lecture 1

UH Department of Chemistry ©CHEM 1331 Professor Geanangel

If you have an incomplete in CHEM 1331, send your name and PS number to me today.

1

CHEM 1331 Fall Semester 2011

Instructor: Prof. Geanangel Office: 123B-Fleming

Please use Blackboard mail to contact me.

Chemistry for Science and Engineering Majors

www.uh.edu/blackboard (Take PA by Fri. 5 PM!)

Office Hours: MWF 11:00 AM - 12:00 PM

Read your syllabus! Keep it for reference.

Sections: MWF 10:00 -11:00 AM & 1:00-2:00 PM 160F


CHEM 1331 Fall Semester 2011Chapter 2: The Components of Matter

2.1 Elements, Compounds & Mixtures: An Atomic Overview*

2.2 The Observations That Led to an Atomic View of Matter

2.3 Dalton’s Atomic Theory

2.4 The Observations That Led to the Nuclear Atom Model

2.5 The Atomic Theory Today

2.6 Elements: A First Look at the Periodic Table

2.7 Compounds: Introduction to Bonding

2.8 Compounds: Formulas, Names, and Masses

2.9 Mixtures: Classification and Separation*

* Required for self study - not covered specifically in lecture

2

UH Department of Chemistry ©CHEM 1331 Professor Geanangel 3

Observations that led to an atomic view of matter

The total mass of substances does not change during chemical reactionsThe number of substances may change, but the total quantity of matter remains constant

Conservation of Mass (nothing is lost, unless...)

reactant 1 + reactant 2 +... product(s)

total mass reactants total mass products=calcium oxide + carbon dioxide calcium carbonate

56.08g + 44.00g ! 100.08g

CaO + CO2 ! CaCO3


Example: When 0.2250 g Mg was heated with 0.5331 g N2 gas, all the Mg was consumed forming 0.3114 g Mg3N2. What mass of N2 is left over?

Mg(s) + N2(g) ! Mg3N2(s) ( not balanced)

Start 0.2250g 0.5331g 0 g

End 0g X g 0.3114 g


Definite Composition (What’s in it and how much?)A chemical compound is always composed of the same elements in the same fractions by mass.Experimental analysis of the elemental mass composition of 20.0 g calcium carbonate:Mass Analysis (grams/20.0 g) 8.0 g calcium 2.4 g carbon 9.6 g oxygen 20.0 g total

Mass Fraction(parts/1.00 part) 0.40 calcium 0.12 carbon 0.48 oxygen 1.00 part by mass

Percent by Mass(parts/100 parts) 40% calcium 12% carbon 48% oxygen 100% by mass

Lab measurement

MF x 100% = % by mass

8.0 g Ca 20.0 g total

= 0.40MF =


Mass of an Element X in Compound XYZ

Galena, a mineral, consists of Pb and S. A 1.27 gquantity of Galena contains 1.10 g of Pb. How many grams of S are present in a 1 kg sample of Galena (G)?

!

mass of X in sample = mass fraction x mass of sample

!

mass fraction X in XYZ = mass of X in XYZmass of XYZ


Problem: A 1.27 g sample of Galena contains 1.10 g of Pb. What is the mass percent of S in Galena?


The Law of Multiple Proportions If elements A and B react to form two compounds, masses of B that combine with a fixed mass of A are found to be in a ratio of small whole numbers. Consider two compounds of elements carbon (C) and oxygen (O); call them carbon oxides I and II.

Carbon oxide I: 57.1 mass %O and 42.9 mass %C Carbon oxide II: 72.7 mass %O and 27.3 mass %C

Use %C and %O to find masses of C and O


From %, find masses of C and O in 100 g of I and II Carbon Oxide I Carbon Oxide IIg O/100g compound 57.1 72.7 g C/100g compound 42.9 27.3g oxygen/g carbon 57.1/42.9 72.7/27.3 = 1.33 = 2.66 Dividing the g O/g C ratio in II by that in I gives a ratio of small whole numbers:

2.66 g O/g C in II = 21.33 g O/g C in I 1


Four postulates help explain the emprical “laws”

1. Matter consists of atoms, tiny, indivisible particles of an element that cannot be created or destroyed.

2. Atoms of one element cannot be converted into atoms of another element.

3. Atoms of an element are identical in mass and other properties and are different from atoms of other elements.

4. Compounds result from the chemical combination of a specific ratio of atoms of different elements.

Dalton’s Atomic Theory (1808)


Dalton’s postulates help explain mass lawsMass conservation Atoms cannot be created, destroyed (post.1) or converted into other types of atoms (post. 2)

Definite composition Compounds are combinations of a specific ratio of different atoms (post. 4), each of which has a particular mass (post. 3).

Atoms have fixed masses (post. 3). In chemical rxns., atoms are just rearranged, - mass doesn’t change.

Multiple proportions Different numbers of B atoms combine with each A atom in the two oxides giving a small, whole-number ratio. A simple arrangement consistent with the mass data has 1 atom of O combined with 1 atom of C in I.

Two atoms of O combine with one atom of C in II.


Workshop on Empirical Mass Laws What set of the three mass laws are illustrated by this reaction?


Workshop What set of these statements contains all that correctly describe the change shown below and no others? Assume A = orange atoms and B = blue atoms

i) A mixture of A2 and B2 molecules reacts to form the compound ABii) The Law of Conservation of Mass is obeyed during the changeiii) Since 3 A2 and 3 B2 combine to form 6 AB, the change illustrates the Law of Multiple Proportions

13


Review: Static Electric Charges and Coulomb’s Law

Neutral matter contains equal numbers of + and - chargesFriction can separate some charge between objects.

According to Benjamin Franklin:Rubbing glass with silk gives the glass a positive charge.Rubbing plastic with fur gives plastic a negative charge.

14


The S.I. unit of charge is the Coulomb (C).

