北京化工大学 Model of Final Examination of Inorganic Chemistry) (bi-lingual course) Course code C HM 2 7 0 T 课程代码 Class No.: Name and ID: Items(题号) 二三四五六 Total score(总分) Score(得分) 一、是非题：（判断下列叙述是否正确，正确的在括号中画√，错误的画×。不必 写在答题纸上。)（本大题共10小题，每题1分，共10分） 1x;2x;3x;4V:5V:6V:7x;8x:9x:10x 二、选择题：（在下列各题中，选择出符合题意的答案，将其代号填入括号内。） (本大题共20题，每题1.5分，共30分) 1B:2D:3B:4A:5C:6B:7D,8A:9D:10B:11B:12B:13B:14C:15A:16C: 17B;18B;19B;20D. Please fill in the blank with correct words,symbol or data according to the requirements(10 questions,2 score/question,20 score in total) 1.六氟合硅(IV)酸四氨合镍(II):[Co(H2O)2NH)4]Cl.2.(a:(b) 3.12.39:0.011.4.HNO3;He2";H2",B4Hio:H3PO4.5.Formation of new matter; break and form of new chemical bonds;change of energy.6.Same;minium energy principle:Pauling incompatibility principle and Hund rule.7. 1.5:(a);paramagnetic.8.4.3x105;8.63.9.NazCO3>CaCO3>MgCO;>CuCO3: 第1页
北京化工大学 Model of Final Examination of 《Inorganic Chemistry》 (bi-lingual course) Course code 课程代码 C H M 2 1 7 0 T Class No.: Name and ID: Items (题号) 一 二 三 四 五 六 Total score(总分) Score(得分) 一、是非题：（判断下列叙述是否正确，正确的在括号中画 √ ，错误的画×。不必 写在答题纸上。）（本大题共 10 小题，每题 1 分，共 10 分） 1 x; 2 x; 3 x; 4 √; 5 √; 6 √; 7 x; 8 x; 9 x ; 10 x. 二、选择题：（在下列各题中，选择出符合题意的答案，将其代号填入括号内。 ） （本大题共 20 题，每题 1.5 分，共 30 分） 1 B; 2 D; 3 B; 4 A; 5 C; 6 B; 7 D, 8 A; 9 D; 10 B; 11 B; 12 B; 13 B; 14 C; 15 A; 16 C; 17 B; 18 B; 19 B; 20 D. 三、Please fill in the blank with correct words, symbol or data according to the requirements (10 questions, 2 score/question, 20 score in total) 1．六氟合硅（IV）酸四氨合镍(II)；[Co(H2O)2(NH3)4]Cl3. 2．(a); (b); 3．12.39; 0.011. 4．HNO3; He2 + ; H2 + , B4H10; H3PO4. 5．Formation of new matter; break and form of new chemical bonds; change of energy. 6．Same; minium energy principle; Pauling incompatibility principle; and Hund rule. 7．σ1s 2 σ1s* 2 σ2s 2 σ2s* 2 π2p 3 ； 1.5；(a); paramagnetic. 8．4.3x10-6; 8.63. 9． Na2CO3 > CaCO3>MgCO3>CuCO3； 第 1 页
Al2O;>AIF3>AICl3.10.HgO+2e+H2O>Hg+20H:Zn-2e+20H>Zn(OH)2 NaO:(-)Zn(s)IZn(OH)2(s)/OH(aq)/HgO(s)/Hg(s)(+). Please balance the following chemical equations(5 equations,2 points for each one,10 points in total) 1. PbCl2+ H20 →Pb(OHCI+HCI B2H6(g)+6HO0) →2HB03+6H2 3. 2 Ni(OH)2+CIO →2 NiO(OH+CI+H0 2 KMnO4+5 H202+3 H2S04>2MnSO4+502+K2S04+8H2O Hg22++ 4I (excess) →Hg+HgL2 Simple questions(two questions,5 points for each one,10 points in total) 1.Anwser:O,as its valence electron shell is 2s22p4 When one e is removed,more energy is required to remove the second ewhen the p orbital is half-filled which is stable in view of energy. 