Transcript chapter 5-Chemical Periodicity

5
Chemical Periodicity
化學週期性
The reaction of H2O and Li (left) to produce
LiOH and H2(g) is much slower than the
analogous reaction between H2O and Na
(right).
Chapter Goals
5-1 More About the Periodic Table
Periodic Properties of the Elements 元素的週期特質
5-2 Atomic Radii 原子半徑
5-3 Ionization Energy游離能
5-4 Electron Affinity電子親和力
5-5 Ionic Radii 離子半徑
5-6 Electronegativity 電負性
5-7 Oxidation States
Chemical Reactions and Periodicity
5-7 Hydrogen & the Hydrides
• Hydrogen
• Reactions of Hydrogen and the Hydrides
5-8 Oxygen & the Oxides
•
•
•
•
Oxygen and Ozone
Reactions of Oxygen and the Oxides
Combustion Reactions
Combustion of Fossil Fuels and Air Pollution
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More About the Periodic Table
• Establish a classification scheme of the elements
based on their electron configurations.
– Noble Gases 鈍氣 Group 8A
• All of them have completely filled electron shells.
• Since they have similar electronic structures, their
chemical reactions are similar.
ns2np6 He
Ne
Ar
Kr
Xe
Rn
1s2
[He] 2s2 2p6
[Ne] 3s2 3p6
[Ar] 4s2 3d10 4p6
[Kr] 5s2 4d10 5p6
[Xe] 6s2 4f14 5d10 6p6
3
More About the Periodic Table
– Representative Elements典型元素
• Are the elements in A groups on periodic chart.
• These elements will have their “last” electron in an
outer s or p orbital.
• These elements have fairly regular variations in
their properties.
4
More About the Periodic Table
• d-Transition Elements過渡
金屬元素
– Elements on periodic chart in B
groups.
– Sometimes called transition metals.
• Each metal has d electrons.
– ns (n-1)d configurations
• These elements make the
transition from metals to
nonmetals.
• Exhibit smaller variations
from row-to-row than the
representative elements.
First transition series (4s and 3d orbital occupy): 21Sc through 30Zn
Second transition series (5s and 4d orbital occupy): 39Sc through 48Cd
Third transition series (6s and 5d orbital occupy): 52La and 72Hf to 80Hg
Fourth transition series (6s and 5d orbital occupy): 89Ac and 104Rf to 112
5
More About the Periodic Table
• f - transition metals
– Sometimes called inner
transition metals.
– Electrons are being added to f
orbitals.
Outermost electrons have the greatest influence on the chemical
properties of elements.
• Adding an electron to an s or p orbital
 dramatic change in the physical and chemical properties
• Adding an electron to a d or f orbital
 much smaller effect on properties
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Periodic Properties
of the Elements
Atomic Radii 原子半徑
Ionization Energy游離能
Electron Affinity電子親和力
Ionic Radii 離子半徑
Electronegativity 電負性
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Atomic radii increase
Atomic Radii
原子半徑
同
族
元
素
原
子
序
增
加

原
子
半
徑
愈
大
relative sizes
Atomic radii decrease
同週期典型元素大致隨原子序增加 →原子半徑愈小
Å = 10-10m
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Atomic Radii
• The reason the atomic radii decrease across a period
is due to shielding or screening effect遮罩效應.
– Effective nuclear charge有效核電荷, Zeff, experienced by an
electron is less than the actual nuclear charge, Z.
–最外層電子受原子的吸引力被內層電子對外層電子的排斥力給
抵銷即稱遮罩效應
• Moving across a period, each element has an
increased nuclear charge and the electrons are going
into the same shell (2s and 2p or 3s and 3p, etc.).
– Consequently, the outer electrons feel a stronger
effective nuclear charge.