Coulomb’s Law (Charles Coulomb, 1736-1806)q1 q2

r

q = charge (C)

+ + F > 0, repulsive force

__ F > 0, repulsive force

+ _ F < 0, attractive force

+_ F < 0, attractive force!

F =k " q1 " q2

r2

!

PE =k '"q1 " q2

r

15


Observations that Led to a Nuclear Model of the AtomSir William Crookes built a glass tube fitted with metal electrodes and pumped most of the air from it.

Cathode (-) Ring Anode (+)

Battery

Vacuum pump


With power on, Crookes saw a bright spot due to a “ray” striking the phosphor on the end of the tube.

Cathode “rays” moved from the negative electrode (cathode) to the positive electrode in a straight line.


Cathode rays were deflected by magnetic fields and electric fields. All cathode rays behaved the same, no matter what metal was used for the cathode.

Particles in the beam were attracted to a positively charged plate, so they must be negatively charged.

Identified by J. J. Thomson, (1897) as electrons.

!

masscharge

= "5.7x10"12kg /C


(–)

(+)X-ray source

0

5

1

2

3

46

7

8

9

Observer times droplet’smotion and controlselectric field

5

Electrically chargedplates influencedroplet’s motion

4

X-rays knockelectrons fromsurrounding air,which stick todroplet

3

Oil droplets fallthrough hole inpositivelycharged plate

2

Fine mist of oil issprayed into apparatus

1

Milikan Oil Drop Exp’t (what’s the electron charge?)


Charges on all droplets were some whole-number multiple of a minimum charge.

Oil drops pick up different numbers of electrons; the minimum charge must be that of one electron.

Millikan’s value was within 1% of the modern value of the electron’s charge, -1.602 x 10 -19 C (Coulomb)

From Thomson’s mass/charge ratio value and the charge, Millikan determined the electron’s mass:

!

mass of e_ =mass

chargex charge

Electron mass ! 10-30 kg!


Atoms are electrically neutral. So, what positive charges balance the negative electrons?If electrons have such tiny masses, what accounts for the remainder of an atom’s mass?

positive charge

electron

• Thomson proposed a spherical atom model composed of diffuse, positively charged matter, in which electrons were embedded like “plums in pudding.”


B Experiment

Gold foil

Lead block

Radioactive sample emitsbeam of ! particles

1

A Hypothesis: Expected result basedon “plum pudding” model

Cross section of gold foilcomposed of “plum pudding”

atoms

Major deflectionsof ! particles areseen very rarely

5

Almost nodeflection

Incoming! particles

Beam of ! particlesstrikes gold foil

2

Minor deflections of! particles are seenoccasionally

4

Flashes of light produced when ! particles strike zinc-sulfide screen showthat most ! particles aretransmitted with little orno deflection.

3

Rutherford Scattering Experiment


23

Data showed that most α particles weren’t deflected but that 1 in 20,000 was deflected by more than 90°.Could only happen if nearly all mass and positive charge reside in a tiny region within the atom.


Rutherford’s results suggested: -volume of atoms are mostly occupied by electrons; within that lies a tiny region, the atomic nucleus.

Rutherford’s model explained the charged nature of matter but didn’t account for all the atom’s mass. Later, Chadwick discovered the neutron: an uncharged particle in nucleus; about same mass as proton, accounted for the “missing mass”.

- atomic nucleus contains all the positive charge and almost all the mass of the atom. - positive particles (protons) occupy the nucleus.


Approximately 10–10 m

B Nucleus

Approximately 10–15 m

Electrons, e–

(negative charge)

A Atom

Nucleus

Proton, p+

(positive charge)

Neutron, n0

(no charge)


++


AZ X Atomic

symbol

Mass number(p+ + n0)

AtomicNumber (p+)

6p+

6n0An atom of carbon-12 12

6C

6e–

8p+

8n0An atom of oxygen-16 168O

8e–

An atom of uranium-235235

92U

92e–

92p+

143n0

92p+

146n0An atom of uranium -238238

92U

92e–

Isotopes (Dalton got this wrong)Atoms with the same number of protons (p+) but different numbers of neutrons (no) are called isotopes.

isotopes


Number of neutrons = A - Z

Chlorine-35 atoms have A = 35, Z = 17, and so they have 35 -17 = 18 no

Mass number, A, is sum of protons and neutrons

Problem: An atom of boron-11 has Z = 5. How many p+, e– and no does it contain?

How to determine the number of neutrons AZ X


Ne + e– ! Ne+ + 2e–


Mass Spectrum of NeMass spectra show the abundance (%) of each isotope.

How to calculate the average mass of neon from its isotope masses weighted according to their abundances.

Weighted average mass ="(isotope mass x fract. abund.)

UH Department of Chemistry ©CHEM 1331 Professor Geanangel31

Problem: Silver (Z =47) has only two isotopes, 107Ag and 109Ag. Given the mass spectrometric data below, calculate the atomic mass of silver:

ISOTOPE MASS (amu) ABUNDANCE (%)107Ag 106.90509 51.84109Ag 108.90476 48.16

Find atomic mass contribution of each isotope: mass contrib = isotopic mass x fractional abund. For 107Ag: = 106.90509 amu x 0.5184 = 55.42 amu For 109Ag: = 108.90476 amu x 0.4816 = 52.45 amu

47Ag

107.9


Atomic Mass Scale TodayThe atomic mass standard is the carbon-12 atom; its mass is defined as exactly 12 atomic mass units.So one atomic mass unit (amu or u) is defined as 1/12th the mass of a carbon-12 atom.On this scale, H atoms have a mass of 1.008 amu.

One amu (u) equals 1.661 x 10–24 g (PT/const. sheet) Exercise: What is the mass of one atom of H?