2.Answer:Addition of excess NaOH to the solution of Zn2+,Co2*and Alresults in precipitate of Co(OH)2 and formation of soluble Al(OH)4'and Zn(OH)4 complexes. Separate the precipitate Co(OH)2 by centrifugation from the solution,then add HCI, NH3.H2O,acetone,KSCN subsequently to Co(OH)2.blue color will be observed, indicating the presence of Co2.To the solution of Al(OH)4 and Zn(OH)complexes is added HCI and NH3.H2O,then Al(OH)3 precipitate will form,which is separated and identified by forming red solid after reaction with HAc and Al regeant,the Zn2 is identified by formation of white ZnS after reaction of Zn with NaS. Answer the following questions according to requests(4 questions,20 points in total) 1.(5 points)(1) Anwser:(1)AgCl-Ag+CI' At equilibrium x0.05 第2页
Al2O3>AlF3 > AlCl3. 10．HgO+2e + H2O→Hg + 2OH- ； Zn -2e + 2OH- →Zn(OH)2； NaO; (-) Zn(s)|Zn(OH)2(s)|OH- (aq)|HgO(s)|Hg(s) (+). 四、Please balance the following chemical equations (5 equations, 2 points for each one, 10 points in total) 1． PbCl2＋ H2O ⎯⎯→ Pb(OH)Cl + HCl 2． B2H6(g)＋ 6 H2O(l) ⎯⎯→ 2H3BO3 + 6H2 3． 2 Ni(OH)2＋ ClO- ⎯⎯→ 2NiO(OH) + Cl- + H2O 4． 2 KMnO4＋ 5 H2O2＋ 3 H2SO4 ⎯⎯→ 2MnSO4 + 5 O2 + K2SO4 +8H2O 5． Hg2 2+ ＋ 4 I- （excess） ⎯⎯→ Hg + HgI4 2- 五、Simple questions (two questions, 5 points for each one, 10 points in total) 1．Anwser: O, as its valence electron shell is 2s2 2p4 . When one e- is removed, more energy is required to remove the second e- when the p orbital is half-filled which is stable in view of energy. 2. Answer: Addition of excess NaOH to the solution of Zn2+，Co2+ and Al3+ results in precipitate of Co(OH)2 and formation of soluble Al(OH)4 - and Zn(OH)4 - complexes. Separate the precipitate Co(OH)2 by centrifugation from the solution, then add HCl, NH3.H2O, acetone, KSCN subsequently to Co(OH)2, blue color will be observed , indicating the presence of Co2+. To the solution of Al(OH)4 - and Zn(OH)4 - complexes is added HCl and NH3.H2O, then Al(OH)3 precipitate will form, which is separated and identified by forming red solid after reaction with HAc and Al regeant; the Zn2+ is identified by formation of white ZnS after reaction of Zn2+ with Na2S. 六、Answer the following questions according to requests (4 questions, 20 points in total) 1．(5 points) (1) Anwser: (1) AgCl Ag+ + ClAt equilibrium x 0.05 第 2 页
0.05x=Kp(AgC1)=1.77x1010,x=3.54x109molL Ag" +2NH3 Ag(NH3)2 Initial 0.1 0 At equilibria 3.54 x 10 y-2(0.1-3.54x10%0.1-3.54x109 (0.1-3.54x10913.54x109y-2(0.1-3.54x10=K(Ag(NH2)=1.12x107 y=1.79 molL! the total concentration of NH3 is 1.79molL. (2)AgI Ag"I 3.54x109 0.001 3.54x10x0.001=3.54 x 102>Ksp(Agl)=8.52x107 So AgI precipitates. 2.(5 points) Anwser:p(SbCls)initial nRT/V=0.0125mol x8.314J-K*x 521K/3.50L=15.47KPa SbCls(g)=SbCl3 (g)+Cl2(g) Initial /KPa 15.47 0 0 At equilibria 15.47-x x (X/100)2/[(15.47-x)/100]=Km°=1.07， x=P(SbCl)=P(C2)=13.71Kpa,P(SbCs)=15.47-13.71=1.76kpa Poa=1.76+13.71x2=29.17kpa 3.(5 points) Anwser:Ag2SO(s)=2Ag*+SO2 0.01 X 0.012x=Kp(Ag2S0)=1.4x103X=0.14 CaSO4(s)Ca2++SOa2- 0.01y 0.01y=Kp(CaS04)=9.1x106y=9.1x10 So Caprecipitates first 第3页