– For Li, Zeff ~ +1
For Be, Zeff ~ +2
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有效核電荷
原子核質子數－內層電子數總和
鎂的有效核電荷約為12 – 10 = 2
•原子半徑隨原子序數的增加呈現週期性變化。這與原
子有效核電荷的週期性變化相關
– 因為有效核電荷愈大, 對外層電子的吸引力愈大, 原子半
徑就愈小
•各週期的主族從左到右, 電子層數不變, 有效核電荷增加明
顯, 原子半徑的逐漸減少也就比較明顯
•長週期中的過渡元素原子半徑先是緩慢縮小然後略有增大
•內過渡元素，有效核電荷變化不大，原子半徑幾乎不變
•稀有氣體原子半徑突然增大，因為它是van der Waals半徑
• 同一主族從上到下，由於電子層數增加，使遮罩效應
明顯加大，所以原子半徑遞增。
凡得瓦半徑 為兩個相鄰非鍵結原子間距離的一半
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Atomic Radii
• Example 5-1: Arrange these elements based on
their atomic radii.
Se, S, O, Te
P, Cl, S, Si
Ga, F, S, As
Cs, F, K, Cl
6A
O < S < Se < Te
Period 3
Cl < S < P < Si
F < S < As < Ga
F < Cl < K < Cs
Atomic radii decrease
Atomic radii increase
–
–
–
–
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Ionization Energy 游離能
• First ionization energy (IE1) 第一游離能
– The minimum amount of energy required to remove
the most loosely bound electron from an isolated
gaseous atom to form a 1+ ion
處於基態的氣態原子失去一個電子，成為＋1 價氣態陽
離子，所需吸收的能量
• Symbolically:
Atom(g) + energy  ion+(g) + e-
Mg(g) + 738kJ/mol  Mg+ + e13
Ionization Energy
• Second ionization energy (IE2)第二游離能
– The amount of energy required to remove the
second electron from a gaseous 1+ ion.
• Symbolically:
ion+ + energy  ion2+ + e-
Mg+ + 1451 kJ/mol Mg2+ + e-
•Atoms can have 3rd (IE3), 4th (IE4), etc.
ionization energies.
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Ionization Energy
Periodic trends for Ionization Energy:
1. IE2 > IE1
It always takes more energy to remove a second electron
from an ion than from a neutral atom.
2. IE1 generally increases moving from IA elements to VIIIA
elements.
Important exceptions at Be & Mg, N & P, etc. due to filled
and half-filled subshells.
3. IE1 generally decreases moving down a family.
Decrease
IE1 for Li > IE1 for Na, etc.
隨
原
子
序
增
大
而
遞
減
increase
隨原子序增大漸增趨勢
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同一族元素的游離能
• 隨原子序的增加而變小
• 例如第1族元素的第一游離能大小順序為：Li >
Na > K > Rb > Cs > Fr
• 同族元素的原子序愈大，原子半徑愈大，最外
殼層電子受原子核的引力愈小，故電子愈容易
移除。
游離能的週期性
•同一週期之各元素的第一游離能，隨原子序的增加
，呈現鋸齒狀的增加
•原子半徑愈小，且有效核電荷愈大，所需的IE愈大
•同一週期第一游離
– 2族> 3族，鎂(738 kJ/mol)>鋁(578 kJ/mol)
– 15(5A) >16(6A) ，磷(1012 kJ/mol)>硫(1000 kJ/mol)
•同一週期過渡元素，因原子半徑及有效核電荷彼此
差異不大，故其第一游離能彼此差異相對較小
游離能的規則
同一週期元素的第一游離能從左至右漸增, 但在中間
有些起伏IIA > IIIA, 而VA > VIA:
• IIA(ns2) > IIIA(ns2np1)是因為np能階較ns
高, 因此IIIA原子的游離能較小
• VA (ns2np3) > VIA(ns2np4)
• 因為VA 族的價電子組態為全半填滿,
而VIA族有一個p軌域填入兩個電子,
將增加電子的斥力, 因此較易移去,
故VA > VIA
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First Ionization Energies
of Some Elements
Elements with low ionization energies (IE) easily lose
to form cations (positive charge)
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Ionization Energy
• Example 5-2: Arrange these elements based on
their first ionization energies.
