无机化学作业（中英文）安排
在习题的安排中，化学反应原理部分的习题中，以基本概念
的 练 习 为 纲 ，并 通 过 分 单 元 的 计 算 题 为 学 生 打 好 基 础 ，同 时 利 用
综合分析能力的习题培养学生分析解决问题的能力各方法。
物质结构部分以基本概念题为主，采用形式为是非、选择和
简 答 题 ，通 过 综 合 训 练 题 提 高 学 生 对 原 子 结 构 、分 子 结 构 和 固 体
结构和配合物结构知识的认识。
元素化学的练习题中以是非、选择、简答题形式的基本概念
题 为 主 ，综 合 分 析 计 算 题 为 辅 ，着 重 使 学 生 学 习 掌 握 元 素 性 质 的
周期性变化的规律性问题。
在保证大纲要求的知识点、线和面都兼顾到的前提下，对各
章具体的习题每年进行更改。
第一章：结论
思考题：3 题，简答题：2 题。
第二章：热化学
基本概念题：4 题，热力学每一定律及热化学计算题：5 题。
第三章：化学动力学基础
基本概念题：2 题。
第四章：化学平衡
基 本 概 念 题 ：5 题 ，化 学 平 衡 计 算 题 ：4 题 ，热 力 学 第 二 定 律
及吉布斯函数计算题：4 题，综合分析训练题：2 题。
第五章：酸碱平衡
基本概念题：3 题，弱酸、弱碱电离平衡计算，4 题，多元弱酸
弱碱计算题 2 题，配位平衡计算题：3 题，综合分析训练题：2
题。
第六章：沉淀溶解平衡
基本概念题：3 题，溶解度与溶度积常数：3 题，沉淀平衡：
4 题，综合训练题：3 题。
第七章：氧化还原反应
电化学基础
基本概念与氧化还原反应配平题：8 题，电极电势与能斯特
方 程 ：5 题 ，酸 碱 - 沉 淀 - 配 位 化 学 - 氧 化 还 原 综 合 平 衡 训 练 题 4 题 。
各类化学平衡习题课：综合计算训练题 5 题。
第八章：原子结构
基 本 概 念 题 ：6 题 ，原 子 结 构 与 元 素 性 质 关 系 题 ：6 题 ，综 合
训练题 4 题。
第九章：分子结构：
价 键 理 论 基 本 概 念 题 ：3 题 ，杂 货 轨 道 理 论 概 念 题 ：4 题 ，价
层电子对互斥理论概念题：4 题，分子轨道理论概念题：3 题；
简答题：5 题，综合分析训练题：4 题。
第十章：固体结构
晶体结构和类型：基本概念 3 题，简答题 2 题；金属晶体：
基 本 概 念 2 题 ，简 答 题 2 题 ；离 子 晶 体 ：基 本 概 念 5 题 ，简 答 题
3 题 ；分 子 晶 体 ：基 本 概 念 3 题 ，简 答 题 3 题 ；综 合 训 练 题 2 题 。
第十一章：配合物结构
基本概念题 5 题，简答题 4 题，计算题 2 题，综合训练题 2
题。
第十二章：s 区元素：
基本概念题 5 题；化合物合成制备题 2 题；元素性质特点题
3 题。
第 十 三 章 ： p 区 元 素 (一 )
B、 Al 性 质 特 点 题 6 题 ； 碳 族 元 素 性 质 特 点 题 6 题 ； 化 合 物
合成制备题 4 题。
第 十 四 章 ： p 区 元 素 (二 )
氮族元素性质特点题 8 题；氧族元素性质特点题 5 题；化合
物合成制备题 5 题。
第 十 五 章 ： p 区 元 素 (三 )
卤素元素基本性质特点题 8 题，卤化物合成制备题 4 题，p
区 元 素 性 质 递 变 规 律 及 性 质 特 点 题 4 题 。主 族 元 素 综 合 训 练 题 3
题。
第 十 六 章 ： d 区 元 素 (一 )
d 区 元 素 基 本 概 念 题 4 题 ，铬 的 性 质 特 点 及 其 化 合 物 制 备 题 4
题 ，锰 的 性 质 特 点 及 其 化 合 物 制 备 题 5 题 ，铁 、钴 、镍 性 质 特 点
及其化合物制备题 6 题。
第 十 七 章 ： d 区 元 素 (一 )
铜 族 元 素 基 本 概 念 题 4 题 ，锌 族 元 素 基 本 概 念 题 5 ，d 区 元 素
训练题 2 题，元素化学综合训练题 4 题。
说明：
中文教材上选出的作业习题，一般都能够在随同该教材的<释疑与习
题解析>辅助材料上得到详尽的答案解释，在此不再重新列出； 对于
英文教材上选出的部分习题， 也因为每年都更新，如下给出的部分
习题和答案仅供参考。
Chapter Two
6.18
The work done to compress a gas is 74J.As a result, 26J of heat is given
off to the surroundings. Calculate the change in energy of the gas.