1H

1.008

2He

4.003

3Li

6.941

4Be

9.012

11Na

22.99

12Mg

24.31

19K

39.10

20Ca

40.08

37Rb

85.47

38Sr

87.62

55Cs

132.9

56Ba

137.3

87Fr

(223)

88Ra

(226)

21Sc

44.96

22Ti

47.88

39Y

88.91

40Zr

91.22

71Lu

175.0

72Hf

178.5

103Lr

(260)

104Rf

(261)

23V

50.94

24Cr

52.00

41Nb

92.91

42Mo

95.94

73Ta

180.9

74W

183.9

105Db

(262)

106Sg

(266)

25Mn

54.94

26Fe

55.85

43Tc

(98)

44Ru

101.1

75Re

186.2

76Os

190.2

107Bh

(262)

108Hs

(265)

27Co

58.93

45Rh

102.9

77Ir

192.2

109Mt

(266)

57La

138.9

58Ce

140.1

60Nd

144.2

61Pm

(145)

62Sm

150.4

63Eu

152.0

59Pr

140.9

89Ac

(227)

90Th

232.0

92U

238.0

93Np

(237)

94Pu

(242)

95Am

(243)

91Pa

(231)

28Ni

58.69

29Cu

63.55

46Pd

106.4

47Ag

107.9

78Pt

195.1

79Au

197.0

30Zn

65.39

48Cd

112.4

80Hg

200.6

64Gd

157.3

65Tb

158.9

67Ho

164.9

68Er

167.3

69Tm

168.9

70Yb

173.0

66Dy

162.5

96Cm

(247)

97Bk

(247)

99Es

(252)

100Fm

(257)

101Md

(258)

102No

(259)

98Cf

(251)

31Ga

69.72

32Ge

72.6149In

114.8

50Sn

118.7

81Tl

204.4

82Pb

207.2

33As

74.92

51Sb

121.8

83Bi

209.0

34Se

78.96

35Br

79.90

52Te

127.6

53I

126.9

84Po

(209)

85At

(210)

36Kr

83.80

54Xe

131.3

86Rn

(222)

5B

10.81

6C

12.01

13Al

26.98

14Si

28.09

7N

14.01

15P

30.97

8O

16.00

9F

19.00

16S

32.07

17Cl

35.45

10Ne

20.18

18Ar

39.953

4

5

6

7

3B(3)

4B(4)

5B(5)

6B(6)

7B(7) (9) (10)

1B(11)

2B(12)

3A(13)

4A(14)

5A(15)

6A(16)

7A(17)

8A(18)

2

1

6

7

(8)

1A(1)

2A(2)

8B

Lanthanides

Actinides

TRANSITION ELEMENTS

INNER TRANSITION ELEMENTS

MAIN–GROUPELEMENTS


110 111 112

(269) (272) (277)

Metals (main-group)Metals (transition)Metals (inner transition)MetalloidsNonmetals

114

(285)

PE

RIO

D

Learn names and symbols of the first 36 elements


The Formation of Ionic Compounds

Ionic compounds made of ions, charged particles formed when atom(s) gain or lose electron(s).

Ionic compounds typically form when a metal reacts with a nonmetal.

Each metal atom loses 1, 2 or 3 of its electrons and becomes a cation, a positively charged ion.

Nonmetal atoms gain electrons lost by the metal atoms becoming anions, negatively charged ions.

Compounds: Introduction to Bonding


How can we predict the number of electrons an atom will lose or gain when it forms a monoatomic ion?

Ionic compounds always contain equal numbers of positive and negative charges. Ionic compounds are neutral; i.e., zero net charge. A grain of table salt consists of a large number of Na+ ions and an equal number of Cl– ions.

At this point in the text, the clearest distinction among the elements is theirclassification as metals, nonmetals, or metalloids. The “staircase” line that runsfrom the top of Group 3A(13) to the bottom of Group 6A(16) in Period 6 is adividing line for this classification. The metals (three shades of blue) appear inthe large lower-left portion of the table. About three-quarters of the elements aremetals, including many main-group elements and all the transition and inner tran-sition elements. They are generally shiny solids at room temperature (mercury isthe only liquid) that conduct heat and electricity well and can be tooled into sheets(malleable) and wires (ductile). The nonmetals (yellow) appear in the smallupper-right portion of the table. They are generally gases or dull, brittle solids atroom temperature (bromine is the only liquid) and conduct heat and electricitypoorly. Along the staircase line lie the metalloids (green; also called semimetals),elements that have properties between those of metals and nonmetals. Several

58 Chapter 2 The Components of Matter

1H

1.008

2He

4.003

3Li

6.941

4Be

9.012

11Na

22.99

12Mg

24.31

19K

39.10

20Ca

40.08

37Rb

85.47

38Sr

87.62

55Cs

132.9

56Ba

137.3

87Fr

(223)

88Ra

(226)

21Sc

44.96

22Ti

47.88

39Y

88.91

40Zr

91.22

71Lu

175.0

72Hf

178.5

103Lr

(260)

104Rf

(263)

23V

50.94

24Cr

52.00

41Nb

92.91

42Mo

95.94

73Ta

180.9

74W

183.9

105Db

(262)

106Sg

(266)

25Mn

54.94

26Fe

55.85

43Tc(98)

44Ru

101.1

75Re

186.2

76Os

190.2

107Bh

(267)

108Hs

(277)

27Co

58.93

45Rh

102.9

77Ir

192.2

109Mt

(268)

57La

138.9

58Ce

140.1

60Nd

144.2

61Pm

(145)

62Sm

150.4

63Eu

152.0

59Pr

140.9

89Ac

(227)

90Th

232.0

92U

238.0

93Np

(237)

94Pu

(242)

95Am(243)

91Pa

(231)

28Ni

58.69

29Cu

63.55

46Pd

106.4

47Ag

107.9

78Pt

195.1

79Au

197.0

30Zn

65.41

48Cd

112.4

80Hg

200.6

64Gd

157.3

65Tb

158.9

67Ho

164.9

68Er

167.3

69Tm

168.9

70Yb

173.0

66Dy

162.5

96Cm(247)