Sr, Be, Ca, Mg
Al, Cl, Na, P
B, O, Be, N
Na, Mg, Al, Si
Sr < Ca < Mg < Be
Na < Al < P < Cl
B < Be < O < N
Na < Al < Mg < Si
increase
Decrease
–
–
–
–
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Ionization Energy
• First, second, third, etc. ionization energies exhibit
periodicity as well
• Look at the following table of ionization energies
versus third row elements
– Notice that the energy increases enormously when
an electron is removed from a completed electron
shell Na: 1s22s22p63s1
Al: 1s22s22p63s23p1
Group and
element
IE1 (kJ/mol)
IE2 (kJ/mol)
IE3 (kJ/mol)
IE4 (kJ/mol)
IA
Na
496
4562
6912
9540
IIA
Mg
738
1451
7733
10,550
IIIA
Al
578
1817
2745
11,580
IVA
Si
786
1577
3232
4356
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Ionization Energy
• The reason Na forms Na+ and not Na2+ is that the
energy difference between IE1 and IE2 is so large.
– Requires more than 9 times more energy to remove
the second electron than the first one.
• The same trend is persistent throughout the series.
– Thus Mg forms Mg2+ and not Mg3+.
– Al forms Al3+.
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Ionization Energy
• Example 5-3: What charge ion would be
expected for an element that has these
ionization energies?
IE1 (kJ/mol)
IE2 (kJ/mol)
IE3 (kJ/mol)
IE4 (kJ/mol)
IE5 (kJ/mol)
IE6 (kJ/mol)
IE7 (kJ/mol)
IE8 (kJ/mol)
1680
3370
6050
8410
11020
15160
17870
92040
Notice that the largest increase
in ionization energies occurs
between IE7 and IE8. Thus this
element would form a 1- ion
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Electron Affinity電子親和力
• Electron affinity (EA) is the amount of energy
absorbed when an electron is added to an isolated
gaseous atom to form an ion with a 1- charge.
加一個電子給中性氣體原子而形成負離子時所需的能量即
稱電子親合力
• Sign conventions for electron affinity.
– If electron affinity > 0 energy is absorbed
– If electron affinity < 0 energy is released
對於放熱愈多的反應，我們稱其電子親和力愈大，
表示該原子愈易獲得電子
• Electron affinity is a measure of an atom’s ability to
form negative ions
• Symbolically:
atom(g) + e- + EA  ion-(g)
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Electron Affinity
Two examples of electron affinity values:
Mg(g) + e- + 231 kJ/mol Mg-(g)
EA = +231 kJ/mol
Br(g) + e-  Br-(g) + 323 kJ/mol
EA = -323 kJ/mol 電子親和力大
Most elements have no affinity for an additional electron and thus
have an electron affinity equal to zero
He(g) + e-  He-(g) EA=0 kJ/mol
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電子親和力的性質
•電子親和力的負值愈大, 表示該原子接
受電子的傾向愈強, 所形成的陰離子也
愈加穩定
•若電子親和力為正值, 則所形成的陰離
子較不穩定
Electron Affinity
• General periodic trend for electron affinity is
– the values become more negative from left to
right across a period on the periodic chart
– the values become more negative from bottom
to top up a row on the periodic chart
• Measuring electron affinity values is a difficult
experiment
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Electron Affinity
沒
有
規
律
性
29
Electron Affinity
2A
5A
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Electron Affinity
• Example 5-4: Arrange these elements based on
their electron affinities.
– Al, Mg, Si, Na
Si < Al < Na < Mg
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Ionic Radii 離子半徑
• Cations (positive ions) are always smaller than
their respective neutral atoms
• Anions (negative ions) are always larger than
their neutral atoms.
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Ionic Radii
• Cation (positive ions) radii decrease from left to
right across a period
– Increasing nuclear charge attracts the electrons
and decreases the radius
• Anion (negative ions) radii decrease from left to
right across a period
– Increasing electron numbers in highly charged
ions cause the electrons to repel and increase
the ionic radius
半
徑
增
加
原
子
序
增
加
核電荷增加
半徑減少
33
Ionic Radii
Isoelectronic species: have the same number of electron
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Ionic Radii
• Example 5-5: Arrange these elements based on
their ionic radii.
– Ga, K, Ca
– Cl, Se, Br, S
K1+ > Ca2+ > Ga3+
Cl1- < S2- < Br1- < Se2-
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Electronegativity (EN) 電負度；陰電性
• Electronegativity is a measure of the relative tendency
of an atom to attract electrons to itself when
chemically combined with another element.
