1 / 65

Boron (p102)

Boron (p102). Introduction (p102). Valence electron configuration, n s 2 n p 1 – so to complete octet, the chemistry of boron involves one of the following means (you can suggest): +3 oxidation state is expected to be most stable, yes or no ?

chaman
Télécharger la présentation

Boron (p102)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Boron (p102)

  2. Introduction (p102) • Valence electron configuration, ns2 np1 – so to complete octet, the chemistry of boron involves one of the following means (you can suggest): • +3 oxidation state is expected to be most stable, yes or no ? The ionization enthalpies are so high that the lattice energies or hydration enthalpies cannot offset the energy, Note: boron is a nonmetal element

  3. Introduction (p102) 2. –5 oxidation state, yes or no ? Comparing the electronegativities of carbon and boron: 2.5 and 2.0, you can understand that there are no borides similar as carbides formed with the most electropositive metals. 3. Form three covalent by sp2 hybrid orbitals, which are not octet. These compounds such as BX3 act as strong Lewis acids.

  4. Introduction (p102) • Boron can form many stable compounds, of which there are the problem of electronic deficiency, according to the conventional 2c–2e bonds. • Note: the special chemical bonding in some compounds of boron.

  5. Boron resembles silicon more than aluminum in some ways: • B2O3 and B(OH)3 are acidic rather like SiO2 and Si(OH)4 whereas the Al compounds are weakly amphoteric. • The borates have some features in common with the silicates. • The halide compounds of B and Si are readily hydrolyzed (except BF3) whereas the halide compounds of Al are only partly hydrolyzed. • The B and silicon hydrides are volatile molecular compounds, which inflame in air, while AlH3 is an involatile solid.

  6. Introduction • To understand the chemistry of boron, you should note three aspects: • Form normal covalent bond, such as BF3 and BF4– • Resemblances to silicon • Distinguished bonding and structures

  7. Sources • Ulexite硼钠钙石: NaCa[B5O6(OH)6].5H2O • Borax硼砂: Na2[B4O5(OH)4].8H2O • Colmanite: Ca2[B3O4(OH)3].2H2O • Kenite: Na2[B4O5(OH)4].2H2O • Preparation: reduction of the oxide, halides

  8. Structures of Elements • Boron would like to be covalent, but it ‘wants’ to make “5” bonds to achieve a Lewis octet. • It only has 4 orbitals though (1x2s+3x2p) • B gets around its problem by forming delocalised polyhedra. • Very important in B chemistry.

  9. Oxygen Compounds of Boron (p103)

  10. Monoborate BO3 BO4

  11. Triborate No BO4, hydrate rapidly; stable in solid

  12. Tetraborate Both BO3 and BO4 are present, N(BO4) = Z

  13. Boric Acid and Borate ions in solution (p106) • B(OH)3: • moderately soluble in water • Act as a weak Lewis acid to OH– giving B(OH)4– and H+ • B(OH)4– occurs in minerals. If formed by fusion, borates have more complex structure. • Form very stable complexes with 1,2-diols

  14. Halides (p107) • the boron halides are highly reactive • They are volatile, covalent compounds 2B + 3X2 2BX3 X = F, Cl, Br, I • BF3 and BCl3 (gases); BBr3 (liquid); BI3 (low • melting point solid) • BX3 acts as Lewis acids

  15. Boron halides are highly reactive, volatile compounds. The gas BF3 acts as a Lewis acid with ammonia. Vacant P orbital

  16. General Properties • The compounds which are coordinatively unsaturated (e.g.BCl3) are very strong Lewis acides. • Tetrahedral adducts and anions are common, for example: BF3:O(C2H5)2      BF4-      B(C6H5)4-

  17. Boron Subhalides • BX3 are the most common halides. • Extensive sub-halide chemistry though, e.g. B4Cl4 and B8F12. • B10F12 structurally characterised for the first time last year!