97Bk

(247)

99Es

(252)

100Fm

(257)

101Md

(258)

102No

(259)

98Cf

(251)

31Ga

69.72

32Ge

72.61

49In

114.8

50Sn

118.7

81Tl

204.4

82Pb

207.2

33As

74.92

51Sb

121.8

83Bi

209.0

34Se

78.96

35Br

79.90

52Te

127.6

53I

126.9

84Po

(209)

85At

(210)

36Kr

83.80

54Xe

131.3

86Rn

(222)

5B

10.81

6C

12.01

13Al

26.98

14Si

28.09

7N

14.01

15P

30.97

8O

16.00

9F

19.00

16S

32.07

17Cl

35.45

10Ne

20.18

18Ar

39.953

4

5

6

7

3B(3)

4B(4)

5B(5)

6B(6)

7B(7) (9) (10)

1B(11)

2B(12)

3A(13)

4A(14)

5A(15)

6A(16)

7A(17)

8A(18)

2

1

6

7

(8)

1A(1)

2A(2)

8B

Lanthanides

Actinides

TRANSITION ELEMENTS




Perio

d

110

(281)

111

(272)

112


114

(289)

113

(284)

115

(288)

116

(292)Ds Rg

(285)

placed below the main body of the table but actually fit between theelements indicated. Metals lie below and to the left of the thick“staircase” line [top of 3A(13) to bottom of 6A(16) in Period 6] and in-clude main-group metals ( purple-blue), transition elements (blue), andinner transition elements ( gray-blue). Nonmetals (yellow) lie to the rightof the line. Metalloids ( green) lie along the line. We discuss the place-ment of hydrogen in Chapter 14. As of mid-2007, elements 112–116had not been named.

Figure 2.9 The modern periodic table. The table consists of ele-ment boxes arranged by increasing atomic number into groups (verti-cal columns) and periods (horizontal rows). Each box contains theatomic number, atomic symbol, and atomic mass. (A mass in paren-theses is the mass number of the most stable isotope of that element.)The periods are numbered 1 to 7. The groups (sometimes calledfamilies) have a number-letter designation and a new group number inparentheses. The A groups are the main-group elements; the B groupsare the transition elements. Two series of inner transition elements are

siL48593_ch02_040-088 30:11:07 10:53pm Page 58



1H

1.008

2He

4.003

3Li

6.941

4Be

9.012

11Na

22.99

12Mg

24.31

19K

39.10

20Ca

40.08

37Rb

85.47

38Sr

87.62

55Cs

132.9

56Ba

137.3

87Fr

(223)

88Ra

(226)

21Sc

44.96

22Ti

47.88

39Y

88.91

40Zr

91.22

71Lu

175.0

72Hf

178.5

103Lr

(260)

104Rf

(263)

23V

50.94

24Cr

52.00

41Nb

92.91

42Mo

95.94

73Ta

180.9

74W

183.9

105Db

(262)

106Sg

(266)

25Mn

54.94

26Fe

55.85

43Tc(98)

44Ru

101.1

75Re

186.2

76Os

190.2

107Bh

(267)

108Hs

(277)

27Co

58.93

45Rh

102.9

77Ir

192.2

109Mt

(268)

57La

138.9

58Ce

140.1

60Nd

144.2

61Pm

(145)

62Sm

150.4

63Eu

152.0

59Pr

140.9

89Ac

(227)

90Th

232.0

92U

238.0

93Np

(237)

94Pu

(242)

95Am(243)

91Pa

(231)

28Ni

58.69

29Cu

63.55

46Pd

106.4

47Ag

107.9

78Pt

195.1

79Au

197.0

30Zn

65.41

48Cd

112.4

80Hg

200.6

64Gd

157.3

65Tb

158.9

67Ho

164.9

68Er

167.3

69Tm

168.9

70Yb

173.0

66Dy

162.5

96Cm(247)

97Bk

(247)

99Es

(252)

100Fm

(257)

101Md

(258)

102No

(259)

98Cf

(251)

31Ga

69.72

32Ge

72.61

49In

114.8

50Sn

118.7

81Tl

204.4

82Pb

207.2

33As

74.92

51Sb

121.8

83Bi

209.0

34Se

78.96

35Br

79.90

52Te

127.6

53I

126.9

84Po

(209)

85At

(210)

36Kr

83.80

54Xe

131.3

86Rn

(222)

5B

10.81

6C

12.01

13Al

26.98

14Si

28.09

7N

14.01

15P

30.97

8O

16.00

9F

19.00

16S

32.07

17Cl

35.45

10Ne

20.18

18Ar

39.953

4

5

6

7

3B(3)

4B(4)

5B(5)

6B(6)

7B(7) (9) (10)

1B(11)

2B(12)

3A(13)

4A(14)

5A(15)

6A(16)

7A(17)

8A(18)

2

1

6

7

(8)

1A(1)

2A(2)

8B

Lanthanides

Actinides

TRANSITION ELEMENTS




Perio

d

110

(281)

111

(272)

112


114

(289)

113

(284)

115

(288)

116

(292)Ds Rg

(285)



siL48593_ch02_040-088 30:11:07 10:53pm Page 58

Often, ions are formed with the same number of electrons as in the nearest noble gas [Group 8A]

10

2


Exercise: What monatomic ions do the elements below form? a) iodine (Z= 53) b) strontium (Z = 38) c) aluminum (Z = 13)



1H

1.008

2He

4.003

3Li

6.941

4Be

9.012

11Na

22.99

12Mg

24.31

19K

39.10

20Ca

40.08

37Rb

85.47

38Sr

87.62

55Cs

132.9

56Ba

137.3

87Fr

(223)

88Ra

(226)