– Electronegativity is measured on the Pauling scale.
– Fluorine is the most electronegative element.
– Cesium (Cs) and francium (Fr) are the least
electronegative elements.
電負性是指元素的原子在分子中吸引電子的能力的相對大小，
電負性大，原子在分子中吸引電子的能力強，反之就弱。
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Electronegativity 電負度；陰電性
• 金屬元素的電負度較小，而非金屬元素的電負度較大
• Elements with high electronegativity (nonmetals)
often gain electrons to form anions.
• Elements with low electronegativity (metals) often
lose electrons to form cations.
• For the representative elements, electronegativities
usually increase from left to right across periods and
decrease from top to bottom within groups.
37
惰性氣體無電負度
同族由上而下遞減，同週期由左而右遞增
38
Electronegativity
39
Electronegativity
• Example 6-11: Arrange these elements based on
their electronegativity.
– Se, Ge, Br, As
– Be, Mg, Ca, Ba
Ge < As < Se < Br
Ba < Ca < Mg < Be
40
Periodic Trends
• It is important that you understand and know the
periodic trends described in the previous sections.
– They will be used extensively in Chapter 7 to
understand and predict bonding patterns.
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週期表各元素性質之變化趨勢
Oxidation Numbers氧化數
• The transfer of electrons from one species to
another are called oxidation-reduction reactions 氧
化還原反應, redox reaction.
• Oxidation number or oxidation state of an element
in a simple binary ionic compound is the number of
electrons gained or lost by an atom of that element
when it forms the compound.
• 金屬與非金屬反應形成離子化合物，涉及一個或更多個電子從金
屬(形成一個陽離子) 轉移到非金屬 (形成一個陰離子)。
• 涉及電子轉移的反應稱為氧化－還原反應 (oxidation-reduction
reaction)。例如：金屬鎂與氧的反應。
43
Oxidation Numbers氧化數
• Guidelines for assigning oxidation numbers.
1. The oxidation number of any free, uncombined
element is zero. Such as: H2, O2, P4, S8
2. The oxidation number of an element in a simple
(monatomic) ion is the charge on the ion.
3. In the formula for any compound, the sum of the
oxidation numbers of all elements in the compound
is zero.
4. In a polyatomic ion, the sum of the oxidation
numbers of the constituent elements is equal to the
charge on the ion.
44
Oxidation Numbers
5. Fluorine, F, has an oxidation number of –1 in its
compounds.
6. Hydrogen, H, has an oxidation number of +1 unless it is
combined with metals, where it has the oxidation
number -1.
–Examples – LiH, BaH2
7. Oxygen usually has the oxidation number -2.
Exceptions:
–In peroxides O has oxidation number of –1.
•Examples - H2O2, CaO2, Na2O2
–In OF2, O has oxidation number of +2.
45
Oxidation Numbers
8. Use the periodic table to help with assigning
oxidation numbers of other elements.
a.IA metals have oxidation numbers of +1.
b.IIA metals have oxidation numbers of +2.
c. IIIA metals have oxidation numbers of +3.
•There are a few rare exceptions.
d.VA elements have oxidation numbers of –3 in
binary compounds with H, metals or NH4+.
e.VIA elements below O have oxidation numbers
of –2 in binary compounds with H, metals or NH4+.
• Summary in Table 4-10 (Table 5-4 p193).
46
Oxidation Numbers
47
Oxidation Numbers
• Example5-5: Assign oxidation numbers to each
element in the following compounds:
•K2Sn(OH)6
• NaNO3
Na = +1
(Rule 8)
K = +1
(Rule 8)
O = -2 (Rule 7)
O = -2
(Rule 7)
N = +5
H = +1
(Rule 6)
–Calculate using rule 3
–+1 + 3(-2) + x = 0
–x = +5
Sn = +5
– Calculate using rule 3
– 2(+1) + 6(-2) + 6(+1) + x = 0
– x = +5
48
Oxidation Numbers
• NO2O = -2 (Rule 7)
N = +3
– Calculate using rule 4.
– 2(-2) + x = -1
– x = +3
• HCO3O = -2
(Rule 7)
H = +1
(Rule 6)
C = +4
– Calculate using rule 4.
