as neon (Ne). b. Mg has an electron arrangement 2, 8, which is the same as neon (Ne). c. K has an electron arrangement 2, 8, 8, which is the same as argon (Ar). d. O has an electron arrangement 2, 8, which is the same as neon (Ne). e. has an electron arrangement 2, 8, which is the same as neon (Ne). F 5.6 a. He b. Kr c. Ar d. Ne e. Kr 5.7 Atoms form ions by losing or gaining electrons to achieve the same electron arrangement of their nearest noble gas. Elements in Groups 1A (1), 2A (2), and 3A (13) lose valence electrons, whereas elements in Groups 5A (15), 6A (16), and 7A (17) gain valence electrons to complete octets. a. Mg loses 2 e b. P gains 3 e c. Group 7A (17) gains 1 d. Na loses 1 e e. Al loses a. gain 2 e b. lose 2 e c. gain 1 d. lose 1 e e. gain 3 e 5.9 a. Li (3, 1) b. F (9, 10 1) c. Mg (1, 10 ) d. Fe 3 (26, 23 3) e. Zn (30, 28 ) 5.10 a. 8 protons, 10 electrons b. 19 protons, 18 electrons c. 35 protons, 36 electrons d. 16 protons, 18 electrons e. 38 protons, 36 electrons 5.11 A metal and a nonmetal are most likely to form an ionic compound. a. (Li and Cl) and c. (K and O) would form ionic compounds b. (Mg and Cl) and d. (K and S) will form ionic compounds. e e 33

2 47374_05_p33-42.qxd 2/9/07 7:52 AM Page 34 Chapter a. Potassium loses 1, and chlorine gains 1 e e K + Cl K + + Cl KCl b. Calcium loses 2, and two chlorine atoms each gain 1. e e Ca + Cl + Cl Ca Cl CaCl 2 c. Each of three sodium atoms lose 1 e, and the nitrogen gains 3 e. Na + Na + Na + N 3Na + + N Na 3 N a. Mg + S Mg 2+ + S 2 b. Al + Cl + Cl + Cl Al 3+ + Cl + Cl + Cl c. Li + Li + O Li + + Li + + O 2 0 3() 1(3) 0 2() 1() a. Na 2 O Check: 2Na and O 2() 1() 0 b. AlBr 3 Check: 1Al 3 and 3Br 1(3) 3() 0 c. BaO Check: Ba and O 1() 1() d. MgCl 2 Check: Mg and 2Cl 1() 2(1) 0 e. Al 2 S 3 Check: 2Al 3 and 3S 2(3) 3() a. AlCl 3 b. CaS c. Li 2 S d. K 3 N e. KI 5.17 a. Ions: Na and S : Na 2 S Check: 2Na and S 2() 1() 0 b. Ions: K and N 3 : K 3 N Check: 3K and N 3 c. Ions: A1 3 and I : AlI 3 Check: Al 3 and I 1(3) 3(1) 0 d. Ions: Li and O : Li 2 O Check: 2Li and O 5.18 a. CaCl 2 b. BaBr 2 c. Na 3 P d. MgO 5.19 a. Chlorine in Group 7A (17) gains 1 electron to form chloride ion Cl. b. Potassium in Group 1A (1) loses 1 electron to form potassium ion K. c. Oxygen in Group 6A (16) gains 2 electrons to form oxide ion O. d. Aluminum in Group 1A (1) loses 3 electrons to form aluminum ion 5.20 a. F b. Ca c. Na d. Li 5.21 a. potassium ion b. sulfide ion c. calcium ion d. nitride ion 5.22 a. magnesium b. barium c. iodide d. chloride 5.23 In the name for an ionic compound, the metal ion is named first followed by the nonmetal ion name ending in ide. a. Ions: Al 3 aluminum and O oxide : aluminum oxide b. Ions: Ca calcium and chloride Cl : calcium chloride c. Ions: Na sodium and O oxide : sodium oxide 34