  18. 10B absorbs neutrons strongly. 40 tonnes of boron carbide were dropped on Chernobyl. Boron carbides also used in light-weight protective armour and bullet-proof clothing. Metal borides can be prepared by direct combination of elements: Ca + 6B CaB6 Metal-rich borides (e.g. TiB2) can have higher conductivities than parent metals. Also much more inert and harder. B-Rich borides contain B-clusters Metal Borides

  19. Hydrides (p113) • boron compounds of hydrogen are • called boranes • Molecular, volatile compounds; BnHm • Simplest borane – diborane, B2H6

  20. C has 4 valence e, H has 1, so C2H6 has enough electrons (8+6) for 7 2c2e bonds. B2H6 only has 6+6=12 electrons. This makes an ethane-like structure impossible Boron Hydrides • These form one of the most structurally diverse series of compounds. • Simplest is diborane, B2H6. • Similar formula to ethane, but structurally very different because it is electron deficient. • Gets around the problem by forming delocalised bonds.

  21. BH3 • B2H6 is a dimer of boron trihydride. • This is a fugitive species, present in low concentration in diborane at high T. • Important in mechanisms of reactions of B2H6 at high T.

  22. Boranes have unusual structures and bonding • Notice that there are two types of bonds in • boranes • :- the regular two-center 2e- bond • :- a three-center 2e- bond

  23. Bonding in Diborane • The B-H-B unit is held together by 2e. • This is called a 3 centre - 2 electron bond (3c2e). • The orbital basis can be made up of two sp3 hybrids of the B atoms and two H(1s) orbitals. • The remaining boron orbitals form normal 2c2e bonds to the terminal H’s. H B B H

  24. 3c2e Bonds in Diborane • The two electrons occupy the fully bonding combination, so that the overall bond order between the B and the bridging H is 1/2.

  25. 3c2e Bonds • 3c2e bonds are occasionally shown in structural diagrams like this: • Bond Energies: BH 381 kJ/mol BHB 441 kJ/mol 1.19 Å 1.32 Å

  26. Electron Deficiency • All boranes are electron deficient. • The need to form 3c2e bonds (BHB and BBB) causes the molecules to ‘curl-in’ on themselves. • The more electron deficient the more ‘spherical’ a molecule becomes. • For example [B6H6]2- is more electron deficient than B4H10

  27. Structure-bonding elements (p116) • Terminal (2c–2e) boron-hydrogen bond B––H • (3c–2e) hydrogen bridge bond B B • (2c–2e) boron-boron bond B––B • Open (3c–2e) B-B-B bond B B • Closed (3c–2e) boron bond B B B H B

  28. Lipscomb’s semitopological sheme (p117) s = 0 t = 4 y = 4 x = 2

  29. Electron Counting: Wade’s Rules (p129) • Framework electrons: the number of electrons that are available in the compound for bonding within the polyhedral framework • Framework electrons F F = 3b + 4c + h + x -2n • Suitable for triangulated, regular polyhedron

  30. Electron Counting • Just how electron deficient a borane is can be derived by counting the number of skeletal pairs of electrons. • Each HB has 4 valence electrons. One pairs used for a 2c2e bond (e.g a terminal BH). • The remaining 2e are used for delocalized cluster bonding. • Any remaining H contribute 1e to the cluster

  31. Just how electron deficient a borane is can be derived by counting the number of skeletal pairs of electrons. Each HB has 4 valence electrons. One pairs used for a 2c2e bond (e.g a terminal BH). The remaining 2e are used for delocalized cluster bonding. Any remaining H contribute 1e to the cluster [B6H6]2- write as (BH)62- Each BH unit contributes 2e Plus the 2- charge gives 14 electrons 6 boron atoms in the cluster bonded with 7 pairs (6+1). Electron Counting

  32. Just how electron deficient a borane is can be derived by counting the number of skeletal pairs of electrons. Each HB has 4 valence electrons. One pairs used for a 2c2e bond (e.g a terminal BH). The remaining 2e are used for delocalized cluster bonding. Any remaining H contribute 1e to the cluster B4H10: Write as (BH)4H6 Each BH => 2e (8e in all) Each additional H gives 1e (6e in all) Total number of electrons = 14 4 Borons in cluster bonded by 7 pairs of electrons (4+3). Electron Counting