21Sc

44.96

22Ti

47.88

39Y

88.91

40Zr

91.22

71Lu

175.0

72Hf

178.5

103Lr

(260)

104Rf

(263)

23V

50.94

24Cr

52.00

41Nb

92.91

42Mo

95.94

73Ta

180.9

74W

183.9

105Db

(262)

106Sg

(266)

25Mn

54.94

26Fe

55.85

43Tc(98)

44Ru

101.1

75Re

186.2

76Os

190.2

107Bh

(267)

108Hs

(277)

27Co

58.93

45Rh

102.9

77Ir

192.2

109Mt

(268)

57La

138.9

58Ce

140.1

60Nd

144.2

61Pm

(145)

62Sm

150.4

63Eu

152.0

59Pr

140.9

89Ac

(227)

90Th

232.0

92U

238.0

93Np

(237)

94Pu

(242)

95Am(243)

91Pa

(231)

28Ni

58.69

29Cu

63.55

46Pd

106.4

47Ag

107.9

78Pt

195.1

79Au

197.0

30Zn

65.41

48Cd

112.4

80Hg

200.6

64Gd

157.3

65Tb

158.9

67Ho

164.9

68Er

167.3

69Tm

168.9

70Yb

173.0

66Dy

162.5

96Cm(247)

97Bk

(247)

99Es

(252)

100Fm

(257)

101Md

(258)

102No

(259)

98Cf

(251)

31Ga

69.72

32Ge

72.61

49In

114.8

50Sn

118.7

81Tl

204.4

82Pb

207.2

33As

74.92

51Sb

121.8

83Bi

209.0

34Se

78.96

35Br

79.90

52Te

127.6

53I

126.9

84Po

(209)

85At

(210)

36Kr

83.80

54Xe

131.3

86Rn

(222)

5B

10.81

6C

12.01

13Al

26.98

14Si

28.09

7N

14.01

15P

30.97

8O

16.00

9F

19.00

16S

32.07

17Cl

35.45

10Ne

20.18

18Ar

39.953

4

5

6

7

3B(3)

4B(4)

5B(5)

6B(6)

7B(7) (9) (10)

1B(11)

2B(12)

3A(13)

4A(14)

5A(15)

6A(16)

7A(17)

8A(18)

2

1

6

7

(8)

1A(1)

2A(2)

8B

Lanthanides

Actinides

TRANSITION ELEMENTS




Perio

d

110

(281)

111

(272)

112


114

(289)

113

(284)

115

(288)

116

(292)Ds Rg

(285)



siL48593_ch02_040-088 30:11:07 10:53pm Page 58


Formation of Covalent Compounds

In covalent compounds, atoms share pairs of electrons.

Simplest case: electron sharing occurs between two hydrogen atoms to form H2 molecule.

Usually occurs between nonmetals, C-H, Cl-Cl, etc.

The Formation of Covalent CompoundsCovalent compounds form when elements share electrons, which usually occursbetween nonmetals. Even though relatively few nonmetals exist, they interact inmany combinations to form a very large number of covalent compounds.

The simplest case of electron sharing occurs not in a compound but betweentwo hydrogen atoms (H; Z ! 1). Imagine two separated H atoms approachingeach other, as in Figure 2.14. As they get closer, the nucleus of each atom attractsthe electron of the other atom more and more strongly, and the separated atomsbegin to interpenetrate each other. At some optimum distance between the nuclei,the two atoms form a covalent bond, a pair of electrons mutually attracted bythe two nuclei. The result is a hydrogen molecule, in which each electron nolonger “belongs” to a particular H atom: the two electrons are shared by the twonuclei. Repulsions between the nuclei and between the electrons also occur, butthe net attraction is greater than the net repulsion. (We discuss the properties ofcovalent bonds in great detail in Chapter 9.)

A sample of hydrogen gas consists of these diatomic molecules (H2)—pairsof atoms that are chemically bound and behave as an independent unit—not sep-arate H atoms. Other nonmetals that exist as diatomic molecules at room tem-perature are nitrogen (N2), oxygen (O2), and the halogens [fluorine (F2), chlorine(Cl2), bromine (Br2), and iodine (I2)]. Phosphorus exists as tetratomic molecules(P4), and sulfur and selenium as octatomic molecules (S8 and Se8) (Figure 2.15).At room temperature, covalent substances may be gases, liquids, or solids.


e–

p+

p+

e–

e–

e–

p+

e–

e–

e–

B Attraction begins

C Covalent bond

D Interaction of forces

A No interaction

p+

p+

e–

p+

p+

p+

SAMPLE PROBLEM 2.6 Predicting the Ion an Element FormsPROBLEM What monatomic ions do the following elements form?(a) Iodine (Z ! 53) (b) Calcium (Z ! 20) (c) Aluminum (Z ! 13)PLAN We use the given Z value to find the element in the periodic table and see whereits group lies relative to the noble gases. Elements in Groups 1A, 2A, and 3A lose elec-trons to attain the same number as the nearest noble gas and become positive ions; thosein Groups 5A, 6A, and 7A gain electrons and become negative ions.SOLUTION (a) I" Iodine (53I) is a nonmetal in Group 7A(17), one of the halogens. Likeany member of this group, it gains 1 electron to have the same number as the nearestGroup 8A(18) member, in this case 54Xe.(b) Ca2# Calcium (20Ca) is a member of Group 2A(2), the alkaline earth metals. Likeany Group 2A member, it loses 2 electrons to attain the same number as the nearest noblegas, in this case, 18Ar.(c) Al3# Aluminum (13Al) is a metal in the boron family [Group 3A(13)] and thus loses3 electrons to attain the same number as its nearest noble gas, 10Ne.

FOLLOW-UP PROBLEM 2.6 What monatomic ion does each of the following ele-ments form: (a) 16S; (b) 37Rb; (c) 56Ba?

