Lewis Formulas CHAPTER 7.

Transcription

1 ATER 7. Lewis ormulas 7-1. See the text. Because silicon is below carbon in the periodic table, we can determine the Lewis formula of Si 4 in the same manner as for Because phosphorus is the unique atom in this molecule, we shall assume that it is central and that the four chlorine atoms are attached to it. The total number of valence electrons is (1 5) + (4 7) 1 = 32. We use eight of the valence electrons to form phosphoruschlorine bonds. We now place valence electrons as lone pairs on the four chlorine atoms, accounting for the remaining 24 valence electrons. The completed Lewis formula is 7-3. We first arrange the atoms as The total number of valence electrons is (1 5) + (3 1) = 8. We use six valence electrons to form the bonds. We place the remaining two valence electrons as a lone pair on the atom. The Lewis formula is 7-4. We first arrange the atoms as N 13

2 14 GENERAL EMISTRY, URT EDITIN McQuarrie, Rock, and Gallogly The total number of valence electrons is (1 5) + (2 1) + 1 = 8. We use four valence electrons to form the N bonds and place the remaining four valence electrons as lone pairs on the N atom. The Lewis formula is N 7-5. See the text The Lewis formula for the tetrafluoroborate ion is B The formal charges on the boron atom and fluorine atoms are calculated using Equation 7.1. formal charge on the B atom = (4) = 1 2 formal charge on each atom = (2) = 0 Thus, the Lewis formula for a B 4 ion with the formal charges indicated is B 7-7. The formal charges on the oxygen and hydrogen atoms in Structure I are assigned according to Equation 7.1. formal charge on end atom = (2) = 1 2 formal charge on bonded atom = (6) =+1 2 The formal charges in Structure II are formal charge on each atom = (2) = 0 formal charge on each atom = (4) = 0 formal charge on each atom = (2) = 0 Structure II has zero formal charges everywhere and is the preferred Lewis formula for hydrogen peroxide (a) We first arrange the atoms as University Science Books, All rights reserved.

3 hapter 7: Lewis ormulas 15 The total number of valence electrons is (1 4) + (1 6) + (2 1) = 12. As in Example 7-8, we cannot satisfy the octet rule with just single bonds. Therefore, we form a double bond between the carbon atom and the oxygen atom, form two bonds, and place the remaining four valence electrons as lone pairs on the oxygen atom to write (b) The Lewis formula for the methyl group is Thus, the Lewis formula of ethanal is 7-9. As in Example 7-9, we cannot satisfy the octet rule with all single bonds or a double bond. Therefore, we form a triple bond between the two carbon atoms and two bonds to write We first arrange the atoms as The total number of valence electrons is (1 4) + (3 6) + (2 1) = 24. The resonance formulas are The formal charges indicated are calculated by using Equation 7.1. formal charge on atom = (4) = 0 formal charge on each single bonded atom = (2) = 1 2 formal charge on double bonded atom = (4) = 0

4 16 GENERAL EMISTRY, URT EDITIN McQuarrie, Rock, and Gallogly The superposition of these resonance forms gives the resonance hybrid 2 The three carbonoxygen bonds in a 2 3 ion are equivalent; they have the same bond length and the same bond energy A nitrogen dioxide molecule has (1 5) + (2 6) = 17 valence electrons and so is a free radical. The two resonance forms of a N 2 molecule are N + + N We first arrange the atoms as The total number of valence electrons is (1 5) + (1 6) + (3 7) = 32. We use eight valence electrons to form three bonds and one bond. The remaining 24 valence electrons are lone pairs on the oxygen atom and three chlorine atoms. The Lewis formula is Both the oxygen atom and phosphorus atom have a formal charge. We can eliminate the formal charges by expanding the the octet on the phosphorus atom to form a double bond between the phosphorus and oxygen atoms to obtain the Lewis formula Notice that all the atoms now have a zero formal charge A 6 ion has (1 5) + (6 7) + 1 = 48 valence electrons. Twelve are used to form the six bonds. The remaining 36 valence electrons are placed as lone electron pairs on the six chlorine atoms. The Lewis formula is University Science Books, All rights reserved.

5 hapter 7: Lewis ormulas A 3 4 ion has (1 5) + (4 6) + 3 = 32 valence electrons. The various resonance forms are + one of these four of these The superpositions of these resonance forms give the resonance hybrid 3 The four phosphorusoxygen bonds in the 3 4 ion are equivalent; they have the same bond length and bond energy An oxygen atom is more electronegative than a hydrogen atom, and so we have δ+ 2δ δ The difference in the electronegativities of a hydrogen atom and a silicon atom is = 0.2. Thus, the bonding in a Si 4 molecule is covalent with a small degree of ionic character. The difference in the electronegativities of a beryllium atom and a chlorine atom is = Thus, the bonding in a Si 4 molecule is polar covalent with a fair degree of ionic character.