  33. Just how electron deficient a borane is can be derived by counting the number of skeletal pairs of electrons. Each HB has 4 valence electrons. One pairs used for a 2c2e bond (e.g a terminal BH). The remaining 2e are used for delocalised. cluster bonding. Any remaining H contribute 1e to the cluster B5H9: (BH)5H4 10 + 4 = 14 electrons 5 Boron atoms bonded by 7 electron pairs (5+2). In terms of electron deficiency B6H62- > B5H9 > B4H10 All have 7 e pairs for skeletal bonding (ie cluster bonding). Electron Counting

  34. closo B7H72- nido B6H10 cage nest 闭式 巢式 arachno B5H11 网式 spider’s web

  35. carborane 碳硼烷 Carboranes or carbaboranes are compounds having as the basic structural unit a number of C and B atoms arranged on the vertices of a triangulated polyhedron. Their structure are closely related to those of the isoelectronic boranes.

  36. Wade’s Rules 5+2 n+2 Nido(鸟)巢状 4+3 n+3 Arachno蛛(网)状 6+1 n+1 Closo闭合型(笼型) nest cage spider’s web

  37. Wade’s Rules: Example 1 • B6H10 • (BH)6H4 • 12 + 4 = 16e = 8 pairs • 8 pairs = 6B + 2 • Nido cluster • Remove one vertex from 7-vertex polyhedron.

  38. Wade’s Rules: Example 2 • B5H11 • (BH)5H6 • 10 + 6 = 16 = 8 pairs • 5 B atoms, 8 pairs • n+3 arachno cluster • based on seven vertex polyhedraon via removal of two vertices.

  39. Wade’s Rules: Example 3 • Heteroatoms: • B10C2H12 • BH contribute 2e • CH contribute 3e • (BH)10(CH)2 • 20 + 6 = 26 e • 12 atoms in cluster • 13 pairs • Closo 12-vertex polyhedron

  40. Wade’s Rules: Example 4 • [Sn9]4- • Each Sn has a lone pair and contributes 2e to cluster bonding, • 18 + 4 = 22 e • 9 atoms, 11 pairs • Nido cluster, remove 1 vertex from 10 vertex polyhedron. Bi-capped square anti- prism

  41. Wade’s Rules: Example 5 • [Pb5]2- • Pb has 1 lone pair • 2e/Pb for cluster bonding • 10 + 2 = 12e • 5 Pb, 6 pairs • Closo structure • Mingos developed Wade’s rules for use in transition metal clusters

  42. Synthesis of Boranes: Diborane • Hf = +80 kJ/mol, so direct combination of B and H is not possible. • 2NaBH4 + I2  B2H6 + 2NaI + H2 • 2NaBH4 + 2H3PO4  B2H6 + 2NaH2PO4 + 2H2 • 4BF3 + 3LiAlH4  2B2H6 + 3LiAlF4 • Air and moisture must be rigorously excluded: diborane is highly pyrophoric! • Boranes burn with a characteristic green flash (decay of excited state of BO)

  43. Higher Boranes • Made by controlled pyrolysis of B2H6 • Highly specific and not at all predictable (cf disconnection approach in organic chemistry). 80°C/200 atm/5hr B4H10 B2H6 160-200°C slow hot tube pyrolysis H2/200-240°C/rapid hot tube pyrolysis B10H14 B5H9

  44. Mechanism of Pyrolysis • Key step is dissociation of B2H6 into highly reactive BH3: • For example, synthesis of B4H10: B2H6 2BH3 B2H6 + BH3  B3H7 + H2 BH3 + B3H7  B4H10

  45. Handling Boranes • Boranes are TOXIC and combust EXPLOSIVELY in air. • Lots of shielding, gloves & face protection are necessary. • Only use small quantities, so if there is an accident it can be contained. • Use vacuum techniques for handling and transferring.

  46. Typical Reactions 1: Lewis Base Cleavage • Boranes are electron deficient. • Lewis bases add electrons • Small boranes may cleave: NMe3

More Related