– +1 + 3(-2) + x = -1
– x = +4
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Example 5-6 Oxidation Numbers
Determine the oxidation numbers of nitrogen in the following
species (a) N2O4, (b)NH3, (c)HNO3, (d) NO3-, (e)N2
(a) N2O4
O = -2
2x +(-2)4 = 0
x = +4
N=+4
(b) NH3
H = +1
x +(+1)3 = 0
x = -3
N=-3
(d) NO3O = -2
x +(-2)3 = -1
x = +5
N=+5
(c) HNO3
H = +1 O=-2
1+x+(+2)3 = 0
x = +5
N=+5
(b) N2
The oxidation number of
any free element is
zero
Exercise 42
50
Chemical Reactions & Periodicity
• In the next sections periodicity will be applied to
the chemical reactions of hydrogen, oxygen, and
their compounds.
51
Hydrogen and the Hydrides
Hydrogen
• Element hydrogen is colorless, ordorless, tasteless diatomic
gas with the lowest molecular weight and density of any
known substance
• Hydrogen gas, H2, can be made in the laboratory by
– Displacement (and redox) reaction
3Fe(s)+4H2O(g)Fe

3O4(s)+4H2(g)
Zn(s)+2HCl(aq)ZnCl2(aq)+H2(g)
– Electrolysis of water 2H2O(l)  2H2(g)+O2(g)
– Water gas reaction C(s)+H2O(g)CO(g)+H2(g)
In cock
steam Water gas水媒氣反應可當燃料
– Steam cracking CH4 (g)+H2O(g)

CO(g)+3H2(g)
Ni
蒸氣裂解
synthesis gas
•To produce organic chemicals
like Methanol...
52
Hydrogen and the Hydrides
• Hydrogen reacts with metals and other nonmetals to form
binary compound called hydrides 氫化物.
– Ionic hydrides — all basic
• Hydride ions, H- , formed when hydrogen gains one electron
per atom from an active metal
• All basic because the react with water to form hydroxide
ions. LiH(s) +H2O(l)  LiOH(s)+H2(g)
– Molecular hydrides
• Hydrogen share s electrons with an atom of another
nonmetal
• Many are acidic; their aqueous contain hydrogen cation (H+),
such as HF, HCl, HBr, HI, H2S, H2Se, H2Te
– The ionic or molecular character of the binary compounds
of hydrogen depends on the position of the other elements
in the periodic table
53
Ionic hydrides
1A: 2M(l) + H2(g)  2(M+,H-)(s)
2Li(l) + H2(g)  2LiH(s)
2A: M(l) + H2(g)  (M2+,2H-)(s)
Ca(l) + H2(g)  CaH2(s)
Molecular hydrides
H2(g) + F2(g)  2HF(g) hydrogen fluoride
2H2(g) + O2(g)  2H2O(g)
N2(g) + 3H2(g)  2NH3(g)
54
Reactions of Hydrogen and
the Hydrides
• The ionic hydrides produced in the two previous
reactions are basic.
– The H- reacts with water to produce H2 and OH-.
H- + H2O  H2 + OH-
•For example, the reaction of LiH with water
proceeds in this fashion.
LiH(s) + H2O(l)  H2(g) + OH-(aq) + Li+(aq)
55
Reactions of Hydrogen and
the Hydrides
• Hydrogen reacts with nonmetals to produce
covalent binary compounds.
• One example is the haloacids鹵化酸類produced by
the reaction of hydrogen with the halogens 鹵素.
H2 + X2  2 HX
• For example, the reactions of F2 and Br2 with H2 are:
H2 + F2  2 HF
H2 + Br2  2 HBr
56
Reactions of Hydrogen and
the Hydrides
• Combustion of H2: highly exothermic
2 H2(g) + O2(g)  2 H2O(l) + energy
• Hydrogen reacts with oxygen and other VIA
elements to produce several common binary
covalent compounds.
– Examples of this reaction include the production of
H2O, H2S, H2Se, H2Te.
2 H2 + O2  2 H2O
8 H2 + S8  8 H2S
57
Reactions of Hydrogen and
the Hydrides
• The hydrides of Group VIIA and VIA hydrides are
acidic.