3 47374_05_p33-42.qxd 2/9/07 7:52 AM Page 35 d. Ions: magnesium and nitride e. Ions: K potassium and I iodide f. Ions: barium and fluoride 5.24 a. magnesium chloride b. potassium phosphide c. lithium sulfide d. lithium bromide e. magnesium oxide f. strontium bromide 5.25 A Roman numeral is used to specify the positive charge on the transition metal in the compound, when that transition metal can have more than one cation Because calcium ion only has a 2 charge, the name calcium is sufficient. However, copper ions can have either a 1 or a 2 charge, which requires a Roman numeral to indicate which copper ion is present a. iron(ii) b. copper(ii) c. zinc d. lead(iv) e. chromium(iii) f. manganese(ii) 5.28 a. silver b. copper(i) c. iron(iii) d. tin(ii) e. gold(iii) f. nickel(ii) 5.29 For metal ions with variable charge, determine the ionic charge and use it as a Roman numeral following the name of the metal. a. Sn and 2Cl : tin(ii) chloride b. Fe and O : iron(ii) oxide c. Cu and S : copper(i) sulfide d. Cu and S : copper(ii) sulfide e. Cd and Br : cadmium bromide f. Hg and Cl : mercury(ii) chloride 5.30 a. silver phosphide b. lead(ii) sulfide c. tin(iv) oxide d. gold(iii) chloride e. chromium(iii) oxide f. cobalt(ii) sulfide 5.31 a. b. c. d. Mg Ba? 3(1) 0 2? 3(2) 0? 4(1) 0? 2(1) 0? 3 2? 6? 3? 4? 5.32 a. 2 b. 2 c. 3 d a. Ions: Mg and Cl : MgCl 2 b. Ions: Na and S : Na 2 S c. Ions: Cu and O : Cu 2 O d. Ions: Zn and P 3 : Zn 3 P 2 e. Ions: Au 3 and N 3 : AuN : magnesium nitride : potassium iodide : barium fluoride : Au 3 : Fe 3 : Pb 4 : Sn CuCl 2 : CoCl : Ca N : CrCl a. Fe 2 O 3 b. BaF 2 c. SnCl 4 d. Ag 2 S e a. Ions: Co 3 and Cl b. Ions: Pb 4 and O 3 : PbO 2 c. Ions: Ag and Cl : AgCl d. Ions: Ca and N 3 e. Ions: Cu and P : Cu 3 P f. Ions: Cr and Cl 5.36 a. Ions: Sn 4 and O : SnO 2 b. Ions: Fe 3 and S : Fe 2 S 3 c. Ions: Mn 4 and O : MnO 2 d. Ions: Cr 3 and I : CrI 3 e. Ions: Li and N 3 : Li 3 N f. Ions: Au and O : Au 2 O F N 3 3 Names and Formulas of Compounds 35

4 47374_05_p33-42.qxd 2/9/07 7:52 AM Page 36 Chapter The most common forms of negatively charged polyatomic ions end in ate; one O less end in ite. A hydrogen in front is named as hydrogen or bi. a. bicarbonate HCO 3 b. ammonium NH 4 3 c. phosphate PO 4 d. hydrogen sulfate HSO 4 e. hypochlorite ClO a. NO b. c. OH 2 SO 3 d. PO 3 e a. sulfate b. carbonate c. phosphate d. nitrate e. perchlorate 5.40 a. hydroxide b. hydrogen sulfite (or bisulfite) c. cyanide d. nitrite e. chromate 5.41 NO 2 C 2 H 3 O 2 CO 3 HSO 4 3 PO 4 Li lithium lithium lithium lithium nitrite carbonate hydrogen sulfate phosphate LiNO 2 Li 2 CO 3 LiHSO 4 Li 3 PO 4 Cu copper(ii) copper(ii) copper (II) copper (II) nitrite carbonate hydrogen sulfate phosphate Cu(NO 2 ) 2 CuCO 3 Cu(HSO 4 ) 2 Cu 3 (PO 4 ) 2 Ba barium barium barium hydrogen barium nitrite carbonate sulfate phosphate Ba(NO 2 ) 2 BaCO 3 Ba(HSO 4 ) 2 Ba 3 (PO 4 ) NH 4 NO 3 HCO 3 SO 3 HPO 4 ammonium ammonium ammonium sulfite ammonium NH 4 NO 3 carbonate phosphite nitrate hydrogen (NH 4 ) 2 SO 3 hydrogen NH 4 HCO 3 (NH 4 ) 2 HPO 4 Al 3 aluminum aluminum aluminum sulfite aluminum Al(NO 3 ) 3 carbonate phosphite nitrate hydrogen Al 2 (SO 3 ) 3 hydrogen Al(HCO 3 ) 3 Al 2 (HPO 4 ) 3 Pb 4 lead(iv) lead(iv) lead(iv) sulfite lead(iv) Pb(NO 3 ) 4 carbonate phosphite nitrate hydrogen Pb(SO 3 ) 2 hydrogen Pb(HCO 3 ) 4 Pb(HPO 4 ) a. The polyatomic ion is CO 3 ; the compound is sodium carbonate. b. The polyatomic ion is NH 4 ; the compound is ammonium chloride. 3 c. The polyatomic ion is PO 4 ; the compound is lithium phosphate. d. The polyatomic ion is NO 2 ; the compound is copper(ii) nitrite. e. The polyatomic ion is SO 3 ; the compound is iron(ii) sulfite. f. The polyatomic ion is C 2 H 3 O 2 ; the compound is potassium acetate. 36

