1 / 39

Chemical Bonding I: The Covalent Bond

Chemical Bonding I: The Covalent Bond. Chapter 9. Group. e - configuration. # of valence e -. ns 1. 1. 1A. 2A. ns 2. 2. 3A. ns 2 np 1. 3. 4A. ns 2 np 2. 4. 5A. ns 2 np 3. 5. 6A. ns 2 np 4. 6. 7A. ns 2 np 5. 7. Valence electrons are the outer shell electrons of an

torn
Download Presentation

Chemical Bonding I: The Covalent Bond

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. Chemical Bonding I:The Covalent Bond Chapter 9

  2. Group e- configuration # of valence e- ns1 1 1A 2A ns2 2 3A ns2np1 3 4A ns2np2 4 5A ns2np3 5 6A ns2np4 6 7A ns2np5 7 Valence electrons are the outer shell electrons of an atom. The valence electrons are the electrons that particpate in chemical bonding.

  3. Lewis Dot Symbols for the Representative Elements & Noble Gases

  4. - - - - + Li+ Li Li Li+ + e- e- + Li+ Li+ + F F F F F F The Ionic Bond Ionic bond:the electrostatic force that holds ions together in an ionic compound. 1s22s1 1s22s22p5 1s2 1s22s22p6 [He] [Ne] LiF 形成”離子對”

  5. Ionic bond: Other example • LiO2 • Mg3N2

  6. E = k Q+Q- r Compound Lattice Energy (kJ/mol) MgF2 2957 Q: +2,-1 MgO 3938 Q: +2,-2 LiF 1036 LiCl 853 Electrostatic (Lattice) Energy Lattice energy (U) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions. E is the potential energy 形成離子固體後, 其穩定性如何? Q+ is the charge on the cation Q- is the charge on the anion r is the distance between the ions Lattice energy increases as Q increases and/or as r decreases. r F- < r Cl-

  7. Born-Haber Cycle for Determining Lattice Energy Lattice energy: + 1017 kJ ° ° ° ° ° ° DHoverall = DH1 + DH2 + DH3 + DH4 + DH5

  8. Chemical formula v.s lattice energy MgCl2(1st ionic energy:738+ 2nd ionic energy: 1450 = 2188KJ/mol) NaCl (2nd inoic energy: 4560KJ/mol) >> Lattice Energy so the NaCl2 is not formed! 2+ leading to large attractive force than 1+

  9. + 8e- 8e- 7e- 7e- F F F F F F F F lonepairs lonepairs single covalent bond single covalent bond lonepairs lonepairs A covalent bond is a chemical bond in which two or more electrons are shared by two atoms. Covalent compound. Why should two atoms share electrons? Lewis structure of F2

  10. Lewis structure of water single covalent bonds + + H H H or H H H O O O 2e- 8e- 2e- Octet rule: suitable for 2nd period(2s+ 2p[Ne]) Double bond – two atoms share two pairs of electrons or O O O O C C Ex: C2H4 8e- 8e- 8e- double bonds double bonds Triple bond – two atoms share three pairs of electrons N or N N N Ex: C2H2 8e- 8e- triple bond triple bond

  11. Lengths of Covalent Bonds Bond Lengths Triple bond < Double Bond < Single Bond

  12. Compare the difference between ionic compound and covalent compound • Covalent compound: gas, liquid or Low melting point solid • Low solubility in water(can’t conductive) • Example: NaCl(solid) and CCl4 (liquid)

  13. F H F H Polar covalent bond or polar bond is a covalent bond with greater electron density around one of the two atoms electron rich region electron poor region e- poor e- rich d+ d-

  14. X (g) + e- X-(g) Electronegativity is the ability of an atom to attract toward itself the electrons in a chemical bond. Electron Affinity - measurable, Cl is highest The attraction of single atom to its electrons Electronegativity - relative, F is highest The attraction of multi atoms to its co-bonded electrons

