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无机化学. 第17章 硼族元素 Chapter 17 Boron family elements. 返回. 本章重点要求掌握本族元素单质、氢化物、氧化物的结构与性质,硼酸盐的结构特点;本族元素的缺电子性及对化合物性质的影响; 硼烷结构中五大成键要素,分析硼烷结构. 基本内容和重点要求. 硼族元素通性 单质的结构及重要性质,硼及 B12 的正二十面体结构,硼的 制取方法。 硼烷的结构与性质,硼的五大成键要素,硼烷结构分析。 硼族元素的重要化合物性质,硼的氧化物,硼酸及硼酸盐,铝、镓、铟、铊的氧化物,卤化物。. 返回. 17.1 硼族元素通性.

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4134027

17

Chapter 17 Boron family elements


4134027

  • B12


4134027

17.1

IIIA: B Al Ga In Tl

B

Al

Ga

In

Tl

2s22p1

+1

+3

+3

I1

EA1

3s23p1

4s24p1

5s25p1

6s26p1


4134027

17.2

2.1

1

B I1I2I3

B sp2BX3B(OH)3

sp3BF4- BH4- B(OH)4-

2Bns2np1

4 2s 2px 2py 2pz

3 2s2 2p1

BX3 , B(OH)3

BF3 + F- = BF4-BF3Lewis

+3, ns2

Tl: +1


4134027

17.2

3

B-B-B

B-B-BB-H-B 3c - 2e (3c-2e bond)

4BFO

/kJmol-1 B-O 561~690 Si-O 452

B-F 613 Si-F 565

5BSi

i Be B C

\ \ \ r* * * / r

Na Mg Al Si


4134027

17.2

  • 2.2

  • 1

  • -

    • -

  • B12

B1212B

3 12 = 36

B12B5B5512/2 = 30


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17.2

B12

n 12

n+1 13

2n+2 26

10 + 26 = 36 B12


4134027

-

17.2

B12

6B127121046B1263c-2e203 nm62/3e = 4e

38956113B3B3B6B-B 2C-2e171 nm62/2e = 6e

B124e + 6e = 10 e


4134027

17.2

2.3

1N2,O2,S,X2

R.T.

2B(s) + 3F2(g) 2BF3 (BF)

973K

4B(s) + 3O2(g) 2B2O3

rH298 = - 2887 kJmol-1

rG298= - 2368 kJmol-1

B-O Si-O C-O

/kJmol-1 560-690 > 452 > 358

(BO) B

2B(s) + N2(g) 2BN

B-N


4134027

17.2

2BH2SO4HNO3

2B(s) + 3H2SO4() 2H3BO3 + 3SO2(g)

B(s) + 3HNO3() H3BO3 + 3NO2(g)

3

2B + 2KOH + 3KNO3 3KNO2 + 2KBO2 + H2O

+3

  • 2.4

  • - 12 BI3 B12(C) + 18 I2(g)

  • Na2B4O710H2O+2HCl=4H3BO3+2NaCl+5H2O

  • 2H3BO3 = B2O3 +3 H2O (800K)

  • B2O3 + Mg = 3MgO + 2B (800K)

800~1100


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17.2

2.5

1

BnHn+4

BnHn+6

2

BnHn+4 B2H6B5H9-9B16H20

BnHn+6 B5H11-11

B1~B10

B11

H

B5H9-9B5H11-11

20


4134027

17.2

  • 3

  • 2BCl3 + 6H2 = B2H6 + 6HCl

  • 2BMn+ 6H+ = B2H6 + Mn3+

  • 3NaBH4 + 4BF3 = 2B2H6 + 3NaBF4

  • 4

  • 1960Lipscomb

    • B2H6

  • 32 + 16 = 12

2BH3B6e


4134027

17.2

B2H6

B -H 119pm

B -H -B 3c-2e H

B2H63c-2e bond (3 center-2 electorn bond)


4134027

17.2

  • Lipscomb

  • 1960William N. Lipscomb Harvard University1976 Nobel Prize in Chemistry.

5


4134027

17.2

B5H9

145 + 19 = 29

5B 9H

2 35 + 19 = 24

5B 9H

5B-H 2c-2e: 2e5=10e

4B-H-B 3c-2e: 2e4=8e

1

3c-2e: 2e1=2e

2B-B 2c-2e2e2=4e

24e, 12


4134027

17.2

(5)

,10-6ppm

COCl2 1

HCN 10

B2H6 0.1

B2H6+ O2 = B2O3+ 3H2O

B2H6+ 6H2O = H3BO3+ 6H2

B2H6+ 6Cl2 = 2BCl3+ 6HCl

()

B2H6+2CO =2[H3BCO]

B2H6+2NH3 =2[BH2(BH2(NH3)2]++[BH4]-


4134027

M

O

B

O

+

2

2

3(s)

M

O

2

3

17.2

2.6

1

B2O3( B2O3rH= - 19.2 kJmol-1

  • 1.

  • B2O3(s) + 2 H2O(l) = 2 H3BO3 rH= - 76.6 kJmol-1

  • B2O3(s) + 2 H2O(l) = 2 H3BO2

  • 2.

