无机化学第十四章

最新北师⼤考研⽆机化学复习题第⼗四章第14 章p 区元素（⼆）⼀、教学基本要求1．了解16-18族元素的特点；2．了解重点元素硫、卤素的存在、制备和⽤途；3. 掌握重点元素硫、卤素的单质及其化合物的性质，会⽤结构理论和热⼒学解释它们的某些化学现象；4. 了解第1个稀有⽓体化合物的诞⽣及其对化学发展的贡献。

⼆、要点1．薄膜法(membrane process)⼯业上利⽤离⼦交换电解NaCl⽔溶液⽣产Cl2的⼀种⽅法，隔开阳极室和阴极室的薄膜式⼀带有⽀链（⽀链上有磺酸基或羧基）的聚全氟⼄烯⾻架⾼分⼦离⼦交换膜，这种阳离⼦交换膜允许Na+由阳极室流向阴极室以保持电解过程中两室的电荷平衡，⽽不让OH-按相反⽅向流向阳极室。

2. 臭氧空洞(ozone hole)指的是因空⽓污染物质，特别是氧化氮和卤代烃等⽓溶胶污染物的扩散、侵蚀⽽造成⼤⽓臭氧层被破坏和减少的现象。

经过跟踪、监测，科学家们找到了臭氧空洞的成因：⼀种⼤量⽤作制冷剂、喷雾剂、发泡剂等化⼯制剂的氟氯烃是导致臭氧减少的"罪魁祸⾸"。

另外，寒冷也是臭氧层变薄的关键，这就是为什么⾸先在地球南北极最冷地区出现臭氧空洞的原因了。

3.恒沸溶液(azeotropic solution)恒沸溶液即恒沸混合物。

在⼀定条件下，当某些溶液的组成与其相平衡的蒸⽓组成相同时，溶液在蒸馏时期沸点保持恒定，故称恒沸溶液。

例如，在⼀⼤⽓压下，氯化氢和⽔的恒沸混合物中含氯化氢的重量百分数为20.24，其恒沸点是108.6℃。

4.制备某些重要含硫⼯业产品的途径：化合态硫或天然单质硫2 22SO4 2Cl23H3H 2SO4三、学⽣⾃测练习题1．是⾮题（判断下列各项叙述是否正确，对的在括号中填“√”，错的填“×”）1.1 所有卤素都有可变的氧化数。

( )1.2 实验室中⽤MnO2和任何浓度HCl作⽤，都可以制取氯⽓。

( )1.3 卤素单质的聚集状态、熔点、沸点都随原⼦序数增加⽽呈有规律变化，这是因为各卤素单质的分⼦间⼒有规律地增加的缘故。

第十四章 答案1. 金刚石中碳原子采取sp 3杂化，C 原子与C 原子之间以共价键结合形成原子晶体，∴金刚石的主要特征为高硬度，高熔点，高沸点，不导电。

石墨中碳原子采取sp 2杂化，一个C 原子与三个C 原子以共价键连接形成层状结构，每个C 原子上还有一个垂直于层状平面的p 轨道，这些p 轨道重叠形成离域π键，其p 电子离域在每一层上，另外层与层之间是分子间作用力，∴石墨是混合型晶体，它的主要特征为具有润滑性，导电性和导热性。

2. CO 与N 2物理性质相似之处：具有较低的熔点和沸点，具有相近的临界温度和临界压力：m.p. (℃) b.p. (℃) 临界压力(atm) 临界温度(℃)C O－200 －190 36－140 N 2 －210 －196 33－146 化学性质相似之处：CO 与N 2都不助燃；化学性质不同之处：CO 作还原剂，N 2的还原性很差；CO 可以作为配体，而N 2配位能力弱。

CO 与N 2分子性质之所以不同是因为，在CO 分子中O 原子上的一对电子占有C 原子的2p 轨道，增强了C 的电子密度，∴C 原子的配位能力强于N 2（ ）分子中的N 原子。

CO 2与N 2O 相似性很小，而差异性很大，显然它们并不是等电子体之故。

3. SiCl 4↑(1)Na 2SiO 3(2)←−−Si (3)−−→SiHCl 3 ↑(4)(5)4(6)←−−SiO 2(7)−−→Na 2SiO 3(8)−−→H 2SiO 3(1) Si + 2Cl (2) Si + 2NaOH + H 2O 2SiO 3 + 2H 2↑3 + H 2 (4) SiO 2 + 2Mg(5) 3SiF 4+ 4H 2O 4SiO 4 + 2H 2SiF 6 (6) SiO 2 + 4HF4 + 2H 2O(7) SiO 2 + Na 2CO 32SiO 3+ CO 2↑ (8) SiO 32－ + 2H +2SiO 3↓4. SiO 2晶体属于原子晶体，基本单元为SiO 4，Si 原子为四面体中心，四个O 原子位于正四面体的顶点，原子之间以—O —Si —O —共价键相连，∴以固体状态存在；而CO 2是直线型分子，低温下固体CO 2属于分子晶体，每个CO 2分子通过分子间作用力连接，作用力很弱，温度略微上升，分子间作用力就被破坏而形成单个CO 2分子，∴常温时为气态。