Answer: work of compression is positive and because heat is released by
the gas, q is negative. Therefore, we have △ E=q + w=-26J+74J=48J.
6.56
The standard enthalpy change for the following reaction is 436.4kJ/mol:
H2(g)→H(g)+ H(g)
Calculate the standard enthalpy of formation of atomic hydrogen (H).
Answer:
ΔHθr, m =Δ Hθf, m [H(g)] + Δ Hθf, m[H2(g)]
436.4kJ/mol=2ΔHθf, m [H2(g)]-0
Δ Hθf, m =218.2 kJ/mol
Chapter Three
13.18 consider the reaction
X +Y → Z
From the following data, obtained at 360K, (a) determine the order of the
reaction, and (b) determine the initial rate of disappearance of X when the
concentration of X is 0.30M and that of Y is 0.40M.
Initial
rate
of
[X] (M)
[Y] (M)
0.053
0.10
0.50
0.127
0.20
0.30
1.027
0.40
0.60
0.254
0.20
0.60
0.509
0.40
0.30
disappearance of X
(M/s)
Solution: Assume that the rate law takes the form
Rate = k[X]a [Y]b
Note the data as experiment 1, 2, 3, 4, and 5 in order
(a). Take the ratio of the rates from 2 & 5 and 3 & 4 experiments, because
in these two groups, the concentration of Y are the same, the ratio equals
the ratio of the initial rate of disappearance of X.
rate 5 rate 5( x ) 0.590 M / s k (0.40 M ) a (0.30 M ) b
=
=
=
= 2a ≈ 4
a
b
rate 2 rate 2 ( x ) 0.127 M / s k (0.20 M ) (0.30 M )
rate3 rate3( x )
k (0.40 M ) 2 (0.60 M ) b
1.02 M / s
=
=
=
= 2a ≈ 4
2
b
rate4 rate4 ( x ) 0.254 M / s k (0.20 M ) (0.60 M )
Therefore, 2 a ≈ 4 or a=2
Consider the rate of the reaction can be given as Y as well initially and
the coefficient of Y is the same as X, so the initial rate of disappearance
of Y is the same as X’s.
Then take the ratio of the rates from 2&4 and 3&5 experiments, we also
can get:
rate4 rate4 (Y ) 0.254 M / s k (0.20 M ) a (0.60 M ) b
=
=
=
= 2b ≈ 2
a
b
rate2 rate2 (Y ) 0.127 M / s k (0.20 M ) (0.30 M )
rate3 rate3(Y ) 0.254 M / s k (0.40 M ) a (0.60 M ) b
=
=
=
= 2b ≈ 2
a
b
rate5 rate5(Y ) 0.509 M / s k (0.40 M ) (0.30 M )
Therefore, 2b ≈ 4 or b=1
Hence the rate law is given by
Rate = k[X]2 [Y] ,which shows that it
is a third-order reaction overall.
(b) As we discussed above, the ratio of the rate can be expressed by the
ratio of the rate of X or Y.
Then assume the initial rate of X asked is A, and take the data from
experiment 3:
1.02 M / s k (0.40M ) 2 (0.60 M ) 24
=
=
A
k (0.30M ) 2 (0.40 M ) 9
So
A = 1.02 M / s × 9 / 24 = 0.38M / s
13.52 For the reaction X 2 + Y + Z → XY + XZ it is found that
doubling the concentration of X2 doubles the reaction rate , tripling the
concentration of Y triples the rate ,and doubling the concentration of Z
has no effect.(a)What is the rate law for this reaction? (b) Why it is that
the change in the concentration of Z has no effect on the rate? (c) Suggest
a mechanism for the reaction that is consistent with the rate law.
Solution:
[ ][ ]
(a) rate = k X 2 Y
(b) The elementary reaction about Z is not the determined reaction.
