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EXAMPLE 5.1

EXAMPLE 5.1. Lewis Symbols. Solution. Magnesium is in group 2A, oxygen is in group 6A, and phosphorus is in group 5A. The Lewis symbols therefore have two, six, and five dots, respectively. They are. • Mg •. Exercise 5.1.

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EXAMPLE 5.1

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  1. EXAMPLE 5.1 Lewis Symbols Solution Magnesium is in group 2A, oxygen is in group 6A, and phosphorus is in group 5A. The Lewis symbols therefore have two, six, and five dots, respectively. They are •Mg• Exercise 5.1 Without referring to Table 5.1, give Lewis symbols for each of the following elements. You may use the periodic table. a. Ar b. Ca c. F d. N e. K f. S • • :P • :O: • • Without referring to Table 5.1, give Lewis symbols for magnesium, oxygen, and phosphorus. You may use the periodic table.

  2. EXAMPLE 5.2 Electron Transfer to Form Ions Solution Sodium has one valence electron, and bromine has seven. Transfer of the single electron from sodium to bromine leaves each with a noble gas configuration. : : Na• + •Br:  Na+ +:Br:– : : Exercise 5.2A Exercise 5.2B Use Lewis symbols to show the transfer of electrons from lithium atoms to fluorine atoms to form ions with noble gas configurations. Use Lewis symbols to show the transfer of electrons from rubidium atoms to iodine atoms to form ions with noble gas configurations. Use Lewis symbols to show the transfer of electrons from sodium atoms to bromine atoms to form ions with noble gas configurations.

  3. EXAMPLE 5.3 Electron Transfer to Form Ions Solution Mg Mg + Mg Mg2+ Mg2+ + Mg2+ N N N 3– N 3–  Exercise 5.3 Each of three magnesium atoms gives up two electrons (a total of six), and each of the two nitrogen atoms acquires three (a total of six). Notice that the total positive and negative charges on the products are equal (6+ and 6–). Magnesium reacts with nitrogen to yield magnesium nitride (Mg3N2). Use Lewis symbols to show the transfer of electrons from aluminum atoms to oxygen atoms to form ions with noble gas configurations. Use Lewis symbols to show the transfer of electrons from magnesium atoms to nitrogen atoms to form ions with noble gas configurations.

  4. EXAMPLE 5.4 Determining Formulas by Electron Transfer Solution Sodium is in group 1A; the sodium atom has one valence electron. Sulfur is in group 6A; the sulfur atom has six valence electrons. Sulfur needs two electrons to gain an argon configuration, but sodium has only one to give. The sulfur atom therefore must react with two sodium atoms. The formula of the compound, called sodium sulfide, is Na2S. Na S Na Na Na+ Na+ + S 2– S +  Exercise 5.4A What are the formulas of the compounds formed by the reaction of (a) calcium with fluorine, and (b) lithium with oxygen? Exercise 5.4B Use data in Figure 5.4 to predict the formulas of the two compounds that can be formed from iron (Fe) and chlorine. What is the formula of the compound formed by the reaction of sodium and sulfur?

  5. EXAMPLE 5.5 Determining Formulas from Ionic Charges Solution First, write the symbols for the ions. (We write the charge on chloride ion explicitly as “1–” to illustrate the crossover method. You may omit the “1” when you are comfortable with the process.) Ca2+ Cl1– Then cross over the numbers as subscripts. Then rewrite the formula, dropping the charges. The formula for calcium chloride is Ca1Cl2 or (dropping the “1”) simply CaCl2 Write the symbols for the ions. Al3+ O2– Give the formulas for (a) calcium chloride and (b) aluminum oxide.

  6. EXAMPLE 5.5 Determining Formulas from Ionic Charges continued Cross over the numbers as subscripts. Then rewrite the formula, dropping the charges. The formula for aluminum oxide is Al2O3 Exercise 5.5 Give the formulas for (a) potassium oxide, (b) calcium nitride, and (c) calcium sulfide.

  7. EXAMPLE 5.6 Naming Ionic Compounds Solution From Table 5.2 we can determine that MgS is made up of Mg2+ (magnesium ion) and S2– (sulfide ion). The name is simply magnesium sulfide. From Table 5.2 we can determine that the ions in FeCl3 are Fe3+ Cl– How do we know the iron ion in FeCl3 is Fe3+ and not Fe2+? Because there are three Cl– ions, each 1–, the one Fe ion must be 3+ because the compound FeCl3 is neutral. The names of these ions are iron(III) ion (or ferric ion) and chloride ion. Therefore, the compound is iron(III) chloride (or, by the older system, ferric chloride). Exercise 5.6 What are the names of (a) CaF2 and (b) CuBr2? What are the names of (a) MgS and (b) FeCl3?

  8. EXAMPLE 5.7 Naming Covalent Compounds Solution With one sulfur atom and two clorine atoms, SCl2 is sulfur dichloride. With one sulfur atom and six fluorine atoms, SF6 is sulfur hexafluoride. Exercise 5.7A Exercise 5.7B What are the names of (a) BrF3 and (b) BrF5? What are the names of (a) N2O and (b) N2O5? What are the names of (a) SCl2 and (b) SF6?

  9. EXAMPLE 5.8 Formulas of Covalent Compounds Solution The tetra- indicates four phosphorus atoms, and the hex- specifies six oxygen atoms. The formula is P4O6. Exercise 5.8A Exercise 5.8B Give the formulas for (a) phosphorus trichloride and (b) dichlorine heptoxide. Give the formulas for (a) nitrogen triiodide and (b) disulfur dichloride. Give the formula for tetraphosphorus hexoxide.

