《現(xiàn)代基礎(chǔ)化學(xué)》答案_化學(xué)工業(yè)出版社

1、現(xiàn)代基礎(chǔ)化學(xué)答案 化學(xué)工業(yè)出版社第一章，習(xí)題答案，P44-461. E = -B/n2 (B= 2.18´ 10-18 JDE = -B/52 + B/22 = 4.58 ´ 10-19 Jhv = DE, ln = cl = c ´ h ¸ DE = 434 nm3不存在；存在4Ni 3s2 3p6 3d8 4s2的四個(gè)量子數(shù)是：3s2: n= 3, l=0, m=0, ms = +1/2; n= 3, l=0, m=0, ms = -1/23p6: n= 3, l=1, m=-1, ms = +1/2; n= 3, 1=1, m=-1, ms = -

2、1/2; n= 3, 1=1, m=0, ms = +1/2n= 3, 1=1, m=0, ms = -1/2; n= 3, 1=1, m=1, ms = +1/2; n= 3, 1=1, m=1, ms = -1/23d8: n= 3, 1=2, m=-2, ms = +1/2; n= 3, 1=2, m=-2, ms = -1/2; n= 3, 1=2, m=-1, ms = +1/2n= 3, 1=2, m=-1, ms = -1/2; n= 3, 1=2, m=0, ms = +1/2; n= 3, 1=2, m=0, ms = -1/2n= 3, 1=2, m=1, ms = +1/

3、2; n= 3, 1=2, m=2, ms = +1/24s2: n= 4, 1=0, m=0, ms = +1/2; n= 4, 1=0, m=0, ms = -1/25 3d 4f 2s 3d 1s 3p; 結(jié)果： < < < = < 6n=4；l=1；m=0；ms=1/2或-1/278 n>3；3d l = 1；2p m= -1 /0/+1; 3p ms = +1/2 或-1/2, 4s948Cd 4s24p64d105s2.10有三個(gè)19K；24Cr；29Cu均第四周期, K: s區(qū)，IA族；Cr: d區(qū)VIB族；Cu: ds區(qū)IB族11. (1) Ar:

4、 1s22s22p63s23p6 (2)Fe: 1s22s22p63s23p64s23d6 (3) I: 1s22s22p63s23p63d104s24p64d105s25p5 ; (4) Ag: 1s22s22p63s23p63d104s24p64d105s112(1)Ca: s區(qū), 第四周期，IIA族；(2) Cl: p區(qū)，第三周期，VIIA；(3) Ti: d區(qū)，第四周期，IVB；(4) Hg: ds區(qū), 第六周期, IIB族 Ca: +2，Cl: +7，Ti: +4，Hg: +2 (1)<(3)<(4)<(2)13鐵原子3d64s2；26Fe1448；Kr4d105s

5、2, IIB族；48Cd15甲：3s23p5，VIIA，非金屬，電負(fù)性高；乙：3d24s2，IVB，金屬，電負(fù)性低。16Cl；Cs；Be；Cl；Mn1922Ti3+ Ar3d1 1個(gè)未成對(duì)；24Cr2+ Ar3d4 4個(gè)未成對(duì)；27Co3+ Ar3d6 4個(gè)未成對(duì)；46Pd2+ Kr4d8 2個(gè)未成對(duì)；58Ce3+ Xe4f1 1個(gè)未成對(duì)；57La3+ Xe5d0 無(wú)未成對(duì)；第二章習(xí)題答案，P81-821 BeH2: sp雜化，直線型 SiH4: sp3雜化，正四面體。 BF3: sp2雜化，平面三角形 CO2: sp雜化，直線型。2BF3采用sp2雜化；BF4-采用sp33H2O sp3不等

6、性雜化；BeH2 sp雜化4H2O sp3不等性雜化，104.5度，CH4 sp3雜化，109.5度；CO2 sp雜化，180度。11H2色散O2色散H2O色散誘導(dǎo)取向氫鍵H2S色散誘導(dǎo)取向H2S-H2O色散誘導(dǎo)取向H2O- O2色散誘導(dǎo)HCl-H2O色散誘導(dǎo)取向CH3Cl色散誘導(dǎo)取向12乙醇高，有氫鍵 甲、乙、丙醇分子結(jié)構(gòu)相似，分子量依次增大，色散力遞增，熔沸點(diǎn)依次升高。丙三醇分子量大，熔沸點(diǎn)高，HF有氫鍵，熔沸點(diǎn)高。13CH4,CCl4,CI4分子量遞增，色散力增大，熔沸點(diǎn)依次升高。H2O有氫鍵，CH4無(wú)氫鍵。第三章習(xí)題答案，P1191NaF AgBr NH3 O2離子晶體 過(guò)渡晶體 含氫