(c) X 2 + Y → X + XY (Slow)
X + Z → XZ
13.62
(Fast)
Consider the following mechanism for the enzyme catalyzed
reaction:
k1
E + S ↔ ES
k −1
k2
ES → E + P
Derive an expression for the rate law of the reaction in terms of
the concentration of E and S. (hint: to solve for [ES ] ,make use of the
fact that ,at equilibrium , the rate of forward reaction is equal to the rate
of the reverse reaction )
Solution: The fast equilibrium process:
k1 [E ][S ] = K −1 [ES ]
And the slow process:
rate = k 2 [ES ]
Then we get rate = k 2
k1
[E ][S ] = k [E ][S ]
k −1
Chapter Four
14.28
Question : A 2.5-mole quantity of NOCL was initially in a 1.50-L
reaction chamber at 400℃ .after equilibrium was established , it was
found that 28.0 percent of the NOCL had dissociated:
2NOCL(g)
⇒
2NO(g) + CL2 (g)
Calculate the equilibrium constant KC for the reaction.
Answer: The concentration of NOCL at the beginning of the
reactions
2.5
=1.67 mol/l : when the reaction reached equilibrium
1 .5
CNOCL = 0.72 * 1.67 = 1.20
=
CNO = 0.28 * 1.67 *
2
=0.31
3
CCL2
1
* 1.67 * 0.28 =0.15
3
The equilibrium constant of the reaction is given as follow KC =
[NO]2 [CL2 ] =
[NOCL]2
O.312 * 0.15
= 0.03
1.2 2
14.40
Question : For the synthesis of ammonia
N2(g) + 3H2(g) ⇔ 2NH3(g)
The equilibrium constant KC at 375℃ is 1.2.starting with [H 2 ] 0 =
0.76M , [N 2 ]0 =0.60M , and [NH 3 ] 0 = 0.48M , which gases will
have increased in concentration and which will have decreased in
concentration when the mixture comes to equilibrium ?
Answer : Let x be the depletion in concentration (mol/l) of N2 ,H2 NH3
at equilibrium.
N2(g) + 3H2(g) ⇔ 2NH3 (g)
Initial
0.76
0.6
0.48
Change
-X
-X
+X
Equilibrium
0.76-X
0.6-X
0.48+X
The equilibrium constant is KC =
[NH 3 ]2 substituting , we get
[N 2 ][H 2 ]3
[0.48 + X ]2
,KC
[0.76 − X ][0.6 − X ]3
N2 ，H2 ，have decreased in
1.2 =
φ 0
concentration. NH3 have increased.
14.58 Question:
consider the following equilibrium process :
PCL5(g) ⇔ PCL3(g) +CL2(g)
ΔH Ο =92.5 KJ/mol
Predict the direction of the shift in equilibrium when (a) the temperature
is raised ;(b) more chlorine gas is added to the reaction mixture ; (c)
some PCL3 is removed from the mixture ;(d) the pressure on the
gases is increased ;(e) a catalyst is added to the reaction mixture.
Answer:
(a).to the right (b).to the left (c).to the right (d).to the left (e).no effect
Chapter Five
15.4 Write the formulas of the conjugate bases of the following acids: (a)
HNO2, (b) H2SO4 ,(c) H2S ,(d) HCN ,(e) HCOOH(formic acid).
Answer：(a) HNO2→ NO2-; (B) H2SO4→ HSO4-; (c)
H2S→ HS-; ,(d)
HCN → CN-; e) HCOOH→ HCOO-.
15.9What is the ion-product constant for water?
Answer: KwΘ = [OH-][H+] = 1.0x10-14
15.27 Explain what is meant by the strength of an acid.
It means how an acid is strong or weak by calculating its Ka, i.e., whether
it is easy to release H+ or not, if it is easy then it is a strong acid.
Chapter Six
16.56 Question find the approximate PH range suitable for the separation
of Fe3+ and Zn2+ ions by precipitation of Fe(OH)3 from a solution
that is initially 0.001M in both Fe3+ and Zn2+.
Answer: the solubility equilibrium for Fe(OH)3 is
Fe(OH)3(s) ⇔ Fe3+(aq) + 3OH-(aq)
KSP = [Fe 3+ ] [OH − ]
3
Because [Fe 3+ ] (aq) = 0.010M , the concentration of [OH − ]
that must be exceeded to initiate the precipitation of Fe(OH)3 is
[HO ]
−
1
1.1 *10 −3.6 3
⎧ K ⎫3
= ⎨ SP3+ ⎬ = (
) = 4.8*10-12 M
0.010
⎩ Fe ⎭
1
Thus, OH − φ 4.8*10-12 M is required to start the precipitation of
[Fe(OH ) 3 ] .