HCl  H+(aq)+ Cl-(aq) (a strong acid)
H2S  H+(aq)+ HS-(aq) (a weak acid)
58
Reactions of Hydrogen and
the Hydrides
• There is an important periodic trend evident in the
ionic or covalent character of hydrides.
1. Metal hydrides are ionic compounds and form
basic aqueous solutions.
2. Nonmetal hydrides are covalent compounds and
form acidic aqueous solutions.
59
Example 5-7, 5-8 Predicting Products of Reactions
Predict the products of the reactions involving the reactions shown. Write a
balanced formula unit equation for each. and predict the ionic or molecular
character of the products.
(a) H2(g) + I2(g) 
(b) K(l) +H2(g) 
(c) NaOH(s) +H2O (l)(excess)
(a) H2(g) + I2(g)  2HI(g)
Molecular
(b) K(l) +H2(g)  2KH(s)
Ionic
(c) NaH(s) +H2O (l) NaOH(aq) +H2(g)
Ionic Molecular
Exercise 54,55
60
Oxygen and the Oxides
Oxygen and Ozone
• Joseph Priestley discovered oxygen in 1774 using this
reaction (decomposition) :
2 HgO(s) 
 2 Hg() + O2(g)
•A common laboratory preparation method for oxygen
is:
2 KClO3 (s)  2 KCl (s) + 3 O2(g)
•Commercially, oxygen is obtained from the fractional
distillation of liquid air.
•O2
• an odorless and colorless gas.
• make up 21% by volume of dry air
• only slightly soluble in water
61
Oxygen and the Oxides
• Ozone (O3) is an allotropic form 同素異構物 of oxygen
which has two resonance structures.
• Ozone (O3) is an unstable, pale blue gas, its density is about
1.5 times of O2. it is a very strong oxidizing agent.
• Ozone is an excellent UV light absorber in the
earth’s atmosphere.
2 O3(g) 3 O2(g)
in presence of UV
62
Reactions of Oxygen and
the Oxides
•Oxygen is an extremely reactive element.
– O2 reacts with most metals to produce normal oxides
having an oxidation number of –2.
4 Li(s) + O2(g)  2 Li2O(s) lithium oxide
However, oxygen reacts with sodium to produce a
peroxide having an oxidation number of –1.
2 Na(s) + O2(g)  Na2O2(s) sodium peroxide
63
Reactions of Oxygen and
the Oxides
• Oxygen reacts with K, Rb, and Cs to produce
superoxides having an oxidation number of -1/2.
K(s) + O2(g)  KO2(s) potassium superoxide
Oxygen reacts with IIA metals to give normal oxides.
2 M(s) + O2(g)  2 MO(s)
2 Sr(s) + O2(g)  2 SrO(s)
64
Class
Contains
Ions
Oxidation No. of
Oxygen
Normal oxides
O2-
-2
Peroxides
O22-
-1
superoxides
O2-
-1/2
4Li(s) + O2(g)  2Li2O(s)
lithium oxide
2Na(s) + O2(g)  2Na2O2(g) sodium peroxide
K(s) + O2(g)  KO2 (s) potassium superoxide
65
Reactions of Oxygen and
the Oxides
• At high oxygen pressures the IIA metals can form
peroxides.
Ca(s) + O2(g)  CaO2(s)
Metals that have variable oxidation states, such as the
d-transition metals, can form variable oxides.
 For example, in limited oxygen:
+0
+2
2 Mn(s) + O2(g)  2 MnO(s)

In excess oxygen:
+0
+3
4 Mn(s) + 3 O2(g)  2 Mn2O3(s)
66
Reactions of Oxygen and
the Oxides
• Oxygen reacts with nonmetals to form covalent
nonmetal oxides.
• For example, carbon reactions with oxygen:
– In limited oxygen
2 C(s) + O2(g)  2 CO(g)
+2
– In excess oxygen
C(s) + O2(g)  CO2(g)
+4
67
Reactions of Oxygen and
the Oxides
• Phosphorous reacts similarly to carbon forming two
different oxides depending on the oxygen amounts:
– In limited oxygen
+0
P4(s)
+3
+ 3 O2(g)  P4O6(s)
– In excess oxygen
P4(s) + 5 O2(g)  P4O10(s)
+0
+5

68
Purple: amphoteric oxide兩性氧化物
is one that shows some acidic and some basic properties
69
Reactions of Oxygen and
the Oxides
• Similarly to the nonmetal hydrides, nonmetal oxides
are acidic.