7、鍵20.225 <r+/r- (ZnS) < 0.414, 0.414 < r+/r (NaCl) <0.732, 0.732 < r+/r (CsCl) <1.0離子r+/r-型號(hào)離子r+/r-型號(hào)LiCl0.331ZnSRbCl0.818CsClCaS0.538NaClMgO0.464NaClRbBr0.755CsClLiI0.278ZnS3熔點(diǎn)隨半徑下降低升高，隨離子電荷升高而升高。4SiC是原子晶體，熔點(diǎn)高。水有氫鍵，熔點(diǎn)高。KCl離子晶體，熔點(diǎn)高。MgCl2離子晶體，半徑小熔點(diǎn)高。KI極化力小，熔點(diǎn)高。5Sn4+> Sn2+ > Fe2

8、+ >Sr2+ S2->O2->F- 第四章習(xí)題答案，P1511配合物中心離子配體配位數(shù)配離子電荷名稱Cu(NH3)4(OH)2Cu2+NH3 42+氫氧化四氨合銅（II）CrCl(NH3)5Cl2Cr3+Cl-,NH3 62+二氯化一氯。五氨合鉻（III）CoCl(NH3)(en)2Cl2Co3+Cl-, NH3,en 62+氯化二氯×一氨×二（乙二氨）合鈷(III)Pt Cl2(OH)2(NH3)2Pt4+Cl-,OH-, NH3 60二氯二羥基。四氨合鉑（IV）Ni(CO)4NiCO 40五羰基合鎳K3Fe(CN)5(CO)Fe2+CN-,CO 63

9、-五氰根。一羰基合鐵（II）酸鉀2配合物中心離子配體配位數(shù)配離子電荷名稱(NH4)3 SbCl6Sb3+Cl- 63-六氯合銻（III）酸銨CoCl(NH3)5Cl2Co3+Cl-, NH3 62+氯化一氯×五氨合鈷(III)Pt Cl(OH)(NH3)2Pt2+Cl-,OH-, NH3 40一氯一羥基。二氨合鉑(II)Cr(NH3)2(H2O)43+Cr3+NH3H2O 63+二氨四水合鉻（III）PtCl3(OH)(NH3)2Pt4+Cl-,OH-, NH3 60三氯。一羥基。二氨合鉑(IV)3名稱化學(xué)式內(nèi)界外界單基配體多基配體1CoCl2(NH3)3(H2O)ClCoCl2(N

10、H3)3(H2O)+Cl-Cl-,NH3,H2O2Na2Ca(EDTA)Ca(EDTA)2-Na+EDTA3(NH4) Cr(SCN)4 (NH3)2 Cr(SCN)4 (NH3)2- NH4+NH3，SCN-4Cr(OH)3(H2O)(en)Cr(OH)3(H2O)(en)無(wú)OH-，H2Oen5K2PtCl6PtCl62-K+Cl-4. Co (NH3)6Cl3； CoCl(NH3)5Cl2； CoCl2(NH3)4Cl； CoCl3(NH3)38. Ni(CN)42-為dsp2雜化， Ni(NH3)42+為sp3雜化， Fe(CN)64-為d2sp3雜化，無(wú)單電子；FeF63-為sp3 d

11、2雜化，有五個(gè)單電子。9. 名稱磁矩電子分布空間構(gòu)型內(nèi)外構(gòu)型FeF63-5.90sp3 d2 ­ ­ ­ ­ ­正八面體外軌型Fe(CN)64-0d2sp3 E E E - -正八面體內(nèi)軌型Fe(H2O)62+5.30sp3 d2 E ­ ­ ­ ­正八面體外軌型Co(NH3)63+0d2sp3 E E E - -正八面體內(nèi)軌型Co(NH3)62+4.26sp3 d2 E E ­ ­ ­正八面體外軌型Mn(CN)64-1.80d2sp3 E E ­ - -正八面體內(nèi)

