Chapter 16 The p-block elements (Ⅲ)

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1 Chapter 16 The p-block elements (Ⅲ)
§16.1 The halogens *§16.2 The rare gases §16.3 The property variations for compounds of p-block elements

2 §16.1 The halogens 16.1.1 The halogens 16.1.2 The elemental substances
Hydrogen halides Halides and Polyhalides halogen oxyacids

3 16.1.1 The halogens Properties of the halogens： Elements(VII)FClBr I
22p52s 23p53s 24p54s 25p55s Valence electron configuration Atomic radius/pm 64 2.66 Electronegativity 3.98 Ionization energy -1 /kJ·mol Electron affinity -328z energy /kJ·mol Oxidation number -1 -1， 1， 3， 5， 7

4 relatively more energy is needed to overcome the repulsion.
幻灯片 3 z1 the radii of F is shortest, the latter coming electron will be repulled strongly by the original electrons, therefore relatively more energy is needed to overcome the repulsion. zho,

5 16.1.2 The elemental substances
1.Physical properties of the elemental substances： F2Cl2Br2 I2 g g l s State of aggregation Intermolecular force small large b.p./℃ -188 -34 59 185 m.p. /℃ -220 -102 -7 114 pale yellowred-brown yellow-green dark red Color

6 2. Chemical properties of the elemental substances
• redox properties F2 Cl2 Br2 I2 − E ( X 2 /X ) /V ： 0.5345 X2 oxidizing ability： strong weak - X reducing ability: weak strong Conclusion： Herein, F2: the strongest oxidant, -: theI strongest reductant

7 X2 oxidization ability：
Example: - +Br -+Br2Cl Cl2 2

8 • reacts with water：two types of reactions
Oxidization reaction： 2 + 2H 2 O ⎯ 4HX + O 2 X⎯→ F2(vigorous) > Cl2 (slow under light) > Br(extremely slow) I2 Disproportionation reaction(without base)： X 2 + H 2O HXO + HX Cl 2 > Br2 > I 2 K (Cl 2 ) = 4.2 × 10 −4 K (Br2 ) = 7.2 × 10 −9 K (I 2 ) = 2.0 × 10 −13 We can see that the major constituent of the solution dissolved with Cl2, Br2 and I2 is their elemental substance, respectively.氯水, 溴水, 碘水的主要成分是单质。 Presence of a base can promote dissolution and dispropor- tionation reaction 在碱存在下，促进X2在H2O中的溶解、歧化。

9 Products of disproportionation reaction under basic condition
⎯→X2 + 2OH ⎯ X + XO + H2O − − − → - 3OX - 2X + XO3 - ⎯→3X2 + 6OH ⎯ 5X + XO + 3H2O − − 3 _

10 ClOClOClO PH > 9 low room heated temperature temperature
3 − − BrO (0 C) PH > 6 − O BrO3 BrO3 Br2 − − − PH > 9 I2 IO3 IO3 IO3

11 2KHF2 ⎯⎯→2KF+ H2 + F2 MnO2 + 4HCl⎯ MnCl2 + Cl2 (g) + 2H2O⎯→
3. The preparation of the elemental substances ： •F2 (g) electrolysis： 2KHF2 ⎯⎯→2KF+ H2 + F2 电解 The formal reaction: + HF electrolysis → The real reaction 2HF F2 + H2 •Cl2 (g) Industry method (electrolysis)： 2NaCl + 2H 2 O ⎯⎯→ H 2 + Cl 2 + 2NaOH 电解 Laboratory method： MnO2 + 4HCl⎯ MnCl2 + Cl2 (g) + 2H2O⎯→ Δ (concentrated)

12 氧化电位水是绿色消毒剂 来源于水 回归于水 软化的自来水加入0.05% 的氯化钠通过电解在阳极 侧生成氯气，氯气与水反
应生成盐酸和次氯酸等物 质。另外，水也在阳极侧 电解，变成氧气和氢离 子，使阳极产生的液体pH 下降到3以下，氧化还原 电位上升到1100mV以 上，有效氯浓度达到 30~80mg/L