  15. The Electronegativities of Common Elements 當形成鍵結的2原子之電負度差異很大時，易形成離子鍵。 反之, 當電負度差異小則易形成極性共價鍵。而電負度相近時 ，則形成非極性共價鍵。

  16. Variation of Electronegativity with Atomic Number 過渡金屬有例外~

  17. Increasing difference in electronegativity Covalent Polar Covalent Ionic partial transfer of e- share e- transfer e- Classification of bonds by difference in electronegativity Difference Bond Type 0 Covalent  2 Ionic 0 < and <2 Polar Covalent

  18. Classify the following bonds as ionic, polar covalent, or covalent: The bond in CsCl; the bond in H2S; and the NN bond in H2NNH2. Cs – 0.7 Cl – 3.0 3.0 – 0.7 = 2.3 Ionic H – 2.1 S – 2.5 2.5 – 2.1 = 0.4 Polar Covalent N – 3.0 N – 3.0 3.0 – 3.0 = 0 Covalent

  19. Oxidation number(氧化數) and Electronegativity Actually, electronegativity is the way to define the oxidation number of a compound: the numbers of charge been transferred Ex: NH3 (N 3- / H+ )

  20. Writing Lewis Structures • Draw skeletal structure of compound showing what atoms are bonded to each other. Put least electronegative element in the center. • Count total number of valence e-. Add 1 for each negative charge. Subtract 1 for each positive charge. • Complete an octet for all atoms except hydrogen • If structure contains too many electrons, form double and triple bonds on central atom as needed.

  21. F N F F Write the Lewis structure of nitrogen trifluoride (NF3). Step 1 – N is less electronegative than F, put N in center Step 2 – Count valence electrons N - 5 (2s22p3) and F - 7 (2s22p5) 5 + (3 x 7) = 26 valence electrons Step 3 – Draw single bonds between N and F atoms and complete octets on N and F atoms. Step 4 - Check, are # of e- in structure equal to number of valence e- ? 3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons

  22. O C O 2 single bonds (2x2) = 4 1 double bond = 4 8 lone pairs (8x2) = 16 O Total = 24 Write the Lewis structure of the carbonate ion (CO32-). Step 1 – C is less electronegative than O, put C in center • Step 2 – Count valence electrons C - 4 (2s22p2) and O - 6 (2s22p4) • -2 charge – 2e- 4 + (3 x 6) + 2 = 24 valence electrons Step 3 – Draw single bonds between C and O atoms and complete octet on C and O atoms. Step 4 - Check, are # of e- in structure equal to number of valence e- ? 3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons Step 5 - Too many electrons, form double bond and re-check # of e-

  23. H H C O H C O H formal charge on an atom in a Lewis structure total number of valence electrons in the free atom total number of nonbonding electrons ( total number of bonding electrons ) 1 - - = 2 Two possible skeletal structures of formaldehyde (CH2O) An atom’s formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure. The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion. 對中性分子，所有原子的形式電荷總合為0，對於離子，形式電荷為離子的帶電荷

  24. H C O H C – 4 e- O – 6 e- 2H – 2x1 e- 12 e- formal charge on an atom in a Lewis structure total number of valence electrons in the free atom total number of nonbonding electrons ( total number of bonding electrons ) 1 - - = 2 single bonds (2x2) = 4 2 1 double bond = 4 2 lone pairs (2x2) = 4 Total = 12 -1 +1 formal charge on C = 4 - 2- ½ x 6 = -1 formal charge on O = 6 - 2- ½ x 6 = +1

  25. C – 4 e- H O – 6 e- C O H 2H – 2x1 e- 12 e- formal charge on an atom in a Lewis structure total number of valence electrons in the free atom total number of nonbonding electrons ( total number of bonding electrons ) 1 - - = 2 single bonds (2x2) = 4 2 1 double bond = 4 2 lone pairs (2x2) = 4 Total = 12 0 0 formal charge on C = 4 - 0- ½ x 8 = 0 formal charge on O = 6 - 4- ½ x 4 = 0

  26. H C O H H C O H -1 +1 0 0 Formal Charge and Lewis Structures • For neutral molecules, a Lewis structure in which there are no formal charges is preferable to one in which formal charges are present. • Lewis structures with large formal charges are less plausible than those with small formal charges. • Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms. Which is the most likely Lewis structure for CH2O?