12、軌型第五章習(xí)題答案1. W = -P(V2 - V1)=-P2(P1V1/P2 - V1)=-1×105(5 × 2 × 10-3 2 × 10-3)=-800J2. U=Q+W=Q+p(V2-V1)=6.48+-93.3(150-50)=4.68+(-9.33)=-2.85kJ5. H = Qp = 20 × 858 = 17.16 × 103 J = 17.16 kJW = -PoutV = -101325 × 20 × 607 × 10-6 = -1230J= -1.230 kJU = Qp + W

13、 = 17.16 1.23 = 15.93 kJ7. 2(2) (1) - 5/2 (3)得，fHm (C2H2) = 90.2 × 2 -(-1255.6) - 483.7 × 2.5 = 226.75 kJ/mol8. 1/2+- rHm°= 0.5-234.7-184.6-30.59+285.85= -127.02kJ/mol9. cHm°= 2fHm°(CO2) + 3fHm°(H2O) -fHm°(C2H5OH) = 2(-393.51) + 3(-285.85) (-276.98) = -1367.59 kJ/m

14、ol；92 克乙醇燃燒放出的熱量為因?yàn)?2/46 = 2mol, rHm°=2×(-1366.75)= -2733.5 kJ12. fHm°(C4H10) = 4fHm°(CO2) + 5fHm°(H2O) cHm°(C4H10) = 4(-393.51) + 5(-285.85) - (-2878.51) = -124.78 kJ/mol七 4. rGm°=rHm°-TrSm°= -91.92-298.15(-198.3)×10-3=-32.79kJ/mol rGm°=rHm

15、6;-TrSm°=181.42-298.15×.216=117.0kJ/mol rGm°=rHm°-TrSm°= -890.36-298.15×(-0.243)= -817.9kJ/mol6. rGm°=(-569.6)+2×(-228.59)-(805.0)-4(-95.27)= 159.5kJ/mol rGm°= -394.38-(-137.3)-(×228.59)= -28.49kJ/mol rGm°= 3(-394.38)-(-741.0)-3(-137.30)= -30.24

16、kJ/mol7. K25°=exp-(rHm°-TrSm°)/RT =exp-(-211.5×103-298.15×116.9)/(8.314×298.2)=exp99.384=1.45×1043 K100°=exp-(-211.5×103-373.15×116.9)/(8.314×373.15)=exp82.23=5.175×10358. rGm°= -RTlnK°= -8.314×3500ln8.28= -61510.7J/mol= -61.

17、51kJ/mol10. rGm°=37.22-298.15×0.09496=8.908kJ/mol>0； T1>rHm°/rSm°=37220/94.96= 319.95KrGm°=299.8-298.15×0.3633= 191.48kJ/mol>0； T2>rHm°/rSm°=299.8/0.3633= 825.2K 第一種好。17. rGm°=-402-298.15×0.1891= -345.6kJ/mol <0 正方向進(jìn)行rGm°=rHm°

18、;-TrSm°因rSm°<0，T，rGm°也，降低溫度有利于反應(yīng)進(jìn)行。反應(yīng)正向進(jìn)行的最低溫度：T=rHm°/rSm°=-402×103/-189.1= 2125.9K八. 思6 條件PAP總TCatalyst速率k不變不變6. v=kCaCc2 三級(jí)； k=0.05/(0.01×0.012)=5×104(mol/L)-2.s-1 v=5×104×0.53=62507. C0/C=v0/v=1×10-5/3.26×10-6=3.067; k=(1/t)ln(C0/C)=(

19、1/3600)ln3.067=3.11×10-4/s t1/2=ln2 /k=0.693/k=0.693/3.11×10-4=2226.26s C=v/k=10-5/3.11×10-4=0.03212mol/L10. k=Ae-Ea/RT Ea=RT1T2ln(k2/k1)/(T2-T1)=(8.314×300×310×ln2)/10=53594J=53.59kJ15. k=v/(CA2CB)=2.5×10-3/0.013=2500(mol/L)-2.s-1 v=2500×0.0152×0.030=0.0

20、1688mol/L.s16. Q=176-12=164kJ/mol； Ea+=176kJ/mol；Ea-=12kJ/mol 在曲線最高處有活化絡(luò)合物存在。 是放熱反應(yīng)九. 1ChemicalskindConjugate baseConjugate acidChemicalskindConjugate baseConjugate acidHCOOHCHCl2COOHH2C2O4 H2OH2PO4- AAADD HCOO-CHCl2COO-HC2O4- HO-HPO42-H3O+H3PO4HPO42- HS-CO32- CN-DDBBPO43- S2-H2PO4- H2SHCO3- HCN2. C