Figure 2.14 Formation of a covalentbond between two H atoms. A, The distance is too great for the atoms toaffect each other. B, As the distance de-creases, the nucleus of each atom beginsto attract the electron of the other. C, Thecovalent bond forms when the two nucleimutually attract the pair of electrons atsome optimum distance. D, The H2 mol-ecule is more stable than the separateatoms because the attractive forces(black arrows) between each nucleus andthe two electrons are greater than therepulsive forces (red arrows) between theelectrons and between the nuclei.

3A(13)

4A(14)

5A(15)

6A(16)

7A(17)

8A(18)

2A(2)

1

2

3

4

5

6

7

N2 F2

Cl2

Se8

I2

O2

P4 S8

Br2

H2

1A(1)

Diatomic molecules

Tetratomic moleculesOctatomic molecules

Figure 2.15 Elements that occur asmolecules.

siL48593_ch02_040-088 30:11:07 10:53pm Page 62





e–

p+

p+

e–

e–

e–

p+

e–

e–

e–

B Attraction begins

C Covalent bond


A No interaction

p+

p+

e–

p+

p+

p+




3A(13)

4A(14)

5A(15)

6A(16)

7A(17)

8A(18)

2A(2)

1

2

3

4

5

6

7

N2 F2

Cl2

Se8

I2

O2

P4 S8

Br2

H2

1A(1)

Diatomic molecules



siL48593_ch02_040-088 30:11:07 10:53pm Page 62





e–

p+

p+

e–

e–

e–

p+

e–

e–

e–

B Attraction begins

C Covalent bond


A No interaction

p+

p+

e–

p+

p+

p+




3A(13)

4A(14)

5A(15)

6A(16)

7A(17)

8A(18)

2A(2)

1

2

3

4

5

6

7

N2 F2

Cl2

Se8

I2

O2

P4 S8

Br2

H2

1A(1)

Diatomic molecules



siL48593_ch02_040-088 30:11:07 10:53pm Page 62





e–

p+

p+

e–

e–

e–

p+

e–

e–

e–

B Attraction begins

C Covalent bond


A No interaction

p+

p+

e–

p+

p+

p+




3A(13)

4A(14)

5A(15)

6A(16)

7A(17)

8A(18)

2A(2)

1

2

3

4

5

6

7

N2 F2

Cl2

Se8

I2

O2

P4 S8

Br2

H2

1A(1)

Diatomic molecules



siL48593_ch02_040-088 30:11:07 10:53pm Page 62


Hydrogen gas consists of many diatomic hydrogen molecules, H2.

These behave as independent H2 units -not separate hydrogen atoms.

Other diatomics are N2, O2, F2, Cl2, Br2, I2

Memorize the “7 Famous Diatomics”


Compounds: Formulas, Names, and Masses

Empirical formulas show the simplest ratio of numbers of atoms of each element in the compound.Empirical formula of hydrogen peroxide is HO Molecular formulas show the actual number of atoms of each element in a molecule of a compound. Molecular formula of hydrogen peroxide is H2O2

Chemical formulas: symbols + numeric subscripts H2O CaCO3 H2SO4

Structural formulas show actual number of atoms and the arrangement of the atoms in the molecule. Structural formula of hydrogen peroxide H-O-O-H.


Names of compounds with monatomic ions.

Rules for binary ionic compounds, (MX).

Metal cation name is the same as that of the metal. Ca Ca 2+

Anion takes the root of the nonmetal name and adds the suffix “-ide.” Br Br –

The compound formed from the metal calcium and the nonmetal bromine is “calcium bromide.”

Learn all the monatomic ions in Table 2.3

Ionic compound names give positive ion (cation) first, followed by negative ion (anion).


Exercise (assumes that you learned 1st 36 elements) Name the binary ionic compound that forms from:

a) strontium (Z =38) and N b) S and Zn

c) Al and fluorine d) oxygen and Li



1H

1.008

2He

4.003

3Li

6.941

4Be

9.012

11Na

22.99

12Mg

24.31

19K

39.10

20Ca

40.08

37Rb

85.47

38Sr

87.62

55Cs

132.9

56Ba

137.3

87Fr

(223)

88Ra

(226)

21Sc

44.96

22Ti

47.88

39Y

88.91

40Zr

91.22

71Lu

175.0

72Hf

178.5

103Lr

(260)

104Rf

(263)

23V

50.94

24Cr

52.00

41Nb

92.91

42Mo

95.94

73Ta

180.9

74W

183.9

105Db

(262)

106Sg

(266)

25Mn

54.94

26Fe

55.85

43Tc(98)

44Ru

101.1

75Re

186.2

76Os

190.2

107Bh

(267)

108Hs

(277)

27Co

58.93

45Rh

102.9

77Ir

192.2

109Mt

(268)

57La

138.9

58Ce

140.1

60Nd

144.2

61Pm

(145)

62Sm

150.4

63Eu

152.0

59Pr

140.9

89Ac

(227)

90Th

232.0

92U

238.0

93Np

(237)

94Pu

(242)

95Am(243)

91Pa

(231)

28Ni

58.69

29Cu

63.55

46Pd

106.4

47Ag

107.9

78Pt

195.1

79Au

197.0

30Zn

65.41

48Cd

112.4

80Hg

200.6

64Gd

157.3

65Tb

158.9

67Ho

164.9

68Er

167.3

69Tm

168.9

70Yb

173.0

66Dy

162.5

96Cm(247)

97Bk

(247)

99Es

(252)

100Fm

(257)

101Md

(258)

102No

(259)

98Cf

(251)

31Ga

69.72

32Ge

72.61

49In

114.8

50Sn

118.7

81Tl

204.4

82Pb

207.2

33As

74.92

51Sb

121.8

83Bi

209.0

34Se

78.96

35Br

79.90

52Te

127.6

53I

126.9

84Po

(209)

85At

(210)