13 Cl ⎯→ Br2 + 2Cl + Cl 2 (适量) + 2I ⎯ I 2 + 2Cl⎯→
•Br2(l) Oxidizing agent： 2 + 2Br ⎯ Cl ⎯→ Br2 + 2Cl − − (sea water 海水) 3Br2 + 3CO Purification 2− 3 → 5Br + BrO + 3CO 2 (歧化) − − 3 − BrO 3 + 5Br + 6H ⎯ 3Br2 + 3H 2 O⎯→ − + (reverse disprotionation reaction 逆歧化) •I2 (s) Algae(海藻） is used as reaction material： Cl 2 (适量) + 2I ⎯ I 2 + 2Cl⎯→ −−+6H2O + 5Cl 2(过量) + I2 2IO3 + 10Cl +12H + − 2 MnO2 + 4H + 2IMn + I 2 + 2H2 O − − Chile saltpeter(智利硝石) as reaction material： − − 2− − 2IO3 + 5HSO ⎯ I2 + 2SO4 + 3HSO + H2O⎯→ 34

14 16.1.3 Hydrogen halides μ/ 1. Properties of hydrogen halides
Colorless gases with pungent odor at room temperature HF HCl HBr HI μ/ -30c·m)(10 3.57 2.76 1.40 molecules’ polarity 6.37 m.p./℃ * m.p.熔点 b.p./℃ * 19.52 b.p.沸点 △fHm /kJ·mol -92.3 -36.4 -26.5 Stability 稳定性 decomposition/℃>1500 -1 570 键能/kJ·mol 1000 300 432 366 298 acidity 酸性 weak 弱 strong 强

15 CaF 2+ H2SO4 (浓) ⎯ CaSO 4+ 2HF⎯→
2. The preparation of hydrogen halides • HCl Industry method： direct synthesis method Cl2 + H2 ⎯ 2HCl⎯→ hv Laboratory method：metathesis reaction NaCl + H 2SO 4 (浓 ⎯ HCl + NaHSO 4) ⎯→ Δ 2NaCl + H 2SO 4 (浓 ⎯ ⎯ ⎯ → 2HCl + Na 2SO 4 ) > 500 C O •HF metathesis reaction CaF 2+ H2SO4 (浓) ⎯ CaSO 4+ 2HF⎯→

16 hydrolysis of the halides •HBr and HI
⎧ PBr 3 + 3H 2 O ⎯ 3HBr + H 3 PO 3 ⎯→ ⎪ ⎨ ⎪ PI 3 + 3H 2 O ⎯ 3HI + H 3 PO 3⎯→ ⎩ ⎧ 2P + 3Br 2 + 6H 2 O ⎯⎯→ 6HBr + 2H 3 PO 3⎪ or simplely ⎨ ⎪ 2P + 3I 2 + 6H 2 O ⎯⎯→ 6HI + 2H 3 PO 3 Metatheticalreaction(复分解反应）, No! for(X=Br,I) KX + H 2SO 4（浓）⎯→ HX + KHSO 4⎯ 2HBr + H 2SO 4 (浓 ) ⎯ SO 2 + Br2 + 2H 2 O⎯→ 8HI + H 2 SO 4（浓）⎯→ H 2 S + 4I 2 + 4H 2 O⎯ Can HBr and HI be prepared via metatheticalreaction of other acids? 能否选用其他酸用复分解反应制备HBr和HI? YES!

17 16.1.4 Halides and Polyhalides
Formed by reaction of halogen with other elements of small electro-negativity 卤素与电负性比较小的元素生成的化合物 2.The classification of halides： Metal halides： Ionic halides：CsF, NaCl, BaCl 2 , LaCl3 − electron ⎧AgCl (18e 构型) configuration ⎪ covalent halides⎨AlCl3 , SnCl 4 , FeCl3 , TiCl 4 ⎪ (metals of high oxidation number) (高氧化值金属) ⎩ et Non-metallic halides：BF3 , SiF4 , PCl 5 , SF6 等al

