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Supp. Notes – Hybridization and Resonance

Supp. Notes – Hybridization and Resonance . 雜化及共振 . A. Hybridization 雜化 l When covalent bond is formed, the atomic orbitals of the bonded atoms _________ with each other. 形成共價鍵時,鍵合原子的原子軌態會彼此 ________ 。 l E.g. Hydrogen 氫 . overlap. 重疊。 . H-H. H-H. stronger.

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Supp. Notes – Hybridization and Resonance

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  1. Supp. Notes – Hybridization and Resonance 雜化及共振

  2. A.Hybridization雜化 lWhen covalent bond is formed, the atomic orbitals of the bonded atoms _________ with each other. 形成共價鍵時,鍵合原子的原子軌態會彼此________。 lE.g. Hydrogen 氫 overlap 重疊。 H-H

  3. H-H stronger lThe greater the degree of overlap, the ___________ is the bond. 重疊幅度越高,共價鍵就__________。 lE.g. Hydrogen 越强

  4. E.g. Hydrogen fluoride氟化氫 F -H F:

  5. 1s 2s 2p ground state 1.Molecular shape of methane lIn the ground state of carbon atom, there is only __________________ electrons, hence carbon is expected to form _________. two unpaired CH2 1.甲烷的分子形狀 l碳原子的基態中,只有____________的電子,故此預計碳形成________ 兩顆不成對 CH2

  6. 1s 1s 2s 2s 2p 2p excited state 激發態 ground state 基態 But we find that only __________ is formed, hence the carbon atom should undergo an ____________ of electrons. CH4 excitation CH4 但我們發現形成的只有_______,故此碳原子應該出現電子的_______。 激發

  7. From the different shapes of s- and p- orbitals, we would expect different bondings be formed. But from accurate experiments, the four C-H bonds are _______________. identical 從s-軌態及p-軌態的不同形狀,我們預計不同的共價鍵會形成。但從精確的實驗顯示,四個C-H鍵均是________的 一致

  8. This led chemists to the conclusion that the central carbon atom must have ____________________, each overlaps with the 1s orbital of a hydrogen atom. 4 identical orbitals 這引導化學家推論出中心碳原子必定擁有______________,每個均與氫原子的1s軌態重疊。 四個同樣的軌態

  9. hybridization The concept of ______________ was thus developed. 雜化 於是建立了_________的概念。

  10. Hybridization is the process in which different atomic orbitals are ____________________ to form a set of ______ hybrid orbitals. mixed (hybridized) new 不同原子軌態__________而形成一系列新的雜化軌態,這過程稱為________ 混合(雜化) 雜化

  11. In methane, the 2s and 2p orbitals of carbon atom mix together to give 4 new ____________ hybrid orbitals, each sp3 orbital overlaps with 1s orbital of a hydrogen atom and forms the methane molecule. equivalent 在甲烷,中碳原子中的2s及2p軌態混合產生四個_______雜化軌態,每個sp3軌態與氫原子的1s軌態重疊形成甲烷分子。 對等

  12. lNote: a)The hybrid orbitals are all _______ except for their __________. identical directions l要點: a) 每個雜化軌態除了__________外其他都是_________的。 方向不同 一樣

  13. b)Only atomic orbitals of _______________ can be hybridized. similar energies b)只有_________的原子軌態才能產生雜化。 . 相若能量

  14. c)Hybridization requires energy, but this energy is then __________ after the formation of bonds. compensated c)雜化需要能量,但這能量可以通過形成化學鍵而_________。 抵消

  15. ground state基態 excited state 激發態 sp3 hybridization sp3雜化 2. Different types of hybridization不同類型的雜化 a)sp3 hybridization: sp3雜化 e.g. CH4 C:

  16. ground state基態 excited state 激發態 sp2 hybridization sp2雜化 b) sp2 hybridization: sp2雜化: e.g. BCl3 B:

  17. ground state 基態 excited state 激發態 sp hybridization sp雜化 c) sp hybridization: sp雜化: e.g. BeF2 Be:

  18. The kind of hybridization employed depends on the information observed from the molecule. 從觀察分子而得的資料,可以從推斷雜化的類型。

  19. excited state激發態 3. Formation of multiple bonds a) Double bond – e.g. ethene To form bonds with 3 atoms, each carbon atom forms a set of 3 equivalent _______ hybrid orbitals. sp2 3.複鍵的形成 a)雙鍵-e.g.乙烯 sp2 要與三原子鍵合,每個碳原子要形成3個對等____雜化軌態。 C2H4

  20. C2H4 excited state激發態

  21. C2H4 excited state激發態

  22. C2H4 excited state激發態

  23. Hence, the C=C bond actually consists of 2 different types of bonds, ______________ and _______________. pi (π ) bond sigma (σ ) bond σ鍵 π鍵 故此,C=C鍵實際由兩種不同的鍵,____及____所組成。

  24. lsigma bond – formed by ___________ overlap of orbitals - electron density is concentrated __________ the line joining the bonded nuclei. head-on along 頭與頭 lσ鍵 - 由________重疊軌能而成。 - 電子密度集中在鍵合原子核_________。 之間

  25. lpi bond – formed by ____________ overlap of orbitals - electron density is concentrated ____________________ the line joining the bonded nuclei. sideway above and below 側向重疊 llπ鍵 – 由____________軌態而成。 - 電子密度集中於鍵合原子核_____________。 之上及之下

  26. The C-C bond energy is +346 kJmol-1 and C=C bond energy is +598kJmol-1, this reflects that a πbond is __________ than a σbond, because πbond is formed by an ineffective _________ overlap of orbitals whileσbond is formed by _____________ overlap of orbitals. weaker sideway head-on C-C鍵能是+346kJmol-1而C=C鍵能是+598kJmol-1,這反映出π鍵比σ鍵為_____,因為π鍵是由效能較差的_________重疊軌態而成;但σ鍵是由_______重疊軌態而成。 . 弱 側向重疊 頭與頭

