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 Structures6.3 Drawing Lewis StructuresFollow 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

B

I

II

B

I

II B

I

I IB

I

II

Chapter 7 Bonding Theory 7.1 Molecular Geometry:

valence-shell electron-pair repulsion (VSEPR) model

Electron-Domain Geometry and Molecular Electron-Domain Geometry and Molecular GeometryGeometry

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 orbitalsbond 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 D

)tetrahedral

B) square planar E)

octahedral

C) square pyramidal

Ans:  D

3. What is the hybridization of As in the AsF4– ion?

A) sp B) sp2 C) sp3 D) sp3d E) sp3d2

Ans:  D

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  

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  

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.

6. Use valance bond theory to explain the bonding in the azide ion.(N3

-)

7. Dose the following molecule have a dipole moment? Why?

C C C

Cl

H

H

Cl

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?

Partials

Electron domain

Electron domain

geometry

Number of lone

pair

Hybridization

Molecular geometry

NCl3 4 Tetrahedral 1 sp3 Trigonal pyramidal

SF4 5 Trigonal bipyramidal

1 sp3d Seesaw-Sharp

CHCl3 4 Tetrahedral 0 sp3 Tetrahedral

H3O+ 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

相似相溶原理 :

过程自发进行 G < 0

G = H - TS

应用（一）：

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

乙醇

水

取向力、诱导力色散力、氢键水 - 水

乙醇 - 乙醇

乙醇 - 水

相似相溶原理 :

应用（一）：

取向力 诱导力色散力 氢键

取向力 诱导力色散力 氢键

苯色散力

水 取向力 诱导力色散力 氢键

诱导力

苯 – 水

水 – 水

苯 – 苯

相似相溶原理 :

应用（一）：

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。

堆积： 芳香环在外界极性环境影响下会通过电子间的作用力，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。

相似相溶原理 :

应用（一）：

肥皂和洗涤剂 :

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

疏水相互作用Hydrophobic forces

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）：

洗涤剂去污过程示意图

肥皂和洗涤剂 :

应用（二）：