5 47374_05_p33-42.qxd 2/9/07 7:52 AM Page a. OH ; potassium hydroxide b. NO 3 ; sodium nitrate c. CO 3 ; copper(ii) carbonate d. HCO 3 ; sodium hydrogen carbonate (or sodium bicarbonate) e. SO 4 ; barium sulfate f. ClO ; calcium hypochlorite Cu 2 SO Write the positive and negative ions. Then use charge balance to write the formula. a. Ions: Ba and 2OH : Ba(OH) 2 b. Ions: 2Na and SO 4 : Na 2 SO 4 c. Ions: Fe and 2NO : Fe(NO 3 ) 2 d. Ions: 3Zn and 3 3 2PO 4 : Zn 3 (PO 4 ) 2 e. Ions: 2Fe 3 and 3CO 3 : Fe 2 (CO 3 ) a. Al(ClO 3 ) 3 b. (NH 4 ) 2 O c. Mg(HCO 3 ) 2 d. NaNO 2 e a. This is an ionic compound with Al 3 ion and the sulfate SO 4 polyatomic ion. The correct name is aluminum sulfate. b. This is an ionic compound with Ca ion and the carbonate CO 3 polyatomic ion. The correct name is calcium carbonate. c. This is an ionic compound with Cr 3 and O. Because chromium has variable valences, a Roman numeral is used to indicate the 3 charge. The correct name is chromium(iii) oxide. 3 d. This is an ionic compound with sodium ion Na and the PO 4 polyatomic ion. The correct name is sodium phosphate. e. This ionic compound contains two polyatomic ions, ammonium NH 4 and sulfate SO 4. It is named ammonium sulfate. f. This is an ionic compound containing the variable metal ion Fe 3 and oxide ion O. It is named using the Roman numeral as iron(iii) oxide a. cobalt(iii) phosphate b. magnesium phosphate c. iron(ii) sulfate d. magnesium sulfate e. copper(i) oxide f. tin(ii) fluoride 5.49 The nonmetallic elements that are not noble gases are likely to form covalent bonds A bond forms between Na and Cl when an electron is lost by Na and gained by Cl. The bond that forms is an ionic bond. A bond forms between N and Cl when a pair of electrons, one from each atom, is shared between the two atoms. The bond that formed is a covalent bond a. When two H atoms share, each has 2 valence electrons. In H 2, there is 1 bonding pair and no lone pair. b. The Br atom achieves an octet by sharing a valence electron with one H atom to give 8 valence electrons, 1 bonding pair, and 3 lone pairs on the Br atom. c. Each Br atom achieves an octet by sharing 1 valence electron to give a total of 14 valence electrons, 1 bonding pair between the Br atoms, and 6 lone pairs (3 lone pairs for each Br atom) a. 8 valence electrons, 2 bonding pairs, and 2 lone pairs b. 8 valence electrons, 3 bonding pairs, and 1 lone pair c. 20 valence electrons, 2 bonding pairs, and 8 lone pairs Names and Formulas of Compounds 5.53 When naming covalent compounds, prefixes are used to indicate the number of each atom as shown in the subscripts of the formula. The first nonmetal is named the same as the element; the second nonmetal changes the ending of the element name to ide. a. one phosphorus atom and three (tri) bromine atoms; phosphorus tribromide b. one carbon atom and four (tetra) bromine atoms; carbon tetrabromide c. one silicon atom and two oxygen atoms; silicon dioxide d. one hydrogen atom and one fluorine atom; hydrogen fluoride e. one nitrogen atom and three (tri) iodine atoms; nitrogen triiodide 37