18 3. The properties of halides： Property：ionic mp：high covalent low
Solubility：mostly soluble in soluble in organic watersolvent Conductibility：aqueous, melts no conductibility Non-metallic halides Metal halides Hydrolysis： Those whose hydroxides are not strong facile hydrolysis and base are easily to hydrolyze, and the The product are HX and products are hydroxides or basic salts BX3,SiX4,PCl3, respectively 对应氢氧化物不是强碱的都易水解，产物为氢 氧化物或碱式盐 Remember：Sn(OH)Cl，SbOCl，BiOCl

19 ionic bondcovalent bond AlCl3AlBr3AlI3AlF3 b.p./℃ 1272181253382 e.g.:
Variation trends of types of bonds and properties of halides 卤化物的键型及性质的递变规律： In a row(同一周期)：从左到右，阳离子电荷数增大， 离子半径减小，离子型向共价型过渡，熔沸点下降。 e.g.: NaCl MgCl2 AlCl3SiCl4 b.p./℃ (升华) 57.6 Different halides of a metal(同一金属不同卤素)：AlX3 随 着X半径的增大，极化率增大，共价成分增多。 ionic bondcovalent bond AlCl3AlBr3AlI3AlF3 b.p./℃ e.g.:

20 halides of IA: all ionic bond
随着离子半径的减小，晶格能增大，熔沸点增大 e.g.: NaF NaCl NaBr NaI m.p./℃ 996 801 755 660 different oxidation numbers of a metal： 高氧化值的卤化物共价性显著，熔沸点相对较低 SnCl4 ； SbCl3 e.g.: SnCl2 SbCl5 m.p./℃ 247 -33 73.4 3.5

21 7 +1+1+1 V.P. ==5 2 I I I I 4. Polyhalides : linear -
I 2 + KI ⎯ KI 3⎯→ I: 25P55S V.P. ==5 2 − 3 : linear I - I I I

22 16.1.5 Halogen oxyacids +7 +5 +3 HXO4 HXO3 HClO2
1.The structures of anions of different halogen oxyacids hybridization 2np5X: ns +1 VPN = 3.5 HXO 3 sp +7 HXO4 +5 HXO3 +3 HClO2 高卤酸 perhalic acid 卤酸 halic acid 亚卤酸 chlorous acid 次卤酸 hypohalous acid Structures of the corresponding halogen oxyanions electron pairs configuration Lewis structure

23 2. Hypohalous acids (次卤酸) and the related salts
hypohalous acids：HClO HBrO HIO K a ) 2.8× × ×10-11 weak acid( Strength of the acids↓ − − E (XO / X )/V 1.495 1.341 0.983 Oxidization ability↓ high low Stability： • 次氯酸很不稳定，只能存在于稀溶液中，而不能制得浓酸。 • 次卤酸见光受热均不稳定

24 Ca(ClO) 2 + CaCl 2 ⋅ Ca(OH) 2 ⋅ H 2 O + H 2 O
Important chemical reactions： 光 2HClO ⎯⎯→ O 2 + 2HCl Δ 3HClO ⎯ HClO 3 + 2HCl⎯→ Basic medium 冷 Cl 2 + NaOH ⎯⎯→ NaClO + NaCl + H 2 O 2Cl 2 + 3Ca(OH) 2 ⎯⎯→ Ca(ClO) 2 + CaCl 2 ⋅ Ca(OH) 2 ⋅ H 2 O + H 2 O bleaching powder 漂白粉

25 3. Halic acids (卤酸) and their related salts
HClO3 HBrO3 HIO3 Strength of acid: strong strong medium Strength of the acids↓ - 3 E (XO / X2 )/V 1.458 1.513 1.209 Maximum acid concentration obtained： 40% 50% crystal Stability： inferior good

26 2ClO + I 2 ⎯ 2IO + Cl 2⎯→ 5Cl 2 + I 2 + 6H 2 O ⎯ 2HIO 3 + 10HCl⎯→
Important chemical reactions： • Preparation by oxidization reaction 2ClO + I 2 ⎯ 2IO + Cl 2⎯→ - 3 - 3 5Cl 2 + I 2 + 6H 2 O ⎯ 2HIO HCl⎯→ - - • Identification of I and Br ions Br Cl2 水 Br2 (棕黄) Cl2 水 棕黄色不消失 ⎯⎯⎯ →⎯⎯⎯⎯ →⎯ - I 2 (紫红)紫红色消失 ICCl4 - 5Cl 2 + Br2 + 6H 2 O ⎯ 2HBrO3 + 10HCl⎯→ 只有在高浓度Cl2气才能实现且难度很大。