– Sometimes nonmetal oxides are called acidic anhydrides
(acidic oxides).
• Form acid with no change in oxidation state of nonmetal.
– They react with water to produce ternary acids. 三元酸
• For example:
CO2(g) + H2O ()  H2CO3(aq)
+4
+4
Cl2O7(s) + H2O ()  2 HClO4(aq)
+7
+7
As2O5(s) + 6 H2O()  4 H3AsO4(aq)
+5
+5
70
Reactions of Oxygen and
the Oxides
• Similarly to the hydrides, metal oxides are basic.
– These are called basic anhydrides.
– They react with water to produce ionic metal
hydroxides (bases)
Li2O(s) + H2O()  2 LiOH(aq)
CaO(s) + H2O ()  Ca(OH)2(aq)
Metal oxides are usually ionic and basic.
Nonmetal oxides are usually covalent and acidic.
An important periodic trend.
71
Reactions of Oxygen and
the Oxides
• Nonmetal oxides react with metal oxides to
produce salts.
+1
+4
+1+4
Li2O(s) + SO2(g)  Li2SO3(s)
Cl2O7(s) + MgO(s)  Mg(ClO4)2(s)
72
Combustion Reactions
• Combustion reactions are exothermic redox
reactions
– Some of them are extremely exothermic.
• One example of extremely exothermic reactions is
the combustion of hydrocarbons.
– Examples are butane and pentane combustion.
2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(g)
+0
-2
-2
C5H12(g) + 8 O2(g)  5 CO2(g) + 6 H2O(g)
73
Example 6-8, 6-9 Acidic Character of Oxides
Arrange the following oxides in order of increasing molecular (acidic)
character: SO3, Cl2O7, CaO, PbO2 and arrange these oxides in order of
increasing basicity.
increasing nonmetallic character
Ca < Pb < S < Cl
2A 4A 6A 7A
increasing molecular character
CaO < PbO2 < SO3 < Cl2O7
increasing basic character
Cl2O7 < SO3< PbO2 <CaO
74
Example 6-10 Predicting Reaction Products
Predict the products of the following pairs of reactants.Write a balanced
equation for each reaction.
(a) Cl2O7(l) + H2O(l)
(b) As4(s)+O2(g)(excess) 

(c) Mg (s)+O2(g)

(a) Cl2O7(l) + H2O(l) 2HClO4(aq)
(b) As4(s)+5O2(g)  2As2O5(s)
(c) 2Mg (s)+O2(g) 2MgO2(s)
75
Fossil Fuel Contaminants
• When fossil fuels are burned, they frequently have
contaminants in them.
• Sulfur contaminants in coal are a major source of air
pollution.
– Sulfur combusts in air.
S8(g) + 8 O2(g)  8 SO2(g)
most harmful pollutant
Next, a slow air oxidation of sulfur dioxide occurs.
2 SO2(g) + O2(g)  2 SO3(g)
Sulfur trioxide is a nonmetal oxide, i.e.
an acid anhydride.
SO3(g) + H2O()  H2SO4(aq)
Main cause of acid rain
76
Fossil Fuel Contaminants
• Nitrogen from air can also be a source of significant
air pollution.
• This combustion reaction occurs in a car’s cylinders
during combustion of gasoline.
N2(g) + O2(g)  2 NO(g)
• After the engine exhaust is released, a slow
oxidation of NO in air occurs.
2 NO(g) + O2(g)  2 NO2(g)
77
Fossil Fuel Contaminants
• NO2 is the haze that we call smog 煙霧
(=smoke and fog)
– Causes a brown haze in air.
• NO2 is also an acid anhydride.
– It reacts with water to form acid rain and,
unfortunately, the NO is recycled to form more acid
rain.
3 NO2(g) + H2O()  2 HNO3(aq) + NO(g)
78
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Decrease
increase
80
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