6 47374_05_p33-42.qxd 2/9/07 7:52 AM Page 38 Chapter a. carbon disulfide b. diphosphorus pentoxide c. dichlorine oxide d. phosphorus trichloride e. dinitrogen tetroxide 5.55 When naming covalent compounds, prefixes are used to indicate the number of each atom as shown in the subscripts of the formula. The first nonmetal is named the same as the element; the second nonmetal changes the ending of the element name to ide. a. two (di) nitrogen atoms and three (tri) oxygen atoms; dinitrogen trioxide b. one nitrogen atom and three (tri) chlorine atoms; nitrogen trichloride c. one silicon atom and four (tetra) bromine atoms; silicon tetrabromide d. one phosphorus atom and five (penta) chlorine atoms; phosphorus pentachloride e. one sulfur atom and three (tri) oxygen atoms; sulfur trioxide 5.56 a. silicon tetrafluoride b. iodine tribromide c. carbon dioxide d. sulfur dioxide e. dinitrogen oxide 5.57 The prefixes in the names of covalent compounds indicate the number of each type of atom, which is placed as a subscript in the formula. When there is no prefix there is one atom in the formula. a. carbon (1C) and tetrachloride (4Cl) b. carbon (1C) and monoxide (1O) CCl 4 CO c. phosphorus (1P) and trichloride (3Cl) PCl 3 d. dinitrogen (2N) and tetroxide (4O) N 2 O a. SO 2 b. SiCl 4 c. IF 5 d. N 2 O 5.59 The prefixes in the names of covalent compounds indicate the number of each type of atom, which is placed as a subscript in the formula. When there is no prefix, there is one atom in the formula. a. oxygen (1O) and difluoride (2F) OF 2 b. boron (1B) and trifluoride (3F) BF 3 c. dinitrogen (2N) and trioxide (3O) N 2 O 3 d. sulfur (1S) and hexafluoride (6F) SF a. SBr 2 b. CS 2 c. P 4 O 6 d. N 2 O a. aluminum chloride b. sulfur trioxide c. dinitrogen oxide d. tin(ii) nitrate e. copper(ii) chlorite 5.62 a. nitrogen b. magnesium hypobromite c. silicon tetrafluoride d. nickel (II) sulfate e. iron (III) sulfide 5.63 a. 15 protons make it phosphorus; 18 electrons gives a charge of 3; P 3. b. 8 protons and 8 electrons make it a neutral oxygen (O) atom. c. 30 protons make it zinc; 28 electrons gives a charge of ; Zn. d. 26 protons make it iron; 23 electrons gives a charge of 3; Fe a. X is in Group 1A (1); Y is in Group 6A (16) b. ionic c. X, Y d. X 2 Y e. XCl f. YCl H (E) 2. Li (C) 3. Li (A) 4. H (B) 5. N 3 (D) 5.66 calcium ions, Ca ; hydroxide ions, OH ; and phosphate ions, PO