0.05x = Ksp(AgCl) = 1.77 x 10-10 , x = 3.54 x 10-9 molL-1 Ag+ + 2NH3 Ag(NH3)2 Initial 0.1 y 0 At equilibria 3.54 x 10-9 y – 2 (0.1 – 3.54 x 10-9) 0.1 – 3.54 x 10-9 (0.1 – 3.54x10-9) /3.54 x 10-9 [y- 2 (0.1 – 3.54 x 10-9)]2 = Kf(Ag(NH3)2) = 1.12 x 107 y = 1.79 molL-1 the total concentration of NH3 is 1.79molL-1. (2) AgI Ag+ + I- 3.54x10-9 0.001 3.54x10-9 x 0.001 = 3.54 x 10-12 > Ksp(AgI) = 8.52 x 10-17, So AgI precipitates. 2．(5 points) Anwser: p(SbCl5)initial = nRT/V = 0.0125mol x8.314J⋅K-1 x 521K/3.50L = 15.47KPa SbCl5(g) SbCl3 (g) + Cl2 (g) Initial /KPa 15.47 0 0 At equilibria 15.47-x x x (X/100)2 /[(15.47-x)/100] = KP Θ = 1.07, x = P(SbCl3) = P(Cl2) =13.71Kpa, P(SbCl5) = 15.47-13.71=1.76kpa Ptotal = 1.76 + 13.71 x 2 = 29.17kpa 3. (5 points) Anwser: Ag2SO4(s) 2Ag+ + SO4 2- 0.01 x 0.012 x = Ksp(Ag2SO4) = 1.4 x 10-5 X = 0.14 CaSO4(s) Ca2+ + SO4 2- 0.01 y 0.01y = Ksp(CaSO4) = 9.1 x 10-6 y = 9.1 x 10-4 So Ca2+ precipitates first. 第 3 页
When Ag'begins to precipitate, CaSO4(s)Ca2++SO42 z0.14 z2x0.14=Kp(CaS04)=9.1x106z=6.5x103M 4.(5 points) Anwser:(①)Cu2*(aq)+2e=CuS)1)）△，Gn8,=-zFE°=-2Fx0.3394Y Cu'(aq)+e=Cu(s)2) △Gnm91=-zFE29=-1Fx0.5180 1)-2)x2=3yCu2*(aq)+Cu(s)=2Cu(aq)3)△，Gm°3=-RTlnK0 △Gm91-2△Gm2=△，Gme3,-2Fx0.3394+2x1Fx0.5180=-RTlnKO. 2x96485x0.3394+2x96485x0.5180=-8.314x2981n9,Ke=9.1x107 (2) Cu2(aq)+2e =Cu(s)1) △，Gn91=-zFE,°=-2Fx0.3394V Cu(aq)+e==Cu(s)2) △，Gm91=-zFE2°=-1Fx0.5180V Cu*+2Cl=CuCl2"3) △，Gnme3=-RTlnK,9 1)-2x2)+2X3)=4) Cu2"(aq)+Cu(s)+4CT(aq)---2CuClz(aq)4)A.G.=-RTlnK △Gm91-2△，Gnm92+2△Gm93=△Gm°4. -2Fx0.3394+2x1Fx0.5180-2x8.314x298ln6.91x10=-RTlnKO, e=4.34x103 第4页
When Ag+ begins to precipitate, CaSO4(s) Ca2+ + SO4 2- z 0.14 z 2 x 0.14 = Ksp(CaSO4) = 9.1 x 10-6 z = 6.5 x 10-5 M 4. (5 points) Anwser: (1) Cu2+(aq)＋2e－ === Cu(s) 1) ΔrGm Θ 1 = -zFE1 = -2F x 0.3394V; Cu+ (aq)＋e － === Cu(s) 2) ΔrGm Θ 1 = -zFE2 = -1F x 0.5180V; ____________________________________ 1) -2) x 2 = 3): Cu2+(aq)＋Cu(s) ===2 Cu+ (aq) 3) ΔrGm Θ 3 = -RTlnKΘ ΔrGm Θ 1 -2ΔrGm Θ 2 =ΔrGm Θ 3, -2F x 0.3394 + 2 x 1F x 0.5180 = -RTlnKΘ, 2 x 96485 x 0.3394 + 2 x 96485 x 0.5180 = -8.314 x 298 ln KΘ, KΘ = 9.1 x 10-7 (2) Cu2+(aq)＋2e－ === Cu(s) 1) ΔrGm Θ 1 = -zFE1 = -2F x 0.3394V; Cu+ (aq)＋e － === Cu(s) 2) ΔrGm Θ 1 = -zFE2 = -1F x 0.5180V; Cu+ + 2Cl- = CuCl2 - 3) ΔrGm Θ 3 = -RTlnKf Θ 1）- 2 x 2) + 2 x 3) = 4): Cu2+(aq)＋Cu(s)＋4Cl- (aq)===2CuCl2 - (aq) 4) ΔrGm Θ 4= -RTlnKΘ ΔrGm Θ 1 -2ΔrGm Θ 2 + 2ΔrGm Θ 3 = ΔrGm Θ 4, -2F x 0.3394 + 2x 1F x 0.5180 – 2 x 8.314 x 298 ln6.91x104 = -RTlnKΘ, KΘ = 4.34 x 103 第 4 页