36Kr

83.80

54Xe

131.3

86Rn

(222)

5B

10.81

6C

12.01

13Al

26.98

14Si

28.09

7N

14.01

15P

30.97

8O

16.00

9F

19.00

16S

32.07

17Cl

35.45

10Ne

20.18

18Ar

39.953

4

5

6

7

3B(3)

4B(4)

5B(5)

6B(6)

7B(7) (9) (10)

1B(11)

2B(12)

3A(13)

4A(14)

5A(15)

6A(16)

7A(17)

8A(18)

2

1

6

7

(8)

1A(1)

2A(2)

8B

Lanthanides

Actinides

TRANSITION ELEMENTS




Perio

d

110

(281)

111

(272)

112


114

(289)

113

(284)

115

(288)

116

(292)Ds Rg

(285)

placed below the main body of the table but actually fit between theelements indicated. Metals lie below and to the left of the thick“staircase” line [top of 3A(13) to bottom of 6A(16) in Period 6] and in-clude main-group metals (purple-blue), transition elements (blue), andinner transition elements (gray-blue). Nonmetals (yellow) lie to the rightof the line. Metalloids ( green) lie along the line. We discuss the place-ment of hydrogen in Chapter 14. As of mid-2007, elements 112–116had not been named.


siL48593_ch02_040-088 30:11:07 10:53pm Page 58metal nonmetal nm m

nmm nm m


Calcium bromide is composed of Ca2+ ions + Br – ions.Two Br – ions are needed to balance each Ca2+, so the formula is CaBr2. (empirical formula)

Deduce other ionic formulas by the same method.

How to Predict Formulas of Ionic CompoundsPositive charges of cations in a formula must bebalanced by the negative charges of the anions.


Problem: Write the empirical formulas for the compounds named in the previous problem.

Plan: Write the ions and find the smallest number of each that gives a neutral formula. Solution: a) strontium nitride Sr 2+ and N 3– ; three Sr 2+ (6+) balance two N 3– (6 –) ! Sr3N2

b) zinc iodide Zn 2+ and I– ; one Zn 2+ ion (2+) balances two I– ions (2–) ! ZnI2

c) aluminum fluoride Al 3+ and F– ; one Al 3+ ion (3+) balances three F– ions (3–) ! AlF3

d) lithium oxide Li + and O2– ; two Li+ ions (2+) balance one O2– ion (2–) ! Li2O


Transition elements (B groups), often form more than one ion, with different charges.

Naming their compounds: give metal’s ionic charge in Roman numerals after the metal ion’s name.

For example, iron forms both Fe 2+ and Fe 3+ ions.The two iron-chlorine compounds are:FeCl2, named iron(II) chloride, and FeCl3, named iron(III) chloride.

Learn the transition element ions given in Table 2.4


45

Problem: Give systematic names for the formulas or formulas for the names of the following compounds:

(b) CrF3

(c) iron(III) oxide

(d) MnS

(a) tin(II) bromide (tin, Z = 50)


Compounds formed from polyatomic ionsLearn the formulas, charges and names of the common polyatomic ions given in Table 2.5 (NO3

-)If two or more of a given polyatomic ion are present, put the ion in parentheses with subscript following.

Calcium nitrate contains one Ca2+ and two NO3– ions;

formula = Ca(NO3)2.

Hydrates are ionic compounds having a number of water molecules associated with each formula unit.Epsom salt has the formula MgSO4

.7H2O and the name magnesium sulfate heptahydrate.

Learn numerical prefixes in Table 2.6.


Families of oxoanions exist that differ only in the number of oxygen atoms.

If only two oxoanions are in the family: -the ion with more O atoms takes the nonmetal root and the suffix “-ate.”-the ion with fewer O atoms takes the nonmetal root and the suffix “-ite.”Examples: SO4

2– = sulfate ion; SO32– = sulfite ion.

Families of OxoanionsMost polyatomic ions are oxoanions, in which an element is bonded to one or more oxygen atoms.

Naming convention for these ions:SO3

2– SO42–


Examples: the four chlorine oxoanions,• ClO4

– is perchlorate • ClO3

– is chlorate

With four oxoanions in the family:-the ion with most O atoms has the prefix “per-,” the nonmetal root, and the suffix “-ate.”

-the ion with one less O atoms has the suffix “-ate.”-the ion with two less O atoms has the suffix “-ite.”-the ion with three less O atoms has the prefix “hypo-” and the suffix “-ite.”

• ClO2– is chlorite

• ClO– is hypochlorite


Problem: Give systematic names for the formulas or formulas for the names of the following compounds:

a) ClO3– is chlorate; since it has a 1- charge, the

cation must be Fe 2+. Name is iron(II) chlorate.

b) Sodium is Na+ ; carbonate is CO32–. Two Na+ ions

balance one CO32– ion. Formula is Na2CO3 .

c) Ba2+ is barium; OH– is hydroxide. There are eight water molecules in each formula unit. Name is barium hydroxide octahydrate.

a) Fe(ClO3)2 b) Sodium carbonate c) Ba(OH)2.8H2O


Naming acidsAcids are a type of hydrogen-containing compound.To name acids, treat them as “anions” connected to number of H+ ions needed for electrical neutrality.Two main types are 1. binary acids and 2. oxoacids:1. Gaseous hydrogen chloride (HCl) dissolved in water forms a solution called hydrochloric acid. - name consists of: prefix hydro- + anion nonmetal root + suffix -ic + word acid.Exercise: Name H2S(aq) as a binary acid

(H+)2(S2-) ! hydro + sulfur + ic + acid

hydrosulfuric acid

!


2. Oxoacid names are based on those of the oxoanions, except for two suffix changes:

Anion “-ate” suffix becomes an “-ic” suffix in acid.

Anion “-ite” suffix becomes an “-ous” suffix in acid.

Oxoanion prefixes “hypo-” and “per-” are retained.