21、hemicalsConjugated acid equilibrium constantsConjugated base equilibrium constants S2-Ka2(H2S) = 1.26 ´ 10-13HCO3-Ka1(H2CO3) = 4.47 ´ 10-7Kw / Ka2 = 2.14 ´ 10-4H2PO4-Ka1(H3PO4) = 6.92 ´ 10-3Kw / Ka2 = 1.62 ´ 10-7HPO42-Ka2(H3PO4-) = 6.17 ´ 10-8Kw / Ka3 = 2.09 ´ 10-2

22、Ac-1.74 ´ 10-53.x = 4.03 x 10-2H+ = SO42- = 4.03 x 10-2 mol / LHSO4- = 0.2 x = 0.16 mol/L4. H+=HCO3-=(0.025×4.36×10-7)1/2=1.06×10-4 mol/L CO32-=4.68×10-11; H2CO3=0.0256. pKa1 = 2.16, pKa2 = 7.21, pKa3 = 12.32 pH = 1, pH < pKa1, H3PO4 > H2PO4- >HPO42- >PO43- pH =

23、5, pKa1 < pH < pKa2, H2PO4- > H2PO4- >H3PO4 >PO43- pH = 10, pKa2 < pH < pKa3, HPO42- > PO43- >H2PO4- >PO43- pH = 14, pH > pKa3, PO43- > HPO42- >H2PO4- >H3PO4 7. HCl=0.025, pH=1.61 mol/L NH4Cl=0.05; H+=(0.05×10-14/1.74×10-5)1/2=5.35×10-6; pH=5.

24、27HAc=0.05; NaAc=0.05; H+=KCa/Cb=1.75×10-5; pH=4.76HAc=0.05; NaAc=0.05; H+=KCa/Cb=1.75×10-5; pH=4.7610. H+=KCa/Cb=10-4.5=3.16×10-5; CNaAc=0.45mol/L; 0.45×0.5×136 = 30.6g;HAc: 0.82×500/6 = 68.3ml11. H+=1.75×10-5×0.02/0.8=4.38×10-7; pH=6.36HAc=(0.1×10/

25、55)+0.1×5/55=0.027; NaAc=0.8-0.1×5/55=0.718; H+=1.75×10-5×0.027/0.718=6.58×10-7; pH=6.18 pH=0.18HAc=(0.1×10/55)-0.1×5/55=0.009; NaAc=(0.1×40/55)+0.1×5/55=0.736; H+=1.75×10-5×0.009/0.736=2.148×10-7; pH=6.67 pH=0.3112. pH = pKa + lg(Ac- / HAc

26、) = 2.68(1) pH = pKa + lg(Ac- / HAc) = 2.38(2) pH = pKa + lg(Ac- / HAc) = 2.8613. K = Ka2/Kw = 4.68 ×103; K = Ka2/Kw = 6.17 × 106;K = Ka1/Kw = 1.41 × 1012;K = Ka2/Ka3 = 1.29 × 10515. 1.37×10-4 / 811.5 = 1.688×10-7; Ksp = (3 × 1.688 × 10-7)3 × (2 × 1.

27、688 × 10-7)2 = 1.48 × 10-3216. PbSO4, Pb2+ = = = 1.59 ´ 10-4 mol / L PbS, Pb2+ = = = 2.83 ´ 10-14 mol / L17. Pb2+=0.0050; I-=0.005; Pb2+I-2 = (0.0050)3 = 1.3 × 10-7 > Ksp(9.8 × 10-9)；有沉淀。Ba2+ = 0.020; CO32- = 0.30; Ba2+CO32- = 0.020×0.30 = 6.0 ×10-3 > Ks

28、p(2.58×10-9)；有沉淀。Ag+ = 0.010; Cl- = 0.10; Ag+Cl- = 0.010×0.100 = 1.0×10-3 > Ksp(1.77×10-10)；有沉淀。OH-=5.0×10-5; Mg2+OH-2=10-7× (5.0×10-5)2=2.5×10-16<Ksp(5.61×10-12)；無(wú)沉淀。19. (1) AgCl 溶解度S1 = = 1.33 ´ 10-5 mol/L Ag2CrO4 溶解度S2 = = 6.5 ´ 10-5 mo