6 ATER 8. rediction of Molecular Geometries 8-1. The Lewis formula of a 3 molecule is There are four valence-shell electron pairs about the central carbon atom, so it is tetrahedral. The bond angles are The Lewis formula of a B 4 molecule is B There are four valence-shell electron pairs about the central boron atom, and so the ion is tetrahedral The Lewis formula of an Al 3 6 molecule is Al 3 An Al 3 6 ion contains six covalent bonds, and so it is an octahedral ion with 90 bond angles The Lewis formula of a S 2 molecule is S.ItisanAX 2 E 2 molecule, and so it is bent with bond angles less than the tetrahedral bond angle of The Lewis formula of a S 2 molecule is S S.ItisanAX 2 molecule, and so it is linear. 18

7 hapter 8: rediction of Molecular Geometries The Lewis formula of a 2 ion is.itisanax 2 E 2 ion, and so it is bent. The resonance formulas of a 3 ion are It is an AX 3 E ion, and so it is trigonal pyramidal The Lewis formulas of a S 2 molecule are + S + S It is an AX 2 E molecule, and so is bent with a bond angle of less than 120 because of the two lone pairs The Lewis formula of a Xe 2 molecule is Xe.ItisanAX 2 E 3 molecule, and so it is linear The Lewis formulas of a Xe 3 molecule are Xe Xe Xe (or) Xe Thus, it is an AX 3 E molecule, and so is trigonal pyramidal (a) (b) δ+ δ δ+ δ Br δ δ tetrahedral with one chlorine atom; polar T-shaped; polar (c) δ δ δ δ Br δ δ+ square pyramidal; polar δ (d) δ See the text. δ As δ+ δ δ (symmetric) trigonal bipyramidal; nonpolar

8 ATER 9. ovalent Bonding 9-1. See the text Bond order = = 0. See the text Using igure 9.13, we have N + : 12 electrons (σ 1s ) 2 (σ1s )2 (σ 2s ) 2 (σ2s )2 (π 2p ) 4 bond order = 8 4 = 2 2 N: 13 electrons (σ 1s ) 2 (σ1s )2 (σ 2s ) 2 (σ2s )2 (π 2p ) 4 (σ 2p ) 1 bond order = 9 4 = N : 14 electrons (σ 1s ) 2 (σ1s )2 (σ 2s ) 2 (σ2s )2 (π 2p ) 4 (σ 2p ) 2 bond order = 10 4 = 3 2 The species with the largest bond order and thus the largest bond energy is N See the text See the text See the text See the text See the text See the text. 20

9 ATER 10. hemical Reactivity See the text See the text See the text See the text The oxalate ion has a charge of 2 (Table 10.1), and so we predict that it is a diprotic acid Using Tables 10.2 and 10.4 as a guide, we get the answers in the text The unreactive gas is N 2 (g) and the metallic mirrorlike deposit is Na(s). Thus, we predict that the decomposition reaction is described by See the text. 2 NaN 3 (s) 2Na(s) + 3N 2 (g) Aluminum replaces the hydrogen atoms in 2 S 4 (aq), giving us See the text. 2Al(s) S 4 (aq) Al 2 (S 4 ) 3 (aq) (g) Mg(s) combines with S 2 (g) to yield MgS 3 (s) as described in Section hlorine is more reactive than bromine and so liberates Br 2 (l) from solutions of bromide salts, and so (a) Soluble by Rule 1 (b) Insoluble by Rule 6 (c) Soluble by Rule 2 (d) Soluble by Rule (a) g 2 Br 2 is insoluble by Rule 3. 2Br (aq) + 2 (g) 2 (aq) + Br 2 (l) g 2 (N 3 ) 2 (aq) + 2 NaBr(aq) g 2 Br 2 (s) + 2 NaN 3 (aq) (b) No reaction. All possible products are soluble by Rule 1. 21