7 47374_05_p33-42.qxd 2/9/07 7:52 AM Page 39 Names and Formulas of Compounds 5.67 Electron-dot Formula of Period Symbols Compound Name of Compound 2 X and Y Li 3 N lithium nitride 4 X and Y CaBr 2 calcium bromide 3 X and Y Al 2 S 3 aluminum sulfide 5.68 Electron-dot Formula of Period Symbols Compound Name of Compound 2 X and Y Be 3 N 2 beryllium nitride 3 X and Y Al 2 S 3 aluminum sulfide 5 X and Y SrI 2 strontium iodide 5.69 Electron-dot Formula of Electron Arrangements Symbols Cations Anions Compound Name of Compound 2, 8, 2 2, 5 Mg N Mg N 3 Mg 3 N 2 magnesium nitride 2, 8, 8, 1 2, 6 K O K O K 2 O potassium oxide 2, 8, 3 2, 8, 18, 7 Al Br Al 3 Br AlBr 3 aluminum bromide 5.70 Formula of Name of Electron Arrangements Electron-dot Symbols Compound Compound 2, 1 2, 8, 6 Li S Li 2 S lithium sulfide 2, 8, 8, 2 2, 8, 5 Ca P Ca 3 P 2 calcium phosphide 2, 8, 1 2, 8, 7 Na Cl NaCl sodium chloride 39

8 47374_05_p33-42.qxd 2/9/07 7:52 AM Page 40 Chapter a. Ne b. Ne c. Ar d. Ne e. He 5.72 a. Ne b. Kr c. Ar d. Ne e. Ar 5.73 a. An element that forms an ion with a charge would be in Group 2A (2). b. The electron-dot symbol for an element in Group 2A (2) is X. c. Be is the Group 2A (2) element in Period 2. d. Three ions, each with a charge, are needed to balance two N 3, which gives a formula X 3 N a. X is in Group 1A (1); Y is in Group 6A (16). b. X, Y c. X 2 Y d. XCl 5.75 a. Tin(IV) is Sn 4. b. The Sn 4 ion has 50 protons and 46 electrons. c. The Sn 4 ion is balanced with two O ions; SnO 2. 3 d. Three Sn 4 ions are needed to balance four PO 4 ions; Sn 3 (PO 4 ) Au (SO ) a. Gold (III) is Au b. The Au 3 ion has 79 protons and 76 electrons. c. Two Au 3 ions are balanced with three SO 4 ions; d. One Au 3 ion balance three Cl ions; AuCl a. X as a X 3 ion would be in Group 3A (13). b. X as a X ion would be in Group 6A (16). c. X as a ion would be in Group 4A (14) a. 3A (13) b. 6A (16) c. 4A (14) 5.79 a. Fe 3 is the iron(iii) ion; iron(iii) chloride. b. calcium phosphate c. aluminum carbonate d. Pb 4 is the lead(iv) ion; lead(iv) chloride. e. magnesium carbonate f. Sn is the tin(ii) ion; tin(ii) sulfate. g. is the copper(ii) ion: copper(ii) sulfide. Cu X a. tin(ii) sulfate b. barium nitrate c. manganese(ii) sulfide d. lithium perchlorate e. chromium(iii) phosphite f. sodium hydrogen phosphate g. calcium chloride N 3 ; Cu N. ; PbS a. Copper(I) is Cu and nitride is 3 b. Potassium ion is K and hydrogen sulfite is HSO ; KHSO 3. c. Lead(IV) is Pb 4 and sulfide is S 3 2 d. Gold(III) is Au 3 and carbonate is CO 3 ; Au 2 (CO 3 ) 3. e. Zinc ion is Zn and perchlorate is ClO ; Zn(ClO 4 ) a. b. c. Fe(NO 3 ) 3 Cu(HCO 3 ) 2 Sn(SO 3 ) 2 d. Ba(H 2 PO 4 ) 2 e. Cd(ClO) 2 : AuCl 5.83 a. Ions: Au 3 and Cl b. Ions: Pb 4 and O 3 : PbO 2 c. Ions: Ag and Cl : AgCl 40

9 47374_05_p33-42.qxd 2/9/07 7:52 AM Page 41 Ca N 3 d. Ions: and : Ca 3 N 2 e. Ions: Cu and P 3 : Cu 3 P f. Ions: Cr and Cl : CrCl a. SnO 2 b. Fe 2 S 3 c. PbS 2 d. CrI 3 e. Li 3 N f. Au 2 O 5.85 a. magnesium oxide b. Cr(HCO 3 ) 3 is chromium(iii) hydrogen carbonate or chromium (III) bicarbonate. c. manganese(iii) chromate 5.86 a. copper(i) sulfide b. iron(ii) phosphate c. calcium hypochlorite 5.87 a. 1N and 3Cl : nitrogen trichloride b. 1S and 2Cl : sulfur dichloride c. 2N and 1O : dinitrogen monoxide d. 2F : fluorine (named as the element) e. 1P and 5Cl : phosphorus pentachloride f. 2P and 5O : diphosphorus pentoxide 5.88 a. 1C and 4Br : carbon tetrabromide b. 1S and 6Fl : sulfur hexafluoride c. 2Br : bromine d. 2N and 4O : dinitrogen tetroxide e. 1S and 2O : sulfur dioxide f. 1C