The solubility equilibrium for Zn(OH)2 is
Zn(OH)2 (s) ⇔ Zn2+ (aq) + 2OH- (aq)
KSP = [Zn 2+ ] [OH − ]
2
So that [OH − ] =
(
1
K SP 12
1.8 *10 -14 2
)
=
(
)
= 1.34*10-6
2+
0.010
Zn
[
]
To precipitate the Fe3+ as Fe(OH)3 without precipitating [OH − ] must be
greater than 4.8*10-12 M and laver than 1.34 *10-6 ,with also mean the PH
is greater than -Lg
10 −14
= 8.13
1.34 * 10 −16
16.68 Question : calculate weather or not a precipitate will form if 2.00
ml of 0.60M NH3 are added to 1.0 L of 1.0 * 10-3 M FeSO4 .
Answer ： [HO − ] =
Kb[NH 3 ] =
-4
1.8 * 10 −5 * 1.2 * 10 −3 =1.47*10
Q = [OH − ] 02 [Fe 2+ ]0 = （1.47*10-4）2 *1.0*10-3 = 2.16*10-11 φ KSP =
1.6*10-14
Therefore a precipitate will form.
16.60 Question :the solubility product of PbBr2 is 8.9*10-6 .determine the
molar solubility (a)in pure water , (b) in 0.2 M solution ,(c) in 0.2
Pb(NH3)2 solution .
Answer ：(a). consider the dissociation of PbBr2 in water . let us be the
molar solubility (in mol/l) of PbBr2
PbBr2 (s) ⇔ Pb2+ (aq) + 2Br-(aq)
Initial (M)
Change(M)
-S
Equilibrium(M)
0
0
+S
+2S
S
2S
The solubility product for PbBr2 is KSP = [Pb 2+ ] [Br − ]2 =4S3
Now ,we can calculate the S
S=(
(b) .
1
K SP 13
8.9 * 10 −6 3
) =(
) = 1.3*10-2 M .
4
4
Step 1 : the relevant species in solution are Br- ions
(from both PbBr2 and KBr ) and Pb2+ .the K+ ions are spectator ions .
Step 2 : because KBr is a soluble strong electrolyte , it
dissociates completely :
KBr (s)
+
2O
⎯H⎯
⎯
→ K (aq) +
0.20M
Br-(aq)
0.20M
Let us be the molar solubility of PbBr2 in KBr solution. We summery the
changes in concentrations as follow.
PbBr2 (s) ⇔ Pb2+ (aq) + 2 Br- (aq)
Initial(M)
Change(M)
0
-S
Equilibrium(M)
+S
S
0.20
+2S
0.2+S
Step 3 : KSP = [Pb 2+ ] [Br − ]2
8.9*10-6 = S(0.2+2S)2
Because PbBr2 is quite insoluble and the present of Br- ions from KBr
further lowers the solubility of PbBr2 , S must be very small compared
with 0.20M . therefore ,applying the approximation 0.20+2S equal 0.20.
we obtain
8.9 * 10-6 =S*0.202
S = 2.2 * 10-4 M
(c) . step 1 . the relevant species in solution are Pb2+ ions
from both PbBr2 and Pb(NO3)2 and Br- ions. The NO3- ions are spectator
ions.