  27. - - + + O O O O O O O O O C C C O O O - - - - O O O - - A resonance structure is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure. What are the resonance structures of the carbonate (CO32-) ion?

  28. The Benzene (C6H6)structure The measured bonding length is 140 pm, less than C-C(154 nm) while longer than C=C(133 pm) 分子的表現行為可以看成是共振結構的加成結果，而非 兩種結構的來回切換。

  29. Be – 2e- 2H – 2x1e- H Be H 4e- B – 3e- 3 single bonds (3x2) = 6 3F – 3x7e- F B F 24e- Total = 24 9 lone pairs (9x2) = 18 F Exceptions to the Octet Rule The Incomplete Octet BeH2 BF3

  30. Coordinate covalent bond(配位共價鍵): 價電子完全由一原子提供來配對 NH3+BF3H3N-BF3

  31. N – 5e- S – 6e- N O 6F – 42e- O – 6e- 11e- 48e- F 6 single bonds (6x2) = 12 F F Total = 48 S 18 lone pairs (18x2) = 36 F F F Exceptions to the Octet Rule (奇墊子分子: 自由基radical) Odd-Electron Molecules NO EX:NO2 The Expanded Octet (central atom with principal quantum number n > 2) SF6 EX: SCl2 still satisfy the Octet rule

  32. Expanded Octet XeF4: leave 4 lone pairs to the Xe atom

  33. 2nd 週期後的元素因填入3d軌域，價電子會超過八個 Margin Art 9.14 Octet Expanded Octet

  34. DH° = 436.4 kJ H2 (g) H (g) + H (g) DH° = 242.7 kJ Cl2 (g) Cl (g) + Cl (g) DH° = 431.9 kJ HCl (g) H (g) + Cl (g) DH° = 498.7 kJ O2 (g) O (g) + O (g) O DH° = 941.4 kJ N2 (g) N (g) + N (g) O N N The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond enthalpy. Bond Enthalpy Bond Enthalpies Single bond < Double bond < Triple bond

  35. DH° = 502 kJ H2O (g) H (g) + OH (g) DH° = 427 kJ OH (g) H (g) + O (g) Average OH bond enthalpy = 502 + 427 2 Average bond enthapy in polyatomic molecules = 464 kJ

  36. Bond Enthalpies (BE) and Enthalpy changes in reactions Imagine reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products. DH° = total energy input – total energy released = SBE(reactants) – SBE(products) exothermic endothermic

  37. Type of bonds broken Number of bonds broken Bond enthalpy (kJ/mol) Enthalpy change (kJ/mol) 1 436.4 436.4 1 156.9 156.9 Type of bonds formed Number of bonds formed Bond enthalpy (kJ/mol) Enthalpy change (kJ/mol) H H F H F F 2 568.2 1136.4 Use bond enthalpies to calculate the enthalpy change for: H2 (g) + F2 (g) 2HF (g) DH° = SBE(reactants) – SBE(products) DH° = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ/mol

  38. Ch9 HW 9.25 Use the Born-Haber cycle outlined in Section 9.3 for LiF to calculate the lattice energy of NaCl. [The heat of sublimation of Na is 108 kJ/mol and (NaCl) = −411 kJ/mol. Energy needed to dissociate mole of Cl2 into Cl atoms = 121.4 kJ]. 9.17 Use Lewis dot symbols to show the transfer of electrons between the following atoms to form cations and anions: (a) Na and F, (b) K and S, (c) Ba and O, (d) Al and N. 9.29 How many lone pairs are on the underlined atoms in these compounds? HBr, H2S, CH4. 9.49 Write Lewis structures for these species, including all resonance forms, and show formal charges: (a) (b) Relative positions of the atoms are as follows:

  39. 2H2 (g) + O2 (g) 2H2O (g) Use bond enthalpies to calculate the enthalpy change for:

More Related