29、l/L(2) 在0.10 mol/L AgNO3溶液中，AgCl 溶解度S1 = = 1.77 ´ 10-9 mol/L Ag2CrO4 溶解度S2 = 1.12 ´ 10-10 mol/L23Fe3+完全沉淀時(shí); Ksp = Fe3+OH-3 OH- = 6.53×10-12; pOH = 11.8; pH = 2.82Cu2+ 開始沉淀時(shí)，Ksp = Cu2+OH-2 OH-=2.10 ×10-10; pOH=9.68; pH = 4.32; pH = 2.82-4.32; 可以使兩種離子完全分離。24K=KZnS/KAg2S=2.5×10

30、-22/6.3×10-50=4.0×1027; K=KZnS/KPbS=2.5×10-22/8.0×10-28=3.1×105;K=KPbCl2/KPbCrO4=1.7×10-5/2.8×10-13=6.1×107;25Cr2O42- = 0.010 mol/LPb2+ = Ksp / Cr2O42- = 4.8 ´ 10-8I- = = 0.45 mol/LNaI = 0.45 + 0.02 = 0.47 mol/L26. Ac-= KHAc HAc/ H+=1.75×103;Ag+=Ksp/

31、Ac-=1.94×10-3/1.75×10-3=1.3mol/L; 1.3×170=221g27. 加入等體積NH3水后，OH- = = 9.43 ´ 10-4 mol/LJ = Mg2+OH-2 = 2.22 ´ 10-7 > Ksp, 產(chǎn)生Mg(OH)2沉淀不產(chǎn)生Mg(OH)2沉淀的條件是Mg2+OH-2Ksp, OH- < 4.74 ´ 10-6 mol/LpOH = pKb + lg, NH4+ = 0.19 mol/LNH4Cl 的質(zhì)量 = 0.19 ´ 53.5 = 10.2 g29. S2-=K1K

32、2H2S/H+2=1.07×10-7×1.26×10-13×0.10/0.12=1.35×10-19Pb2+S2-=0.001×1.35×10-19=1.35×10-22>Ksp,PbS(8×10-28); 有沉淀。十. 1 N = 0.0406 mol; t = nZF / I = (0.0406X2X96485/5) = 1567 s = 0.44 h4. 2KMnO4+5K2SO3+3H2SO4=6K2SO4+2MnSO4+3H2O 5NaBiO3(s) + 2MnSO4 + 16HNO3 =

33、 2HMnO4 + 5Bi(NO3)3 + 2Na2SO4 + NaNO3 + 7H2O 4Zn + NO3- + 10H+ = 4Zn2+ + NH4+ + 3H2O 3Ag+NO3-+4H+=3Ag+NO+2H2O Cl2+2OH-=Cl-+ClO-+H2O 8Al+3NO3-+5OH-+18H2O=8Al(OH)4-+3NH35 (a) 氧化劑：Ag+, 還原劑：Cu -)Cu½Cu2+(c1) |Ag+ (c2) ½Ag (+(b) 氧化劑：Pb2+, 還原劑：Cu -) Cu,CuS½S2-(c1) |Pb2+ (c2) ½Pb (+(c) 氧

34、化劑: H+,還原劑: Pb 原電池: -) Pb , PbCl2 | Cl- (c1) | H+ (c2)| H2 , Pt (+6 最強(qiáng)氧化劑最強(qiáng)還原劑1.MnO4-(H+) 1.5VFe2+ 0.771V2.Cr2O72- 1.23Cr(OH)3 -0.133.Fe3+ 0.771Fe -0.4477 能共存，不能共存8 ECu2+/Cu=0.342+0.0296lg0.5=0.333V E=0.342+0.0296lgKsp1/2=-0.179VECuS/Cu=0.342+0.0296lgKsp=-0.700V9. E = -0.0592V E = -0.414 V E = -0.17

35、 V10. 正極：Cu, 負(fù)極：Zn正極：Cu2+ + 2e ® Cu, 負(fù)極：Zn - 2e ® Zn2+電池反應(yīng)不：Zn + Cu2+ = Zn2+ + CuE=0.342-0.760+0.0296lg0.001 = 0.342-(-0.849) = 1.19 V正極：Pb, 負(fù)極：Cu正極：Pb2+ + 2e ® Pb, 負(fù)極：Cu + S2- - 2e ® CuS電池反應(yīng)：Cu + Pb2+ + S2- = Pb + CuSE=-0.126+0.0296lg0.1-0.342+0.0296lg(6.3X10-36/0.1)=-0.156-(-0.