Step 2. because Pb(NO3)2 is a soluble electrolyte it
dissociates completely :
2+
O
⎯
→ Pb (aq) + 2NO3 (aq)
Pb(NO3)2 (s) ⎯H⎯
2
0.20
0.40
Let us be the molar solubility of PbBr2 in Pb(NO3)2 solution . we
summarize the changes in concentrations as follow :
PbBr2(s) ⇔ Pb2+ (aq) + Br- (aq)
Initial(M)
Change(M)
-S
Equilibrium(M)
0.2
0
+S
+2S
0.2+S
2S
Step 3 . KSP = [Pb 2+ ] [Br − ]2 =(0.2 + S) (2S)2
Because the PbBr2 is quite insoluble and the presence of Pb2+ ions
from Pb(NO3)2 further lowers the solubility of PbBr2 .S must be very
small compared with 0.20 M. therefore, applying the approximation
0.2+S ≈ 0.2
We obtain 8.9*10-6 =0.2*4*S2
S = 3.3*10-3 M
Chapter Seven
19.2 Balance the following redox equations by the ion electron method:
(a)
Mn2 + H 2 O2 → MnO2 + H 2 O
(b)
Bi (OH )3 + SnO 22− → SnO32 − + Bi
(c)
Cr2 O72− + C 2 O42− → Cr 3+ + CO2
−
(d) ClO 3 + Cl
−
→ Cl 2 + ClO 2
Solution:
Mn2 + H 2 O2 → MnO2 + H 2 O
1
(a) ○
2 Oxidation: Mn
○
2+
(in basic solution)
+4
→ Mn O2
−1
−2
Reduction: H 2 O2 → H 2 O
3 We balance each half-reaction for number and type of atoms and
○
charges
Oxidation half-reaction: Mn
Reduction half-reaction:
2+
+ 2 H 2 O → MnO2 + 4 H + + 2e −
H 2O2 + 2 H + + 2e − → 2 H 2O
4 We add the Oxidation and Reduction half-reactions to give the
○
overall reaction:
Mn2 + H 2O2 → MnO2 + 2 H +
1
(b) ○
Bi (OH )3 + SnO22− → SnO32− + Bi (in basic solution)
+2
+4
2−
2−
2 Oxidation: Sn O2 → Sn O3
○
+3
0
Bi (OH ) 3 → Bi
Reduction:
3 We balance each half-reaction for number and type of atoms and
○
charges
2−
2−
+
Oxidation half-reaction: SnO2 + H 2 O → SnO3 + 3H + 2e
Reduction half-reaction:
−
Bi 3+ + 3e − → Bi
4 We add the Oxidation and Reduction half-reactions to give the
○
overall reaction:
2 Bi (OH )3 + 3SnO22− → 3SnO32− + 2 Bi + 3H 2O
2−
1 Cr2O7
(c) ○
+ C2O42 − → Cr 3+ + CO2
(in basic solution)
2−
2 Oxidation: C2O4 → CO2
○
+6
2−
7
Reduction: Cr 2 O
+3 3+
→ Cr
3 We balance each half-reaction for number and type of atoms and
○
charges
2−
+
Oxidation half-reaction: C2O4 → 2CO2 + 2 H + 2e
−
2−
3+
−
Reduction half-reaction: Cr2O7 + 14H + + 6e → 2Cr + 7 H 2O
4 We add the Oxidation and Reduction half-reactions to give the
○
overall reaction:
Cr2O72− + 3C2O42− + 14 H + → 2Cr 3+ + 6CO2 + 7 H 2O
−
−
1 ClO3 + Cl → Cl 2 + ClO2 (in basic solution)
(d) ○
−1
−
0
2 Oxidation: Cl → Cl 2
○
+5
−
3
+4
Reduction: Cl O → Cl O2
3 We balance each half-reaction for number and type of atoms and
○
charges
−
Oxidation half-reaction: 2Cl → Cl2 + 2e
−
ClO3− + 2 H + + e − → Cl O2 + H 2O
Reduction half-reaction:
4 We add the Oxidation and Reduction half-reactions to give the
○
overall reaction:
2ClO3− + 2Cl − + 4 H + → Cl 2 + 2ClO2 + 2 H 2O
19.26 Given that E0=0.52V for the reduction
Cu(+aq ) + e − → Cu ( s)
Calculate E0, ΔG 0 ,and k for the following reaction at 25℃:
2Cu(+aq) → Cu(s) + Cu(2aq+ )
+
2+
Solution: As we know, the equation is: 2Cu( aq) → Cu( s) + Cu( aq)
The half-cell reaction is:
Anode (Oxidation):
Cu(+aq ) → Cu(2aq+ ) + e −
Cathode (Reduction):
Cu(+aq ) + e − → Cu ( s)
0
0
0
0
0
Ecell
= Ecathode
− Eanode
= ECu
− ECu
= 0.52V − .15V = 0.37V
+
2+
/ Cu
/ Cu +
ΔG 0 = − nFE 0 = −96500 J / V .mol × 0.37V = −35.71KJ / mol
nE 0
1 × 0.37V
ln K =
=
= 14.4
0.0257V 0.0257V
K = e14.4 = 1.78 × 10 6
2+
−
2+
19.86 Given that 2 Hg ( aq ) + 2e → Hg 2 ( aq )
E 0 = 0.92V
Hg 2 + ( aq ) + 2e − → 2 Hg (l )
E 0 = 0.85V
Calculate ΔG 0 and k for the following process at 25℃:
Hg 22+ ( aq ) → Hg 2+ (aq ) + Hg (l )
Solution: From the reaction we see that
0
0
0
Ecell
= Ecathode
− Eanode
= 0.85V − 0.92V = −0.07V
ΔG 0 = − nFE 0 = −96500 J / V .mol × ( −0.07)V = 6.76 KJ / mol
ΔG 0 = − RT ln K , K = 0.065
19.124 Calculate the equilibrium constant for the following reaction at
298K:
Zn( s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s)
Solution:
0
0
0
Ecell
= Ecathode
− Eanode
= 0.34V − (− 0.76V ) = 1.10V
E = 0.0257
ln k
37
, therefore, K = 1.5 × 10
2
Chapter Eight
7.30 The first line of the Ballmer series occurs at a wavelength of 656.3
nm.What is the energy difference between the two energy levels in what
in the emission that results in this spectral line?