27 3ClO + Br ⎯ BrO + 3Cl⎯→ 5Cl 2 + Br2 + 6H 2 O ⎯ 2HBrO3 + 10HCl⎯→
• Oxidization ability (BrO /Br2 ) = 1.513V EA EA (Cl 2 /Cl ) = 1.36V - 3 - 5Cl 2 + Br2 + 6H 2 O ⎯ 2HBrO3 + 10HCl⎯→ High concentration of Cl2 is necessary to oxidize Br2 - Normally, we prepare BrO3 under basic condition： 3ClO + Br ⎯ BrO + 3Cl⎯→ - - - 3 - EB (ClO /Cl ) = 0.89V EB (BrO / Br ) = 0.613V - - - 3 -

28 cat. 4KClO 3 ⎯⎯ ⎯ → 3KClO 4 + KCl⎯ An important salt ：KClO3
MnO 2 2KClO 3 ⎯⎯ → 2KCl + 3O 2⎯ cat. 小火加热 4KClO 3 ⎯⎯ ⎯ → 3KClO 4 + KCl⎯ Strong oxidizing agent： explode and fire when mixed with various tinder(易燃物) An oxidizing agent in the head of safety matches cane sugar 蔗糖 A flame of the mixture of KClO3 and C12H22O11 (KClO3)

29 4. Perhalic acids and the related salts
Perhalic acids： HClO4 HBrO4 H5IO6 Strength of acid：strongest strong weak −4(Ka1 = 4.4× 10 ) − 4 − 3 EA (XO /XO )/V 1.226 All are strong oxidizing agents, pure solid available metaperiodic acid 偏高碘酸 HIO4 3d2I:sp periodic acid H5IO6

30 5H 5 IO6 + 2Mn ⎯ 2MnO 4⎯→ + 5IO3 + 7H 2 O + 11H
The important chemical reactions： 2+ 5H 5 IO6 + 2Mn - ⎯ 2MnO 4⎯→ - + 5IO3 + + 7H 2 O + 11H

31 An important salt：perchlorate salts
Most perchlorate salts are soluble, but perchlorate +、NH +、Cs+、Rb+ dissolve to a smallsalts of K 4 extent in water. KClO4 is relatively more stable than KClO3 . 610°C KClO 4 ⎯⎯ ⎯ → KCl + 2O 2⎯ Mg(ClO4)2 , Ca(ClO4)2 can be used as desiccant(干燥剂) NH4ClO4：the propellent in modern rockets

32 5. The comparison of various chlorine oxyacids
HClO HClO2 HClO3 HClO4 雨林木风1 Strength of acid：↑. Stability： unstable unstable relative stable stable 40% solution solid Θ E (oxidized form/ -)/VX 1.495 1.55 1.45 1.409 Oxidization ability ↓ ( except for HClO2)

33 Cl的氧化数增加， H-O键被Cl极化引起的变形增加，在水分子作用下H+易被解离出来
幻灯片 31 雨林木风1 Cl的氧化数增加， H-O键被Cl极化引起的变形增加，在水分子作用下H+易被解离出来 webuser,

34 The rare gases 16.2.1 The discovery of the rare gases
*§16.2 The rare gases The discovery of the rare gases Properties and uses of the rare gases The existence and recovery of the rare gases The rare gas compounds

35 Ar 16.2.1 The discovery of the rare gases Rare gases：He Ne Ar Kr Xe Rn
－：ns2np6e Valence “第三位小数的胜利” -1 g•L fractionate N2 from air：1.2572 Ar -1 ：1.2505g•L Chemical preparation of N2

36 16.2.2 Properties and uses of the rare gases
Physical properties of the rare gases： He Ne Ar Kr Xe Rn -1 I1/kJ·mol m.p./℃ S/ml/kg in H2O 8.6 108 230 Critical temp to get liquefied /K

