发布时间 : 星期三 文章第七章习题答案更新完毕开始阅读52465285998fcc22bcd10dd6
第七章习题
MgO的晶格能U = 3824 kJ?mol；Mg的升华热?H?s =146.4 kJ?mol；
?1?1
Mg(g)的电离能 I1=737.7 kJ?mol，I2=1451 kJ?mol； 解：
Mg(s) + 1/2O2(g) MgO(s)
?H?s(Mg)
1/2D(O2) ?fH?m(MgO)
?1?1
Mg(g) O(g) ?rH?m = ?U
A1+A2 I1+ I2
2+2?
Mg + O?fH?m(MgO) = ?H?s(Mg) + I1+ I2 + 1/2D(O2) + A1+A2 + ?rH?m A1+A2 = ?fH?m ? (?H?s(Mg) + I1+ I2 + 1/2D(O2) + ?rH?m)
= [? 601.7 ? 146.4 ? 737.7 ? 1451 ? 497/2 + 3824] kJ?mol = 638.7 kJ?mol?1
37．据理论计算公式，计算KBr的晶格能。 解： U??138840Z?Z?A?1??1??kJ?moldn???1?1
KBr晶体：r+/r?=133/196=0.679 为NaCl型晶体，A =1.748，
Z+ = Z? = 1， n = ( 9 + 10 ) /2 = 9.5， d = (133+196) pm = 329 pm U??138840?1?1?1.748?1?1???kJ?mol3299.5???1?1 ??660kJ?mol38．试分析温度对导体和半导体的导电性的影响。 解： 温度升高，导体内质点的热运动加快，阻止了电子在电场中的定向运动，因而电
阻增大，所以，温度升高导体的导电能力下降； 温度升高，半导体价带中的更多电子获得能量跃迁进入空带，使空带成为导带，
同时在价带中留下空穴，使价带也成为导带，因而，温度升高，半导体的导电能力增强。 39．试说明石墨的结构是一种多键型的晶体结构。利用石墨作电极或作润滑剂各与它的哪
一部分结构有关？ 解： 石墨是一种层状晶体，层于层之间靠分子间力结合在一起；而同一层内的C原子
互相以sp2杂化轨道形成共价键；同时同一层内每个C原子上还有一个垂直于sp2杂化平面的2p轨道，每个未杂化的2p轨道上各有一个自旋方向相同的单电子，这些p
轨道互相肩并肩重叠，形成大?键，因而石墨晶体中既有共价?键和?键又有分子间作用力，为多键型分子。
利用石墨作电极与石墨晶体中同一层内C原子的大? 键有关，同一层内每个C
原子上未参与杂化的一个2p轨道各有一个单电子，平行自旋形成大?键，大?键上的电子属整个层的C原子共有，在外电场作用下能定向流动而导电，因而石墨可作电极。 石墨作润滑剂则与石墨晶体中层与层之间为分子间作用力有关，由于分子间作用
力要比化学键弱得多，因此石墨晶体在受到平行于层结构的外力时，层与层之间很容易滑动，这是石墨晶体用作固体润滑剂的原因。
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无机及分析化学学习指导
40．一价铜的卤化物CuF、CuCl、CuBr、CuI按r+ / r?比均应归于NaCl型晶体，但实际上
都是ZnS型，为什么？ 解：由于Cu+ 是18电子构型，极化能力和变形性均很大，X?又有较大极化率，易变形，
因此CuX的离子极化作用较强，带有较大的共价成分，使之成为具有较大共价成分的ZnS型结构。
41．(1) Write the possible values of l when n=5.
(2) Write the allowed number of orbitals (a) with the quantum numbers n = 4, l = 3; (b) with the quantum numbers n = 4, (c) with the quantum numbers n = 7, l = 6, m = 6; (d) with the quantum numbers n = 6, l = 5.
Solution: (1) When n =5, the possible values of l is 0, 1, 2, 3 and 4 .
(2) (a)The allowed number of orbitals with the quantum numbers n = 4 , l = 3 is 7.
(b) The allowed number of orbitals with the quantum numbers n = 4 is 16.
(c) The allowed number of orbitals with the quantum numbers n = 7, l = 6, m = 6
is 1.
(d) The allowed number of orbitals with the quantum numbers n = 6, l = 5 is 11.
42．How many unpaired electrons are in atoms of Na, Ne, B, Be, Se, and Ti? Solution:
atom Na Ne B Be Se Ti unpaired electrons 1 0 1 0 2 2
43．What is electronegativity? Arrange the members of each of the following sets of elements in
order of increasing electronegativities:
(1) B, Ga, Al, In; (2) S, Na, Mg, Cl; (3) P, N, Sb, Bi; (4) S, Ba, F, Si,
Solution: Electronegativity is a ability that a element’s atom attracted electron in a molecular. The order of increasing electronegativities are
(1) In < Ga < Al < B ; (2) Na < Mg < S < Cl; (3) Bi < Sb < P < N; (4) Ba < Si < S < F;
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44．Write the electron configuration beyond a noble gas core for (for example, F, [He]2s2p) Rb,
La, Cr, Fe2+, Cu2+, Tl, Po, Gd, Sn2+ Ti3+ and Lu. Solution:
Rb [Kr]5s
Po
[Xe] 4f5d6s6p
10
14
10
2
4
1
La [Xe]5d6s
1
2
Cr [Ar]3d4sGd
5
1
Fe
2+
6
Cu
2+
9
Tl
[Xe]4f5d6s6p
Lu
1
14
10
2
1
[Ar]3d Sn2+
[Ar]3d Ti3+
[Xe]4f5d6s
712
[Kr]4d5s5p
1022
[Ar]3d[Xe]4f5d6s
1412
第七章习题
45．Predict the geometry of the following species (by VSEPR theory): SnCl2, I3, [BF4], IF5, SF6,
SO42?, SiH4, NCl3, AsCl5, PO43?, ClO4?. Solution:
n VP geometry n VP SiH4 0 4 NCl3 1 4 AsCl5 0 5 PO43? 0 4 ClO4? 0 4 SnCl2 1 3 angular
I3? 3 5 linear
[BF4]? 0 4
IF5 1 6
SF6 0 6
SO42? 0 4
??
tetrahedron square pyramidal octahedron tetrahedron
geometry tetrahedron trigonal pyramidal trigonal bipyramidal tetrahedron tetrahedron 46．Use the appropriate molecular orbital enerry diagram to write the electron configuration for
each of the following molecules or ions, calculate the bond order of each, and predict
which would exist. (1) H2+, (2) He2, (3) He2+, (4) H2?, (5) H22?. Solution: (1) H2+[(?1s)1], the bond order is 0.5; (2) He2[(?1s)(?1s)], the bond order is 0;
(3) He2+[(?1s)2(??1s)1], the bond order is 0.5;
(4) H2? [(?1s)2(??1s)1], the bond order is 0.5;
(5) H2 [(?1s)(?1s)], the bond order is 0; The (1), (3) and (4) would exist.
47．Which of these species would you expect to be paramagnetic? (a) He2+, (b) NO, (c) NO+, (d) N22+, (e) CO, (f) F2+, (g) O2.
Solution: The species to be paramagnetic are He2+, NO, F2+ and O2.
48．The boiling points of HCl, HBr and HI increase with increasing molecular weight. Yet the
melting and boiling points of the sodium halides, NaCl, NaBr, and NaI, decrease with increasing formula weight. Explain why the trends opposite.
Solution: HCl, HBr and HI are all molecular crystal. There is a force in moleculars. The
boiling points of HCl, HBr and HI increase with increasing molecular weight because
the force in moleculars increase with increasing molecular weight.
The melting and boiling points of the sodium halides, NaCl, NaBr, and NaI,
decrease with increasing formula weight, because the ionicity of NaCl, NaBr, and NaI decrease with increasing formula weight.
49．For each of the following pairs indicate which substance is expected to be: (a) More covalent :
MgCl2 or BeCl2 CaCl2 or ZnCl2 CaCl2 or CdCl2 TiCl3 or TiCl4 SnCl2 or SnCl4 CdCl2 or CdI2 ZnO or ZnS NaF or CuCl FeCl2 or FeCl3
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2?
2
?
2
2
?
2
无机及分析化学学习指导
(b) higher melting point:
NaF or NaBr Al2O3 or Fe2O3 Na2O or CaO Solution: (a) More covalent :
MgCl2 < BeCl2 CaCl2 < ZnCl2 CaCl2 < CdCl2 TiCl3 < TiCl4 SnCl2 < SnCl4 CdCl2 < CdI2 ZnO < ZnS NaF < CuCl FeCl2 < FeCl3 (b) higher melting point:
NaF > NaBr Al2O3 > Fe2O3 Na2O < CaO
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