Solution:
(6.63 × 10 −34 J .s )(3 × 10 8 m / s )
E = hν = h =
= 3.03 × 10 −19 J
−9
λ
1 × 10 m
656.3nm ×
1nm
c
From equation (17.6), we write
⎛ 1
1 ⎞
1
1
ΔE = RH ⎜ 2 − 2 ⎟ = −2.18 × 10 −18 J ( 2 − 2 ) = 3.027 × 10 −19 J
⎜n
⎟
3
2
⎝ i nf ⎠
There is only a little difference between the two energy levels in
what in the emission that results in this spectral line.
7.32 Calculate the Frequency and wavelength of the emitted photon when
an electron drops from the n=4 to the n=2 level in a hydrogen atom.
Solution:
⎛ 1
1 ⎞
1
1
ΔE = RH ⎜ 2 − 2 ⎟ = 2.18 × 10 −18 J ( 2 − 2 ) = −4.09 × 10 −19 J
⎜n
⎟
4
2
⎝ i nf ⎠
The negative sign indicates that is energy associated with an emission
process, because Δe = hν ,we can calculate the frequency of the photo
by writing:
ΔE
4.09 × 10 −19 J
ν=
=
= 6.17 × 1014 Hz
−34
h
6.63 × 10 J .s
3 × 10 8 m / s
1nm
λ= =
= 4.86 × 10 −7 m = 4.86 × 10 −7 m ×
= 486nm
14 −1
ν 6.17 × 10 s
1 × 10 −9 m
c
7.60 What is the difference between
α 2 px
and
α 2 py
orbital?
Solution: Differ in orientation only.
7.62 List all the possible sub shells and orbital associated with the
principal quantum number n, if n=6?
Solution: when n=6, l can be 5, 4,3,2,1 and 0.
The orbital responding to the l are 6s, 6p, 6d, 6f, 6g, and 6h.
Chapter Nine
10.10
Predict the geometry of the following molecules and ion using the
VSEPR model: (a) CH3I, (b) ClF3, (c) H2S,(d) SO3,(e)SO42-.
Solution: (a) Tetrahedral (b)T-shaped (c)Bent (d )Trigonal planar (e)
Tetrahedral
10.36
Consider the reaction BF3+Cl- → F3B-NH3
Describe the changes in hybridization (if any) of the Band N atoms as a
result of this reaction.
Solution；B: sp2 to sp3; N: remain sp3
10.104
For each pair listed here, state which one has higher first ionization
energy and explain your choice: (a) H or H2. (b) N or N2, (c) O or O2 (d)
F or F2.
Solution:
(a)
H2
The electron is removed from the more stable bonding
molecular orbital.
(b)
N2 Same as (a)
(c) O The atomic orbital in O is more stable than the anti-bonding
molecular orbital in O2
(d) F The atomic orbital in F is more stable than the anti-bonding
molecular orbital in F2
Chapter Ten
11.10 List the types of intermolecular forces that exist between molecules
(or basic units ) in each of the following species: (a)benzene (C6H6), (b)
CH3Cl, (c) PF3, (d) NaCl, (e) CS2
11.136 Use the concept of intermolecular forces to explain why the far
end of a walking cane rises when one raises the handle.
Chapter Eleven
22.26 Draw structures of all the geometric and optical isomers of each of
the following cobalt complexes: (a) [Co(NH3)4Cl2]+, (b) [Co(en)3]3+.
