Transcript 习题课2
第六、七章复习课 Inorganic Chemistry 2012 内容回顾： Chapter 6 Presenting Molecular 6.1 The Octet rule Lewis Structure ( valence electron covalent bonding shared electron pairs lone pairs ) 6.2 Electronegativity and Polarity 6.3 Drawing Lewis Structures Follow these steps when drawing Lewis structure for molecules and polyatomic ions. 1)Draw the skeletal structure of the compound. The least electronegative atom is usually the central atom. Draw a single covalent bond between the central atom and each of the surrounding atoms. 2) Count the total number of valence electrons present; add electrons for negative charges and subtract electrons for positive charges. 3) For each bond in the skeletal structure, subtract two electrons from the total valence electrons. 4)Use the remaining electrons to complete octets of the terminal atoms by placing pairs of electrons on each atom. Complete the octets of the most electronegative atom first. 5)Place any remaining electrons in pairs on the central atom. 6)If the central atom has fewer than eight electrons, move one or more pairs from the terminal atoms to form multiple bonds between the central atom and terminal atoms. 6.4 Formulary Charge： Formal charge = valence electrons – associated electrons BF3 BI3 Chapter 7 Bonding Theory 7.1 Molecular Geometry: valence-shell electron-pair repulsion (VSEPR) model Electron-Domain Geometry and Molecular Geometry The steps to determine the electron-domain and molecular geometries are as follows: Step 1: Draw the Lewis structure of the molecule or polyatomic ion. Step 2: Count the number of electron domains on the central atom. Step 3: Determine the electron-domain geometry by applying the VSEPR model. Step 4: Determine the molecular geometry by considering the positions of the atoms only. No. of e in bonding orbitals No. of e in antibonding orbitals bond order = 2 HF CO C O 1. What is the molecular shape of the thiocyanate anion, SCN–, as predicted by the VSEPR theory? (Carbon is the central atom.) A) linear B) bent C) angular D) trigonal E) none of these choices is correct Ans: A 2. What is the molecular shape of ClO3F as predicted by the VSEPR theory? A) trigonal pyramidal B) square planar C) square pyramidal Ans: D D) tetrahedral E) octahedral 3. For which one of the following molecules is the indicated type of hybridization not appropriate for the central atom? A) BeCl2 sp2 D) C2H2 sp B) SiH4 sp3 E) H2O sp3 C) BF3 sp2 Ans: A 3. What is the hybridization of As in the AsF4– ion? A) sp B) sp2 C) sp3 D) sp3d E) sp3d2 Ans: D 5. Which of the following is not true of molecular orbitals? A) The number of molecular orbitals formed is always equal to the number of atomic orbitals combined. B) A molecular orbital can accommodate up to two electrons. C) When electrons are added to orbitals of the same energy, the most stable arrangement is predicted by Hund's rule. D) Low-energy molecular orbitals fill before high-energy molecular orbitals fill. E) For any substance, the number of electrons in molecular orbitals is equal to the sum of all the valence electrons on the bonding atoms. Ans: E 6. Use valance bond theory to explain the bonding in the azide ion.(N3-) Since nitrogen is a second row element, it cannot exceed an octet of electrons. Since there are no lone pairs on the central nitrogen, the molecule must be linear and sp hybrid orbitals must be used. The 2py orbital on the central N atom overlaps with the 2py on the terminal N atom, and the 2pz orbital on the central N overlaps with the 2pz orbitals on the terminal N atoms to form delocalized molecular orbitals. 7. Dose the following molecule have a dipole moment? Why? 8. The BO+ ion is paramagnetic. Determine (a) whether the order of molecular-orbital energies is like that in B2 or O2; (b) the bond order; and (c) the number of unpaired electrons in the ion. 9. Carbon monoxide (CO) is a poisonous compound due to its ability to bind strongly to Fe2+ in the hemoglobin molecules. The molecular orbitals of CO have the same energy order as those of the N2 molecules.(a) Draw a Lewis structure of CO and assign formal charges. Explain why CO has a rather small dipole moment of 0.12D. (b) Compare the bond order of CO with that from molecular orbital theroy.(C) Which of the atoms (C or O) is more likely to form bonds with Fe2+ ion in hemoglobin? Partial s Electron domain Electron domain geometry Number of lone pair Hybridi zation Molecular geometry NCl3 4 Tetrahedral 1 sp3 Trigonal pyramidal SF4 5 Trigonal bipyramidal 1 sp3d Seesaw-Sharp CHCl3 4 Tetrahedral 0 sp3 Tetrahedral H 3 O+ 4 Tetrahedral 1 sp3 Trigonal pyramidal NH4+ 4 Tetrahedral 0 sp3 Tetrahedral PCl6- 6 Octahedral 0 sp3d2 Octahedral IF3 5 Trigonal 2 sp3d T-sharped 分 子 间 作 用 力 —— 应 用 篇 应用（一）： 相似相溶原理: 极性分子易溶于极性分子所组成的溶剂，非极性 分子易溶于非极性分子所组成的溶剂。 系统 乙醇 - 水 苯-四氯化碳 苯 - 水 混合状况 互溶 互溶 分层 过程自发进行 DG < 0 DG = DH - TDS 应用（一）： 相似相溶原理: 乙醇 - 乙醇 乙醇 取向力 诱导力 色散力 氢键 乙醇-水 取向力、诱导力 水-水 水 取向力 诱导力 色散力 氢键 色散力、氢键 应用（一）： 相似相溶原理: 苯 苯–苯 色散力 苯–水 水–水 水 诱导力 取向力 诱导力 色散力 氢键 为了保持系统能量最低，水会避免与苯接触，以保存分子间较强的氢键。 表现出的就好像水对苯的“排斥”作用。 应用（一）： 相似相溶原理: pp 堆积： 芳香环在外界极性环境影响下会通过p电子 间的作用力，分子平面将垂直于中心轴以面对 面的方式形成堆积结构。这种作用被称为pp 堆积力。 它是系统中芳香环之间的作用力和水保 持自身氢键而对芳香环“排斥” 协同作用的结 果。它不是一种特殊的分子间相互作用力。 应用（二）： 肥皂和洗涤剂: 疏水相互作用 Hydrophobic forces 非极性基团在水相中为了避开 水而发生的自聚集作用。 系统中溶质和溶剂水分子之间 为了达到系统能量最低的综合表现 pp堆积是一种特殊的疏水相互作用 应用（二）： 肥皂和洗涤剂: 应用（二）： 肥皂和洗涤剂: 洗涤剂去污过程示意图