Mn+


4134027

17.2

3. BN

B2O3(s) + 2NH3(s) 2BN(s) + 3H2O(g)

rG= +74.78 kJmol-1

rS= +158.7 Jmol-1K-1

T > 733K,

(BN)x(CC)x

(BN)xB-N (BN)x

(BN)x m.p.3000


4134027

17.2

2

T

B(OH)3 HBO2 H2B4O7 B2O3

500

120

140-160

B(OH)3

P78416-11

B sp2

B(OH)3

R.T.T

T/ 25 50 100

S/g/100gH2O 5.44 10.24 27.53


4134027

17.2

B(OH)3

  • Lewis

  • B(OH)3H2O = B(OH)4 + H3O+ Ka = 5.810-10

    H+H2OOH-H+

Ka =10-6

  • B(OH)3+ 3ROH==== B(OR)3 + 3H2O


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17.2

  • H2B4O7 H3BO3

  • Ka =1.510-7 > Ka =5.810-10

  • Pauling XOm (OH)n

  • + H+ H3BO3

3

BO33- BO45-

[B4O5(OH)4]2-

Na2B4O710H2O

NaBO2

Mg2B2O5 H2O

Na2B4O5(OH)4 8H2O


4134027

17.2

  • [B4O5(OH)4]2-

  • 2B sp2 BO3

  • 2B sp3BO4

  • [B4O5(OH)4]2-

[B4O5(OH)4]2-

  • 1 ()

  • [B4O5(OH)4]2- + 5H2O = 2H3BO3 + 2B(OH)4-

  • +OH- +H+

  • 1:1

H+OH-pH 20 pH=9.24


4134027

17.2

2. (BN)x

Na2B4O710 H2O + 2 NH4Cl

= 2NaCl +B2O3(g) + 4H2O+ 2BN

3

B2O3B(OH)3

Na2B4O7+CoO = Co(BO2)22NaBO2

3Na2B4O7+Cr2O3 = 2Cr(BO2)36NaBO2

Cu(BO2)2

CuBO2

Fe(BO2)2

Fe(BO2)3

Ni(BO2)2

MnO22B2O3


4134027

17.2

  • 4

  • Mg2B2O5H2O + NaOH = 2Mg(OH)2 + 2NaBO2

  • CO2pH

  • 4NaBO2 + CO2 + 10H2O = Na2B4O7 10H2O + Na2 CO3

  • 2.7 BX3X = FClBrI

  • Lewis

  • BF3 BCl3 BBr3 BI3

  • 3 + 1 46

  • /pm B-F 132 (B-F150)

  • B-X 613.3 456 377 263.6

  • BCl3BBr346BI346


4134027

17.2

1Lewis

BX3Lewis

BF3 + :NH3 = F3BNH3

BF3 + HF = HBF4

H2SiF6

BX3 + X- = BX4-

sp2sp3

2 Lewis BF3 < BCl3 < BBr3 > BI3

3 Lewis

BF3BCl3AlCl3GaCl3Friedel-Craft


4134027

17.3

3.1

Al Ga In Tl

3s23p1 3d104s24p14d105s25p1(4f105d10)6s26p1

6s2

+3 +3+1 +3+1 +1

IVAVA

3.2

1

i Be B C

\ \ \ r* * * / r

Na Mg Al Si


4134027

17.3

2

4 Al(s) + 3O2(g) 2Al2O3(s)

rH298= - 3356 kJmol-1

2Al + 3X2 2 AlX3

2Al + N2 2AlN

H2

2Al + 6H+ = 2Al3+ + 3H2

2Al + 2OH- + 6H2O = 2Al(OH)4 - + 3H2

Al Zn

Al HNO3H2SO4


4134027

17.3

3.3

1

AlF3 AlCl3 AlBr3 AlI3

  • AlCl3(s)

  • Al+3Cl2(g)= AlCl3

  • Al+3HCl (g) = AlCl3+H2g)

  • Al2O3+3C + 3X2 = 2AlX3 +CO

  • 2Alsp3

  • 2Cl

23c-4e

AlCl3: Al2Cl6


4134027

17.3

(2)

Al2O3

-Al2O3

-Al2O3

Al(OH)3


4134027

17.3

3.3

2Al + 3I2 2AlI3

Ga + I2

1 B < Al > Ga > In > Tl

3s23p13d104s24p1

M /pm Al (143.2) > Ga (122.1)

Z*, r* = Z* / r Al<Ga

M3+/pm Al (51) < Ga ( 62)

Ga+In+ Ga3+In3+

2Ga(OH)3Al(OH)3

3Tl(III)

Al 3 Ga 3 In 3 Tl 3

TlCl3 TlCl + Cl2


4134027

17.3

TlBr3

Tl(III) I3 Tl(I) I3

TlCl TlBr TlI

TlF

GaCl2InCl2 GaCl2Ga(I) [Ga(III)Cl4]

InCl2In(I)[In(III)Cl4]


4134027

1 Mg2B2O5H2OCO2

Mg2B2O5H2O+2 NaOH=2 Mg(OH)2+2 NaBO2

2 NaBO2+CO2+10H2O=Na2B4O710 H2O+ Na2CO3

2

B4O5(OH)42- + 5 H2O = 2H3BO3 + 2B(OH)4-

H3BO3B(OH)4-Na2[H4B4O9] 8 H2O


4134027

17.4

4.1

s2p0-6s+(n-2)+n

s

Tl() 6s2Tl()6s0

4.2

Li Be BC N O F

NaMg Al SiP S Cl


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