  10. EXAMPLE 5.9 Covalent Bonds from Lewis Structures Solution a. b. bonding pair F + F F F bonding pair H + F H F Exercise 5.9 Use Lewis structures to show the formation of a covalent bond between (a) two bromine atoms, (b) between a hydrogen atom and a bromine atom, and (c) between an iodine atom and a chlorine atom. Use Lewis structures to show the formation of a covalent bond (a) between two fluorine atoms and (b) between a fluorine atom and a hydrogen atom.

  11. EXAMPLE 5.10 Covalent Bonds from Lewis Structures Solution • Two H atoms have exactly the same electronegativity; the electronegativity difference is 0; the bond is nonpolar covalent. • b. The electronegativity difference is 3.5 – 2.1 = 1.4; the bond is polar covalent. • c. The electronegativity difference is 2.5 – 2.1 = 0.4; the bond is nonpolar covalent. Exercise 5.10A Use data from Figure 5.5 to classify bonds between each of the following pairs of atoms as nonpolar covalent, polar covalent, or ionic: a. H, Br b. Na, O c. C, C Exercise 5.10B Use a periodic table to classify the following bonds as nonpolar covalent or polar covalent: a. C—N b. C—O c. C C — — Use data from Figure 5.5 to classify bonds between each of the following pairs of atoms as nonpolar covalent, polar covalent, or ionic: a. H, H b. O, H c. C, H

  12. EXAMPLE 5.11 Formulas Using Polyatomic Ions Solution Ammonium ion is found in Table 5.4; sulfide ion is a sulfur atom (group 6A) with two additional electrons. The ions are NH4+ S2– Crossing over, we get (NH4+)2S12– The parentheses with a subscript 2 indicate that the entire ammonium unit is taken twice; there are two nitrogen atoms and eight (4 x 2 = 8) hydrogen atoms. What is the formula for ammonium sulfide?

  13. EXAMPLE 5.11 Formulas Using Polyatomic Ions continued Exercise 5.11B Exercise 5.11A How many (a) nitrogen atoms are in the formula for ammonium nitrate; and (b) carbon atoms are in the formula for calcium acetate? What are the formulas for (a) calcium acetate, (b) ammonium nitrate, and (c) potassium permanganate?

  14. Solution The ions are Na+, sodium ion, and CN–, cyanide ion (found in Table 5.4). The compound is sodium cyanide. Exercise 5.12A Exercise 5.12B What are the names of (a) CaCO3 and (b) Mg3(PO4)2? What are the names of (a) K2CrO4 and (b) (NH4)2Cr2O7? EXAMPLE 5.12 Naming Compounds with Polyatomic Ions What is the name of the compound NaCN?

  15. EXAMPLE 5.13 Lewis Formulas Solution a. We start by following the preceding rules: 1. The total number of valence electrons is 4 + (4 x 1) + 6 = 14. 2. The skeletal structure must have all the H atoms on the outside. That means the C and H atoms must be bonded to each other. A reasonable skeletal structure is 3. Now, we count five bonds with two electrons each, making a total of ten electron. Thus four of the 14 valence electrons are left to be assigned. They are placed (as two lone pairs) on the oxygen atom. (The remaining steps are not necessary; both carbon and oxygen have octets of electrons.) H H C O H H H H C O H H Give Lewis formulas for (a) methanol, CH3OH, (b) the BF4– ion, and (c) carbon dioxide, CO2.

  16. EXAMPLE 5.13 Lewis Formulas continued b. Again, we start by applying the preceding rules: 1. There are 3 + (4 x 7) + 1 = 32 electrons. 2. The skeletal structure is 3. Place three lone pairs on each fluorine atom. 4. We have assigned 32 electrons. None remain to be assigned. A negative sign is added to show that this is the structure for an anion, not a molecule. F F B F F – F F B F F

  17. EXAMPLE 5.13 Lewis Formulas continued • c. Again, we start by applying the rules: • There are 4 + (2 x 6) = 16 valence electrons. • The skeletal structure is O—C—O. • Place three lone pairs on each oxygen atom. O—C—O • We have assigned 16 electrons. None remain to be placed. • The central carbon atom has only four electrons. It needs to form two double bonds in order to have an octet. Move a lone pair from each oxygen atom to the space between the atoms to form a double bond on each side of the carbon atom. O C O Exercise 5.13A Exercise 5.13B Give Lewis formulas for (a) oxygen difluoride, OF2, and (b) methyl chloride, CH3Cl. Give Lewis formulas for (a) azide ion, N3–, and (b) nitryl fluoride, NO2F (O—N—O—F skeleton).

  18. EXAMPLE 5.14 Shapes of Molecules Solution H We follow the preceding rules, starting with the Lewis structure. 1. The Lewis structure is 2. There are three sets to consider: two C—H single bonds and one C O double bond. 3. The three sets get as far apart as possible, giving a triangular arrangement of the sets. 4. All the sets are bonding pairs; the molecular shape is triangular, the same as the arrangement of the electrons. C O H 120° H C O H What are the shapes of (a) the H2CO molecule and (b) SCl2 molecule?

  19. EXAMPLE 5.14 Shapes of Molecules continued b. Again, we follow the rules, starting with the Lewis structure. 1. The Lewis structure is 2. There are four sets on the sulfur atom to consider. 3. The four sets get as far apart as possible, giving a tetrahedral arrangement of the sets about the central atom. 4. Two of the sets are bonding pairs and two are LP. Ignore the LP; the molecular shape is bent, with a bond angle of about 109.5°. Exercise 5.14A Exercise 5.14B What are the shapes of (a) the PH3 molecule and (b) the nitrate ion (NO3–)? What are the shapes of (a) the carbonate ion (CO32–) and (b) the hydrogen peroxide (H2O2) molecule? S Cl Cl

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