10 22 GENERAL EMISTRY, URT EDITIN McQuarrie, Rock, and Gallogly (c) ds is insoluble by Rule 5. Na 2 S(aq) + d(n 3 ) 2 (aq) 2 NaN 3 (aq) + ds(s) Using the solubility rules, we find that the resulting double-replacement reaction produces a silver chromate precipitate. The complete equation for the reaction is The net ionic equation is 2 Ag 4 (aq) + Na 2 r 4 (aq) 2 Na 4 (aq) + Ag 2 r 4 (s) 2Ag + (aq) + r 2 4 (aq) Ag 2r 4 (s) This is essentially an acid-base reaction, where Li 2 (s) dissolves in water to yield a base (see Section 10.3), according to Thus, or The net ionic equation is Li 2 (s) + 2 (l) 2 Li(aq) 2 S 4 (aq) + 2 Li(aq) Li 2 S 4 (aq) (l) 2 S 4 (aq) + Li 2 (s) Li 2 S 4 (aq) + 2 (l) 2 + (aq) + Li 2 (s) 2Li + (aq) + 2 (l) The chemical formula for sodium hydrogen carbonate is Na 3 (Table 10.1) and that of acetic acid is 3 (Table 10.3). Thus, the equation for the reaction is Na 3 (s) + 3 (aq) Na 3 (aq) + 2 (g) + 2 (l) (a) The ionic charges on iron in iron metal, e(s), and chlorine in chlorine gas, 2 (g), are zero. owever, the ionic charge on a chloride ion,,is 1, and so that on the iron atom in a e 3 formula unit must be +3. Because the ionic charges of the iron and chlorine atoms are changing, this is an oxidation-reduction reaction. (b) Because the nitrate ion, N 3, has an ionic charge of 1, the silver atom in a AgN 3 formula unit must have an ionic charge of +1. The ionic charges of the Na and S atoms in ana 2 S formula unit are +1 and 2, respectively. Similarly, we find for the products that the ionic charges of the Ag and S atoms in a Ag 2 S formula unit are +1 and 2, respectively, and the ionic charge of the Na atom in a NaN 3 formula unit is +1. Thus, we conclude that this is not an oxidation-reduction reaction because the ionic charges of each atom in the reactants are identical to their ionic charges in the products. (c) The ionic charge on zinc in zinc metal, Zn(s), is zero, and that of the zinc atom in a Zn 2 formula unit is +2. Similarly, the ionic charge on the mercury atom in a g 2 formula unit is +2, and that in mercury metal is 0. Thus, we conclude that this is an oxidation-reduction reaction The ionic charge of the aluminum atom changes from 0 in aluminum metal to +3 in aluminum oxide. Similarly, the ionic charge of the manganese atom changes from +3 in manganese oxide to 0 in manganese metal. Thus, in the chemical equation, two aluminum atoms are oxidized (0 +3) and two manganese atoms are reduced (+3 0), and so Al(s) is the reducing agent and Mn 2 3 (s) is the oxidizing agent. University Science Books, All rights reserved.

11 ATER 11. hemical alculations S 4 (l) 1000 kg 1000 g number of moles = ( mol 2 S 4 metric ton) 1 metric ton 1kg g 2 S 4 N 2 (g) = mol 2 S kg 1000 g number of moles = ( mol N 2 metric ton) 1 metric ton 1kg g N (g) = mol N kg 1000 g number of moles = ( mol 2 4 metric ton) 1 metric ton 1kg g (g) = mol kg 1000 g number of moles = ( mol 2 metric ton) 1 metric ton 1kg g 2 2 (g) = mol kg 1000 g number of moles = ( mol 2 metric ton) 1 metric ton 1kg g 2 = mol 2 n a mole basis, 2 (g) is produced in the greatest quantity annually in the United States. 23

12 24 GENERAL EMISTRY, URT EDITIN McQuarrie, Rock, and Gallogly The mass of a 2 molecule is ( g 2 mass of one molecule = 1 mol 2 = kg The mass of a S 6 molecule is ( g S6 mass of one molecule = 1 mol S 6 = kg ) ( )( 1kg 1000 g ) ( )( 1kg 1000 g 1 mol mol The number of each is given by 1000 ml 1.00 g number = ( mol 2 L) 1L 1mL g 2 ( ) molecule 1 mol = molecules = atoms = atoms As usual, take a 100-gram sample and write 52.1 g 13.2 g 34.7 g Divide each value by the corresponding atomic mass to get Dividing by 2.17, the smallest number, we get or The amount of oxygen is given by Thus, we have mol mol 2.17 mol 2.00 mol 6.04 mol 1.00 mol mass of oxygen = 3.37 g Sc 2.18 g Sc = 1.16 g 2.18 g Sc 1.16 g Dividing each value by the corresponding atomic mass gives mol Sc mol ) ) University Science Books, All rights reserved.