无机化学第四版第十四章习题答案（Inorganic chemistry, Fourth Edition, fourteenth chapter, exercises, answers）Fourteenth chapter halogen14-1 why do not use KF water solution when making fluorine by electrolysis? Why liquid hydrogen fluoride does not conduct electricity while potassium fluoride anhydrous hydrogen fluoride solution can conduct electricity?1 solution: because F2 reacts with water, there is no free ion in the liquid HF molecule, so it can not conduct electricity. In the anhydrous HF solution of KF, K+ and HF2- exist;14-2 what are the special characteristics of fluorine in this element? What are the characteristics of hydrogen fluoride and hydrofluoric acid?2, solution: (1) as the radius of the F is very small, so F2 can dissociation is very small, the F- hydration heat of other halogen ions.(2) hydrogen bonds exist between HF molecules, so the boiling point and heat of vaporization of HF molecules are especially high.(3) AgF is a compound easily soluble in water.(4) the product of F2 reaction with water is complex.(5) HF is a weak acid with increasing acidity and increasingconcentration.(6) HF can react with SiO2 or silicate to form gaseous SiF4;14-3 (1) the reaction tendency of Cl2 was generated by comparing KMnO4, K2Cr2O7 and MnO2 with hydrochloric acid (1mol.L-1) according to the electrode potential.(2) what is the minimum concentration of hydrochloric acid if MnO2 is reacted with hydrochloric acid to cause Cl2 to occur successfully?3, Xie: (1) according to the relationship between electrode potential, we can see the reaction trend:KMnO4>K2Cr2O7>MnO2;14-4 according to the potential map, the reaction equilibrium constant of Br2 in alkaline aqueous solution is reduced to Br- and BrO3- at 298K.4. Solution: by formula: -ZFE=-RTlnKGet: K=exp (ZFE/RT)=2.92 * 103814-5, three, fluorinated nitrogen NF3 (boiling point -129 degrees Celsius) does not show Lewis alkaline, and relatively low molecular mass compounds NH3 (boiling point -33 degrees Celsius) is a well known Lewis base. (a) to explain the reasonwhy their volatility is so great; (b) explain why they differ in alkalinity.5. Xie: (1) NH3 has a higher boiling point because of the hydrogen bond between the molecules.(2) the NF3 atom has a large radius of F atoms, which makes it difficult to match Lewis acid because of steric hindrance.In addition, the electronegativity of the F atom is larger, which weakens the electronegativity of the central atom N.14-6 from bittern making Br2 available chlorine oxidation method. But from a thermodynamic point of view, Br- can be oxidized to Br2 by O2. Why not use O2 to make Br2?14-7 pass Cl2 can be used in slaked lime to obtain bleaching powder, and hydrochloric acid can be added to Cl2 in the solution of bleaching powder. The two phenomena are explained by using electrode potential.7, Xie: because Cl2 pass into the hydrated lime is in alkaline medium, and because of, so Cl2 in alkaline conditions prone to disproportionation reaction.When the hydrochloric acid is added to the bleach solution, the following reaction can be carried to the right:HClO + Cl- + H+ = Cl2 + H2O14-8 which of the following oxides are anhydrides: OF2, Cl2O7,ClO2, Cl2O, Br2O and I2O5? If the anhydride is prepared, the reaction is obtained by the corresponding acid or other method of getting the anhydride.8 solution: Cl2O7 is the anhydride of HClO4. Cl2O, Br2O are HClO, HBrO anhydrides, respectively14-9 how do you identify the three salts of KClO, KClO3, and KClO4?9 、 solution: add a small amount of solid into the dried test tube, and then do the following experimentAdd dilute hydrochloric acid, that is, Cl2 gas release is KClO;KClO+2HCl=KCl+Cl2+H2OAdding concentrated hydrochloric acid has Cl2 and emits, and the solution turns yellow is KClO3;8KC1O3+24HCl (strong) =9Cl2 = +8KCl+60ClO2 (yellow) +12H2OThe other is KClO414-10 the reaction equation for preparing HIO4, KIO3, I2O5 and KIO4 was prepared with I2 as raw material.The solubility of 14-11 (1) I2 in water is very small. The concentration of I2 saturated solution is calculated from the following 2.5 reactions at 298K.I2 (s) + 2e- = 2I-; 0.535V = Phi thetaI2 (AQ) + 2e- = 2I-; 0.621V = Phi theta(2) 0.100mol I2 was dissolved in 1.00L 0.100mol.L-1 KI solution and I3- solution was obtained. The Kc value of the I3- generation reaction is 0.752, and the concentration of I2 in the I3- solution is calculated.14-11, (1) I2 (AQ) =I2 (s)K=exp (ZFE/RT)=812K=1/[I2 (AQ)]I2 (AQ) =1/812=0.00123mol/L(2) I2+I-=I3-;KC=; so [I2]=?The solution is x=0.0934mol/L.Explain the following phenomena: 14-12 by adding a small amount of NaClO in the electrode potential of starch potassium iodide solution, blue A solution, adding excessive NaClO, B get a colorless solution, then acidified and add a small amount of solid Na2SO3 in B solution, A blue reproduction, when the excessive amount of Na2SO3 blue and then faded become colorlesssolution C, A the solution then add NaIO3 solution and blue. Point out A, B, C why each kind of substance, and write out the reaction equation of each step.12, Xie: A:I2;B:NaIO3;C:NaI14-13 write out the equation for the reaction of iodate and excess H2O2, such as adding starch to the system. What do you see?13 solution: HIO3+3H2O2=3O2+HI+3H2O; if the starch is added to the system, the solution turns blue slowly and then fades.14-14 write out the molecular formula of three metal halides with covalent bonds, and explain the common properties of this type of halides.14 and Xie: (AlCl3) 2; (AlBr3) 2; (AlI3) 2; all molecules contain coordinate bonds14-15 what is a multi halide? What is the trend of the formation of Br3- and Cl3- ions in comparison with I3- ions?14-16 what are halogen compounds?(a) write ClF3, BrF3 and IF3 halogen halide complexes, central atom hybridization orbitals, molecular electronicconfigurations and molecular configurations.(b) is there any risk of explosion when the following compounds are exposed to BrF3? Explain why.SbF5; CH3OH; F2; S2Cl2(c) why are halogenated compounds often diamagnetic, covalent, and chemically active than halogen?14-17 laboratory has a calcium halide, soluble in water, try to use concentrated H2SO4 to determine the nature and name of this salt.17. Solution: different phenomena can be identified by the reaction of halide with concentrated sulfuric acid.14-18 please click the following example,The bromine, iodine, and halogen ions containing various conversion and transformation conditions as oxygen acid interaction diagram.。