and 2S : carbon disulfide 5.89 a. 1C and 1O : CO b. di(2) and penta(5) c. di(2) and 1S d. 1S and di(2)cl : P 2 O 5 : H 2 S : SCl a. 1Si and di(2)o b. 1C and tetra(4)br c. 1S and tri(3)o d. 2N and 1O : SiO 2 : CBr 4 : SO 3 : N 2 O 5.91 a. ionic, iron(iii) chloride b. ionic, sodium sulfate c. covalent, 2N and 1O : dinitrogen oxide d. covalent, fluorine (named as the element) e. covalent, 1P and 5Cl : phosphorus pentachloride f. covalent, 1C and 4F : carbon tetrafluoride 5.92 a. ionic, aluminum carbonate b. covalent, sulfur hexafluoride c. covalent, diatomic element : bromine d. ionic, magnesium nitride e. covalent, sulfur dioxide f. ionic, chromium(iii) phosphate Names and Formulas of Compounds 5.93 a. Tin(II) is Sn ; carbonate is CO 3. With charges balanced, the formula is SnCO 3. b. Lithium is Li ; phosphide is P 3. Using three Li for charge balance, the formula is Li 3 P. c. Silicon has 4 valence electrons to share with 4 chlorine atoms to give SiCl 4. 41

10 47374_05_p33-42.qxd 2/9/07 7:52 AM Page 42 Chapter 5 d. Iron(III) is Fe 3 ; sulfide is S. Charge is balanced with 2 Fe 3 and 3 S to write the formula Fe 2 S 3. e. Carbon has 4 valence electrons to form 2 double bonds with 2 oxygen atoms to give the formula CO 2. f. Calcium is Ca ; Bromide is Br. With charges balanced, the formula is CaBr a. Na 2 CO 3 b. NO 2 c. Al(NO 3 ) 3 d. Cu 3 N e. K 3 PO 4 f. PbO Na 2 S has two Na ions to every S ion Na ions 1 S ion 2 Na ions S ions 5.96 Mg has three Mg 3 N 2 ions to every two ions Mg ions 2 N3 ions ions 3 Mg ions S 5.97 The valence electrons are the electrons in the highest energy level lost or gained in the formation of ionic compounds The octet rule states that a representative element will gain or lose electrons until the atom has 8 electrons in its outermost electron level Elements in Group 2A (2) will lose 2 electrons to attain an octet; elements in Group 6A (16) will gain 2 electrons to attain an octet. Both either gain or lose 2 electrons a. The correct name of the compound is calcium nitrate. b. Copper(II) oxide is CuO. Cu 2 O is copper(i) oxide. c. It is not necessary to put the symbol for potassium in brackets; the correct formula is K 2 CO 3. d. Na 2 S is sodium sulfide. Sodium sulfate is Na 2 SO 4. e. The silver ion is Ag. The formula for silver sulfate is Ag 2 SO N 3 Formula of Compound Type of Compound Name of Compound FeSO 4 ionic iron(ii) sulfate SiO 2 covalent silicon dioxide NH 4 NO 3 ionic ammonium nitrate Al 2 (SO 4 ) 3 ionic aluminum sulfate Co 2 S 3 ionic cobalt (III) sulfide a. Ar b. Kr c. Kr Compounds with a metal and nonmetal are classified as ionic; two nonmetals as covalent. a. ionic, lithium oxide b. covalent, dinitrogen oxide c. covalent, carbon tetrafluoride d. covalent, dichlorine oxide e. ionic, magnesium fluoride f. covalent, carbon monoxide g. ionic, calcium chloride h. ionic, potassium phosphate a. iron(ii) chloride b. dichlorine heptoxide c. nitrogen d. calcium phosphate e. phosphorus trichloride f. aluminum nitrate g. lead(iv) chloride h. magnesium carbonate i. nitrogen dioxide j. tin(ii) sulfate k. barium nitrate l. copper(ii) sulfide 42