22.34 Transition metal complexes containing CN- ligands are often
yellow in color, whereas those containing H2O ligands are often green or
blue . Explain.
22.36 The absorption maximum for the complex ion [Co(NH3)6]3+ occurs
at 470nm, (a) Predict the color of the complex and (b) calculate the
crystal field splitting in KJ/mol.
Chapter Twelve
20.30 Calculate the volume of CO2 at 10.0oC and 746mm Hg pressure
obtained by treating 25.0 g of Na2CO3 with an excess of hydrochloric
acid.
20.68 Describe a medicinal or health-related application for each of the
following compounds: NaF, Li2CO3, Mg(OH)2, CaCO3, BaSO4,
Al(OH)2NaCO3
Chapter Thirteen
20.50 The pressure of gaseous Al2Cl6 increases more rapidly with
temperature than predicted by the ideal gas equation even though Al2Cl6
behaves like an ideal gas. Explain.
21.28 Unlike CaCO3, Na2CO3 does not readily yield CO2 when heated .
On the other hand, NaHCO3 undergoes thermal decomposition to produce
CO2 and Na2CO3, (a) Write a balanced equation for the reaction. (b) How
would you test for the CO2 evolved? [Hint: Treat the gas with limewater,
an aqueous solution of Ca(OH)2.]
21.32 Sodium hydroxide is hygroscopic-that is, it absorbs moisture when
exposed to the atmosphere. A student placed a pellet of NaOH on a watch
glass. A few days later, she noticed that the pellet was covered with a
white solid. What is the identity of this solid? (Hint: Air contains CO2. )
Chapter Fourteen
21.42 Write a balanced equation for the formation of urea, (NH4)2CO,
from carbon dioxide and ammonia. Should the reaction be run at a high or
low pressure to maximize the yield?
21.46 Explain why nitric acid can be reduced but not oxidized.
21.74 Explain why SCl6. SBr6, and SI6 cannot be prepared.
Chapter Fifteen
21.84 A 375-gallon tank is filled with water containing 167 g of bromine
in the form of Br- ions. How many liters of Cl2 gas at 1.00 atm and 20oC
will be required to oxidize all the bromide to molecular bromine?
21.92 (a) Which of the following compounds has the greatest ionic
character? PCl5, SiCl4, CCl4, BCl3 (b)Which of the following ions has the
smallest ionic radius? F-, C4-, N3-, O2- (c) Which of the following atoms
has the highest ionization energy? F, Cl, Br, I (d) Which of the following
oxides is most acidic? H2O, SiO2, CO2
21.82 Sulfuric acid is a weaker acid than hydrochloric acid. Yet hydrogen
chloride is evolved when concentrated sulfuric acid is added to sodium
chloride. Explain.
Chapter Sixteen
22.12 Complete the following statements for the complex ion
[Co(C2O4)2(H2O)2]-. (a) The oxidation number of Cr is ______. (b) The
coordination number of Cr is _______. (c) ________is a bidentate ligand.
22.48 Which is a stronger oxidizing agent in aqueous solution, Mn3+ or
Cr3+? Explain your choice.
22.46 In a dilute nitric acid solution, Fe3+ react with thiocyanate ion
(SCN-) to form a dark-red complex:
[Fe(H2O)6]3++SCN- ≒ H2O+ [Fe(H2O)5NCS]2+
The equilibrium concentration of [Fe(H2O)5NCS]2+ may be determined
by how darkly colored the solution is (measured by a spectrometer). In
one such experiment, 1.0 ml of 1.0*10-3 M KSCN and 8.0 ml of dilute
HNO3. The color of the solution quantitatively indicated that the
[Fe(H2O)5NCS]2+ concentration was 7.3*10-5 M. Calculate the formation
constant for [Fe(H2O)5NCS]2+.
Chapter Seventeen
22.56 From the standard reduction potentials listed in table 19.1 for
Zn/Zn2+ and Cu+/Cu2+, calculate ΔGo and the equilibrium constant for the
reaction
Zn(s) + 2Cu2+(aq)→Zn2+(aq) +2Cu+(aq)
22.72 (a) The free Cu(I) ion is unstable in solution and ha a tendency to
dis proportionate :
2Cu+(aq)≒Cu2+(aq)+Cu(s)
Use the information in Table 19.1 to calculate the equilibrium constant
for the reaction. (b) Based on your result in (a), explain why most Cu(I)
compound are insolube