在O３中，中心氧原子的δ＝６－２－６×３畅４４３畅４４＋３畅４４＝１左侧端基氧原子：δ＝６－４－４×１２＝０右侧端基氧原子：δ＝６－６－２×１２＝－１ 当然，对O３这一特例，部分电荷的计算与形式电荷的计算是一致的，这是一种巧合。

通过计算表明O３是极性分子，正电荷的中心靠近中心氧原子，负电荷中心靠近端基氧原子。

必须指出的是：这种计算虽然能说明O３的极性，但仍然很粗糙，实际上两个端基氧原子的电荷密度是相等的。

O３中有Π４３键，端基O 与中心O之间并不是一个双键、一个配键。

这正是Lewis结构的不足所致。

在CO中，δ（C）＝４－２－６×２．５５２．５５＋３．４４＝－０．５５４δ（O）＝６－２－６×３．４４２．５５＋３．４４＝＋０．５５４由此可以推断：在CO中正电荷中心靠近O，负电荷中心靠近C。

这一推断与实验事实是一致的（分子轨道理论也对CO的极性做出了解释）。

通过部分电荷的计算，讨论O３，CO的极性，以便开阔思路，可尝试推广之。

主要参考文献 ［１］Bodner G M．Core Text Chemistry Structure&Dynamics．John Wiley&Sons Inc，１９９６． ［２］Shriver D F．Inorganic Chemistry．２nd ed．Oxford University Press，１９９４． 中译本：高忆慈，译．无机化学．２版．北京：高等教育出版社，１９９７．三、习题解析 1（141）畅试写出下列物质之间的反应方程式（（１），（４）略）。

（２）液氨和钠； （３）浓硝酸和汞； （５）稀硝酸和银；（６）锡和浓硝酸； （７）氯化铵溶液与亚硝酸钠溶液；（８）酸性溶液中碘化钾与亚硝酸钠。

解：（２）２N H３（l）＋２Na（s）NaN H２（am）＋H２（g） （３）H g（l）＋４H N O３（浓）H g（N O３）２（aq）＋２N O２（g）＋２H２O（l） （５）３Ag（s）＋４H NO３（稀）３AgN O３（aq）＋NO（g）＋２H２O（l） （６）Sn（s）＋４H N O３（浓）β－H２SnO３（s）＋４N O２（g）＋H２O（l） （７）N H４Cl（aq）＋NaN O２（aq）△N２（g）＋NaCl（aq）＋２H２O（l） （８）２I－（aq）＋２N O－２（aq）＋４H＋（aq）I２（aq）＋２N O（g）＋２H２O（l） 2（142）畅完成并配平下列反应方程式（（１），（２）略）： （３）Cu（N O３）２△（４）H g（N O３）２△ 解：（３）２Cu（N O３）２（s）△２CuO（s）＋４NO２（g）＋O２（g） （４）H g（N O３）２△H g（g）＋２N O２（g）＋O２（g） 樉熟悉硝酸盐受热分解的规律。