  27. Although the strength ofπbond is weaker than aσbond, the πbond is strong enough to __________________________ about the line joining the bonded atoms. prevent free bond rotation 防止化學鍵自由旋轉 雖然π鍵較σ鍵弱,但足以________________________。

  28. excited state 激發態 b) Trible bond – e.g. ethyne To form bonds with 2 atoms, each carbon atom forms a set of 2 equivalent _______ hybrid orbitals. sp b)三鍵-e.g.乙炔 要與兩顆原子鍵合,每個碳原子需要形成兩個對等______雜化軌態。 sp C2H2

  29. C2H2 excited state 激發態

  30. C2H2 excited state 激發態

  31. C2H2 excited state 激發態 Hence, the CΞC bond consists of _____ and _____bonds. 1 σ 2 π 1 σ 2 π 故此,C≣C鍵由______及_______鍵組成。

  32. B. Resonance structures 共振結構 Note the structure of the compound benzene: C6H6 留意化合物苯的結構:C6H6

  33. lThis Kekule’s structure had been well accepted by chemists at early times because it agreed with some experimental observations: 1.A ______________ compound. 2. _______________. cyclic planar l在早期凱庫勒結構為化學家接受因為它與好些實驗觀察吻合: 1.  ________化合物 2. ________ 環狀 平面的

  34. However, later many serious disagreements were discovered: Three bonds are expected to be 0.154nm (C-C) and the other three are 0.134nm (C=C); however, the real picture is that the six carbon-carbon bonds are all ________________ with length 0.140nm, i.e. __________________ between a single and a double bond. identical intermediate 然而,好些嚴重的不吻合被發現了: 預計三個鍵為0.154nm(C-C),而其餘三個為0.134nm(C=C);然而真實圖像是6個碳-碳鍵均是_______的鍵長為0.140______單鍵與雙鍵間。 一致 介乎

  35. The ‘unsaturated’ structure implies a high reactivity; however, benzene is ______________. not so reactive 並非很活潑 該不飽和結構引致-高活潑性;然而,苯_______________。

  36. 3. The thermochemical data are different: e.g. The enthalpy of hydrogenation of gaseous cyclohexene: 3.熱化學數據不同: e.g.氣態環己烯的氫化焓:

  37. 我們預計氣態苯的氫化焓= We would expect the enthalpy of hydrogenation of gaseous benzene = -120 x 3 = -360 kJ mol-1 From experiment, the enthalpy of this reaction is ______________ only. -208 kJ mol-1 -208 kJ mol-1 從實驗,這反應的焓只有_____________。

  38. Energy 能量 + 3 H2 ? + 3 H2 -360 kJ mol-1 -208 kJ mol-1

  39. e.g. From the following data, calculate the enthalpy of formation of Kekule’s benzene: e.g. 從以下數據,計算凱庫勒苯的生成焓: Bond鍵 Bond enthalpy 鍵焓 (kJmol-1) C-H +413 C-C +348 C=C +612 Reaction反應ΔH (kJmol-1) C(graphite) C(g) +715 ½H2(g) H(g) +218 C6H6(l) C6H6(g) + 31

  40. Δ Hf 6 C(graphite) + 3 H2(g) C6H6(l) 6 C(g) + 6 H(g) C6H6(g) Δ Hf = 715x6 + 218x6 -413x6 -348x3 -612x3 -31 = 209 kJ mol-1

  41. ? From experiment, the enthalpy of formation of benzene is ______. +49 kJmol-1 從實驗,苯的生成焓是_________。 +49 kJmol-1 Energy 能量 209 kJ mol-1 +49 kJmol-1 6 C(graphite) + 3 H2(g) As a result of such discrepancy, chemists are forced to re-examine the structure of benzene. 因着這些差異,化學家必須重新檢示苯的結構。

  42. Resonance structure of benzene: There are two Kekule structures of benzene can be drawn: • 苯的共振結構: 苯的兩個凱庫勒結構: canonical forms: 正則形態:

  43. canonical forms: 正則形態: between The true benzene structure must lie somewhere _________ this two ______________ forms. This is called __________________. canonical resonance hybrid 介乎 正則形態 苯的真正結構應________這兩個___________之間,稱之為____________。 共振雜化體

  44. energy 209 kJ mol-1 ? +49 kJmol-1 6 C(graphite) + 3 H2(g) The resonance structure is energetically __________ than any one of the canonical forms. The difference is called _______________. more stable resonance energy 這共振結構比起任何正則形態,在能學上均____________。這差異稱之為___________。 較穩定 共振能 Resonance energy 共振能 = 160 kJmol-1

  45. lIn terms of orbital overlap, the true benzene structure is pictured as follows: l從軌態重疊看,苯的真正結構如下:

  46. Each carbon atom from __________ bonds with neighbouring carbon atoms and hydrogen atoms (sp2 hybridization), the remaining p orbitals of the six carbon atoms overlap in a side-way manner to form a ___________________________ ___. three delocalized π–electron cloud 三 每個碳原子與相鄰的碳原子及氫原子形成_____個鍵(sp2雜化),六個碳原子餘下的p軌態彼此側向重疊形成_________________。 離域π-電子雲

  47. It is the delocalization of the p-electrons that gives the _______________ of the molecule and gives rise to the ___________________. stability resonance energy 因着p-電子的離域作用,使到分子產生________並帶來___________。 穩定性 共振能

  48. 2. Other aromatic compounds: 其他芳香族化合物: Benzene, C6H6 苯

  49. Naphthalene, C10H8

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