13 hapter 11: hemical alculations 25 Dividing by , we get 1 mol Sc 1.49 mol or, multiplying by 2, Sc The amount of oxygen is given by ( ) grams = grams. Thus, we have g M g The number of moles of oxygen is ( ) 1 mol moles of = (1.053 g ) = mol g The empirical formula of the oxide is M 2 3, so we have ( ) 2 mol M moles of M = ( mol ) = mol 3 mol and so or The metal is gallium As usual, take a 100-gram sample and write g M mol M g M 1 mol M g g g mol mol mol 1 mol 2 mol 1 mol or 2 is the empirical formula. Given that its molecular mass is and the empirical formula mass is 49.48, the molecular formula must be 98.95/49.48 = 2 times the empirical formula. Thus, the molecular formula is We have that ( g ) mass of = (3.710 g 2 ) g 2 = g mass of = (1.013 g 2 ) ( ) g = g g 2 mass % = g 100 = 40.89% g sample

14 26 GENERAL EMISTRY, URT EDITIN McQuarrie, Rock, and Gallogly mass % = As usual, take a 100-gram sample and write g 100 = 4.578% g sample mass % = ( )% = 54.53% g g g mol mol mol 1.00 mol 1.33 mol = 4 mol 1.00 mol 3 3 mol 4 mol 3 mol The empirical formula is Given that the empirical formula is and that its molecular mass is 176, the molecular formula of vitamin is (a) 2 K 3 (s) 2 K(s) (g) ( ) 3 mol 2 (b) moles of 2 = (0.50 mol K 3 ) = 0.75 mol 2 mol K 3 ( )( )( ) 1 mol K3 3 mol g 2 (c) mass of 2 = (30.6 g K 3 ) = 12.0 g g K 3 2 mol K 3 1 mol We have that ( ) ( ) 1g 1 mol ND 3 moles of ND 3 = (7.15 mg ND 3 ) 1000 mg g ND 3 = mol ( )( ) (1000 ) 3 mol mass of D 2 = ( D g D2 mg mol D 2 ) 1 mol ND 3 1 mol D 2 1g = 21.4 mgd 2 The number of millilters of D 2 (l) required is given by ( ) ( ) g 1mL volume = (21.4mg) = ml 1mg g We have that 1 mol S 1 mol S g S 6 mass of S 6 = (5.00 g S) = 22.8 g g S 1 mol S 1 mol S 6 and 1 mol S 3 mol g 2 mass of 2 = (5.00 g S) = 17.8 g g S 1 mol S 1 mol (a) The number of grams of I 3 (s) prepared is given by 1 mol 2 mol I g I 3 mass of I 3 = (1.25 g ) = 16.6 g g 2 mol 1 mol I 3 University Science Books, All rights reserved.

15 hapter 11: hemical alculations 27 (b) The number of grams of I 2 (s) required is given by 1 mol 3 mol I g I 2 mass of I 2 = (1.25 g ) = 15.4 g g 2 mol 1 mol I All the copper in ues 2 (s) ends up as copper metal. Thus, 30% ues g 1 mol ues 2 mass of u = (1000 kg ore) 100% ore 1kg g ues 2 ( ) ( ) 1 mol u g u 1 mol ues 2 1 mol u = g = metric ton Let x be the mass percentage of Na(s) in the mixture. The mass of Na(s) in the mixture is given by ( ) x mass of Na = (2.86 g) 100 and the mass of Ba 2 is given by mass of Ba 2 = (2.86 g) ( 1 x ) 100 The number of moles of each compound is ( )( ) x 1 mol Na moles Na = (2.86 g Na) = x mol g Na moles Ba 2 = [ ( (2.86 g) 1 x ) ] ( ) 1 mol Ba 2 Ba 2 = ( x) mol g Ba 2 The number of moles of Ag(s) is given by ( ) 1 mol Ag moles Ag = (4.81 g Ag) = mol Ag g Ag Because all the chloride in the Ag(s) comes from the Na(s) and Ba 2 (s), from the reaction stoichiometry, we have Therefore, we have moles Ag = moles Na + 2 moles Ba mol = ( x) mol + (2)( x) mol = x or x = 29%. Therefore, the mixture contains 29% Na(s) and 71% Ba 2 (s) The balanced chemical equation is as 4 (s) + 4(s) as(s) + 4(g)

16 28 GENERAL EMISTRY, URT EDITIN McQuarrie, Rock, and Gallogly We must check to see if there is a limiting reactant. ( ) 1 mol as4 moles of as 4 = (125 g as 4 ) = mol g as 4 ( ) 1 mol moles of = (125 g ) = 10.4 mol g Dividing each by its respective stoichiometric coefficient in the balanced equation, we find that the carbon is in great excess. The number of grams of as(s) produced is given by ( ) (72.14 ) 1 mol as g as mass of as = (0.918 mol as 4 ) = 66.2 g 1 mol as 4 1 mol as or a theoretical yield, the number of grams of tin required is given by ( )( ) ( ) 1 mol Sn4 1 mol Sn g Sn mass of Sn = (0.106 g Sn 4 ) = g g Sn 4 1 mol Sn 4 1 mol Sn or a percentage yield of 64.3%, the masss of tin required is ( ) 100% mass of Sn = ( g Sn) = g 64.3% University Science Books, All rights reserved.