36、670)=0.514V正極：Hg, 負(fù)極：H2, Pt正極：Hg2Cl2 + 2e ® 2Hg + 2Cl-, 負(fù)極：H2 - 2e ® 2H+電池反應(yīng)：H2 + Hg2Cl2- = 2Hg + 2 HClE = 0.268 + 0.327 0 = 0.327 V正極：H2, 負(fù)極：Zn正極：2H+ 2e ® H2, 負(fù)極：Zn - 2e ® Zn2+電池反應(yīng)：Zn + 2H+ = Zn2+ + H2E=0.05916lg(cKa)1/2-0.760+0.296lg0.1 = -0.17-(-0.790) = 0.62 V11. E=0.799-0.05

37、92 - (-0.76) - 0.0296lgZn2+ = 1.51 Zn2+ = 0.545 V12. E = -0.1375 + 0.1262 - lgKsp = 0.22; Ksp=1.53 ´ 10-813. 和正向進(jìn)行，反向進(jìn)行14. K=1.41X1041K=0.149 K=2.97X10-5 K=2.14X106216. EBrO-/Be- =0.76; EBrO3-/Br-=0.5217. (3X1.679+2x)/(3+2)=1.507V x=1.249V18. G = -ZFE = -6X96500X(1.23-1.36)X10-3=72.9kJ/molECr2O

38、72-/Cr3+=1.23+(0.0592/6)lg1214; ECl2/Cl-=1.36+0.0296lg12-2=1.29VE=1.38-1.29 = 0.09V19. G = -ZFE=-2X96500X(0.222-0.151)X10-3 = -13.7 kJ/mol20. (1) E=0.236；G=-22.7 kJ/mol;K=9.32X103(2) E=0.236, G=-45.34 kJ/mol;K=8.96X107 十一.1. b6 = 1.0 X 1020, K不穩(wěn)=1.0X10-20 2. Ag+ = 9.92X10-8 mol/L; NH3=0.3 mol/L; Ag(

39、NH3)2+ = 0.1 mol/L3. 使AgCl溶解，NH3=2.25+0.2=2.45mol/L; . 使AgI溶解，NH3= + 0.2 = 3.23 X 103mol/L, 氨水不能使AgI溶解；使AgI溶解，CN- = 3.0 X 10-4 + 0.2 = 0.2 mol/L;4. (1) Ag+ = 2.85 X 10-9 mol/L; NH3 =1.12 mol/L; Ag(NH3)2+ = 0.04 mol/L(2) Ag+Cl- = 2.85 X 10-10 > Ksp, 有AgCl沉淀產(chǎn)生。（3）Ag+ < Ksp/ Cl- , Ag+ < 1.17 X

40、 10-9; NH3 > 1.42 mol/L, NH3始 1.42 + 0.08 = 1.50 mol/L, VNH3 = 1.50 X 100 ¸ 12 = 12.5 mL.5. Cu2+ = 0.05 / (2.84 X Ks) = 3.89 X 10-17 mol/L;Cu(NH3)42+ = 0.05 mol/L; NH3 = 2.8 mol/LOH- = 0.2 ´ 1 ¸ 21 = 9.52 ´ 10-3 mol/LCu2+ OH-2 = 3.89 X 10-17 X (9.52 X 10-3)2 = 3.53 X 10-21 <

41、; Ksp = 2.2 X 10-20, 不產(chǎn)生沉淀。十二。2 鹵素與水作用有以下兩種反應(yīng)：1）. 2F2 + 2H2O = 4HF + O2X2 + H2O = HXO + HX (X = Cl, Br, I)原因：是由于鹵素原子得電子能力隨原子序數(shù)的依次增大而減小，可以從它們的標(biāo)準(zhǔn)電極電位可以看出。8 （1）鍵能：HF > HCl > HBr > HI, F, Cl, Br, I得電子能力依次減弱。（2）解離能：Cl2 > Br2 > I2, 原因：它們的化學(xué)鍵能依次減小。（3）沸點(diǎn)：He < Ne < Ar < Ke < Xe < Rn, 原因：分子晶體，隨原子量的增大，分子間作用力依次增加。（4）還原性: HF <