BrO4– is perbromate, so HBrO4 is perbromic acid

IO2– is iodite, so HIO2 is iodous acid.


Problem: Name the following anions and give the names and formulas of the acid solutions derived from them:

(a) Br –

(b) IO3–

(c) CN–

(d) NO2–


Names and Formulas of Binary Covalent Compounds A few simple covalent compounds, have common names: ammonia (NH3) water (H2O) Most are named in a systematic way using 4 rules:1. The element with the lower group number in the periodic table is named first

Important exception: If the compound contains oxygen and a halogen, the halogen is named first

2. If both elements are in the same group, the one lower in the group is named first 3. The second element is named with its root and the suffix “-ide”

4. Use numerical prefixes (Table 2.6) to indicate the number of atoms of each element in the compound


Problem: (a) What is the formula of carbon disulfide?

(b) What is the name of AsF5 ?

(c) Give the name and formula of the compound formed from two N atoms and five O atoms.



1H

1.008

2He

4.003

3Li

6.941

4Be

9.012

11Na

22.99

12Mg

24.31

19K

39.10

20Ca

40.08

37Rb

85.47

38Sr

87.62

55Cs

132.9

56Ba

137.3

87Fr

(223)

88Ra

(226)

21Sc

44.96

22Ti

47.88

39Y

88.91

40Zr

91.22

71Lu

175.0

72Hf

178.5

103Lr

(260)

104Rf

(263)

23V

50.94

24Cr

52.00

41Nb

92.91

42Mo

95.94

73Ta

180.9

74W

183.9

105Db

(262)

106Sg

(266)

25Mn

54.94

26Fe

55.85

43Tc(98)

44Ru

101.1

75Re

186.2

76Os

190.2

107Bh

(267)

108Hs

(277)

27Co

58.93

45Rh

102.9

77Ir

192.2

109Mt

(268)

57La

138.9

58Ce

140.1

60Nd

144.2

61Pm

(145)

62Sm

150.4

63Eu

152.0

59Pr

140.9

89Ac

(227)

90Th

232.0

92U

238.0

93Np

(237)

94Pu

(242)

95Am(243)

91Pa

(231)

28Ni

58.69

29Cu

63.55

46Pd

106.4

47Ag

107.9

78Pt

195.1

79Au

197.0

30Zn

65.41

48Cd

112.4

80Hg

200.6

64Gd

157.3

65Tb

158.9

67Ho

164.9

68Er

167.3

69Tm

168.9

70Yb

173.0

66Dy

162.5

96Cm(247)

97Bk

(247)

99Es

(252)

100Fm

(257)

101Md

(258)

102No

(259)

98Cf

(251)

31Ga

69.72

32Ge

72.61

49In

114.8

50Sn

118.7

81Tl

204.4

82Pb

207.2

33As

74.92

51Sb

121.8

83Bi

209.0

34Se

78.96

35Br

79.90

52Te

127.6

53I

126.9

84Po

(209)

85At

(210)

36Kr

83.80

54Xe

131.3

86Rn

(222)

5B

10.81

6C

12.01

13Al

26.98

14Si

28.09

7N

14.01

15P

30.97

8O

16.00

9F

19.00

16S

32.07

17Cl

35.45

10Ne

20.18

18Ar

39.953

4

5

6

7

3B(3)

4B(4)

5B(5)

6B(6)

7B(7) (9) (10)

1B(11)

2B(12)

3A(13)

4A(14)

5A(15)

6A(16)

7A(17)

8A(18)

2

1

6

7

(8)

1A(1)

2A(2)

8B

Lanthanides

Actinides

TRANSITION ELEMENTS




Perio

d

110

(281)

111

(272)

112


114

(289)

113

(284)

115

(288)

116

(292)Ds Rg

(285)

placed below the main body of the table but actually fit between theelements indicated. Metals lie below and to the left of the thick“staircase” line [top of 3A(13) to bottom of 6A(16) in Period 6] and in-clude main-group metals (purple-blue), transition elements (blue), andinner transition elements (gray-blue). Nonmetals (yellow) lie to the rightof the line. Metalloids ( green) lie along the line. We discuss the place-ment of hydrogen in Chapter 14. As of mid-2007, elements 112–116had not been named.


siL48593_ch02_040-088 30:11:07 10:53pm Page 58


Hydrocarbons have only H and C atomsAlkanes are one type of hydrocarbon

Methane, CH4, is the simplest alkane

Organic Compounds; Hydrocarbons

Alkane series general formula = CnH2n+2

Ethane, C2H6, (n = 2) is the 2nd member

Learn straight-chain alkanes in Table 2.7

Alkanes with branches have a number for branch location. 2-methylbutane


Specific groupings of atoms attached to a carbon chain

Functional Groups

• Alcohols

• Amines

• Carboxylic Acids


Exercise: What is the structure of 3-pentanol?

Exercise: What is the name of CH3CH2CH2COOH


58

Problem: Using the PT, calculate the molecular mass of the compound tetraphosphorus trisulfide:


Section 2.9 (p76-79) is assigned for self study.

End of Chapter 2


Avogadro’s Number (NA) = 6.02214 x 10 23 mol-1

Atomic Mass Unit (u) = 1.66054 x 10 -27 kg

Electron charge (e) = 1.60218 x 10 -19 C

Faraday’s constant (F) = 9.64853 x 10 4 C/mol

Universal gas const. (R) = 8.20578 x 10 -2 L.atm/(mol.K) = 8.31451 J/(mol.K)

Plank’s constant (h) = 6.62607 x 10 -34 J.s

Rydberg's constant = 1.09678 x 10 m

Speed of light (c) = 2.99792 x 10 8 m/s

(R) 7 -1

60

lecture 1

Documents

g carbon

g oxygen

g of pb

g total mass fractionparts1

g mg3n2

g calcium carbonate

mass of n2

sample of galena g