37 16.2.4 The rare gas compounds 138940 Z1Z 2 A1 U=(1 − ) R0n
1.Synthesis：XePtF6 (red crystal) idea 思路：O2[PtF6] has been synthesized O2 ⎯ O + e⎯→ + 2 − I 1 = kJ ⋅ mol -1 Xe ⎯ Xe + e⎯→ rO + = 201pm + − I1 = kJ ⋅ mol -1 Z1Z 2 A1 U=(1 − ) R0n 2 rXe + = 210pm prediction 预测：Xe + PtF6 ⎯⎯→ Xe[PtF6 ]

38 Question：Describe the molecular shape of
2. Molecular structure VSEPR theory： 1 XeOF4 V.P. = ( ) = 6 square pyramidal 2 XeF6 V.P. = (8 + 6) = 7 distroted octahedral Question：Describe the molecular shape of XeF4 and XeF2 ？

39 §16.3 The property variations of compounds of p-block elements
The property variations for hydrides of p-block elements The property variations for oxides and the related hydroxides of p- block elements

40 16.3.1 The property variations for hydrides of p-block elements
Acidity of aqu.solt. increases Thermostability decreases CH 4 NH 3 H 2O HF Reducibility increases SiH 4 PH 3 H 2S HCl GeH 4 AsH 3 H 2Se HBr SnH 4 SbH 3 H 2Te HI Thermostability↑ 热稳定性增强 Reducibility decreases↓还原性减弱 Acidity of aque. Solt. ↑ 水溶液酸性增强 The electronegativity difference between H and non-H atoms is responsible for these.

41 16.3.2 The property variations for oxides
and the related hydrates of p-block elements Example: hydroxides or oxyacids of the elements th row with highest oxidation numberin 4 KOH Ca(OH)2 Ga(OH)3 强碱 强碱 两性 Ge(OH)4 H3AsO4 H2SeO4 HBrO4 两性偏酸 中强酸 强酸 强酸 acidity increases, basicity decreases with increasing atomic number

42 Pauling’s rule：(qualitative 定性)
The hydroxides or oxyacids can be written as： (OH)mROn m: the number of oxygen in hydroxy group n:the number of oxygen in non-hydroxy group example：HClO4 or HOClO3 m=1，n=3 酸性的强弱取决于羟基氢的释放难易，而 羟基氢的释放又取决于羟基氧的电子密度。 若羟基氧的电子密度小,易释放氢,酸性强。

43 氧电子密度小，酸性强；非羟基氧的数目多， 可使羟基氧上的电子密度小，酸性强。例如： H4SiO4 H3PO4 H2SO4 HClO4 半径
中心原子R的 电负性、半径、氧化值 (OH)mROn 羟基氧的 电子密度取决于 非羟基氧的数目 若 R 的电负性大、半径小、氧化值高则羟基 氧电子密度小，酸性强；非羟基氧的数目多， 可使羟基氧上的电子密度小，酸性强。例如： H4SiO4 H3PO4 H2SO4 HClO4 R电负性 半径 氧化值 非羟基氧 0123 酸性

44 HClO ＜ HClO2 ＜ HClO3＜HClO4 acidity n(非羟基氧) 1 2 3 ＞ acidity HClO4 HNO3 Electronegativity 3.16 3.04 n(非羟基氧) 3 2 H2S2O7 ＞ H2SO4 acidity n(非羟基氧) 2.5 2 Acidity ↑ when polymerization degree ↑ 缩和程度愈大，酸性愈强。

45 Pauling’s rule(半定量)： n=0 弱酸 ≤10 (K a HClO, HBrO n=1 中强酸 ( K a
-5) ≤10 (K a HClO, HBrO n=1 中强酸 ( K a H2SO3,HNO2 -1~103) n=2 强酸 ( K a =10 H2SO4,HNO3 -4~10-2)=10 n=3 特强酸 ( K a 3)>10 HClO4

46 中文教材 pp: 548—550: 1, 4, 5, 15 本章作业(exercises) Textbook in English: