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 Calcium carbonate is one of the most interesting and versatile compounds on the planet.  Roughly 4% of earth’s crust  The major component of rocks.

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Presentation on theme: " Calcium carbonate is one of the most interesting and versatile compounds on the planet.  Roughly 4% of earth’s crust  The major component of rocks."— Presentation transcript:

1  Calcium carbonate is one of the most interesting and versatile compounds on the planet.  Roughly 4% of earth’s crust  The major component of rocks such as limestone and marble  The white cliffs of Dover are one of the most famous natural formations of CaCO 3.

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3  T HE PROPERTIES OF SUBSTANCES & C HEMICAL BONDS  The properties of substances are determined in large part by the chemical bonds (O 2 vs N 2 )  What determines the type of bonding in each substance?  The electronic structures of the atoms are the key to answering the question  We will examine the relationship between the electronic structures of atoms and the chemical bonds they form

4 8.1 C HEMICAL BONDS, L EWIS SYMBOLS, AND THE OCTET RULE  Three general types of chemical bonds Ionic bond: electrostatic forces Covalent bond: the sharing of electrons Metallic bond: free electrons

5  The valence electrons The electrons involved in chemical bonding In most atoms, they reside in the outermost occupied shell  The Lewis symbol Diagrams that show the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule 8.1 C HEMICAL BONDS, L EWIS SYMBOLS, AND THE OCTET RULE  L EWIS S YMBOLS

6  The noble gases have very stable electron arrangements (high ionization energies)  Atoms often gain, lose, or share electrons to achieve the same number of electrons as the noble gas  The octet rule Atoms tend to gain, lose or share electrons until they are surrounded by eight valence electrons The rule provides a useful framework to introduce concepts of bonding although there are many exceptions 8.1 C HEMICAL BONDS, L EWIS SYMBOLS, AND THE OCTET RULE  T HE O CTET R ULE

7 8.2 I ONIC B ONDING  The reaction can be explained by the low ionization energy of Na and the high electron affinity of Cl  We can also explain the reaction by the octet rule  Consider a reaction of Na(s) and Cl 2 (g)

8 8.2 I ONIC B ONDING  F ORMATION OF SODIUM CHLORIDE

9  The formation of sodium chloride is very exothermic  The heat of formation of other ionic substances is also quite negative  The first ionization energy of Na(g) is 496 kJ/mol  The electron affinity of Cl(g) is −349 kJ/mol  If we consider only the electron transfer for the formation reaction, the energy change would be: 496 − 349 = 147 kJ/mol How can we explain this? 8.2 I ONIC B ONDING  E NERGETICS OF IONIC BOND FORMATION

10  The lattice energy is the energy required to completely separate a mole of a solid ionic compound into its gaseous ions  The potential energy of two interacting charged particles  The magnitude of lattice energies depends predominantly on the ionic charges 8.2 I ONIC B ONDING  T HE L ATTICE E NERGY

11 8.2 I ONIC B ONDING  T HE L ATTICE E NERGY

12 Sample Exercise 8.1 Magnitudes of Lattice Energies Without consulting Table 8.2, arrange the following ionic compounds in order of increasing lattice energy: NaF, CsI, and CaO. Answer: CsI < NaF < CaO. Which substance would you expect to have the greatest lattice energy, MgF 2, CaF 2, or ZrO 2 ? Practice Exercise Answer: ZrO 2

13 8.2 I ONIC B ONDING  The lattice energy can not be determined directly by experiment

14  The formation of Na 2+ and Cl 2− are energetically very unfavorable  Groups 1A, 2A, and 3A atoms form 1+, 2+, and 3+ ions, respectively  Groups 5A, 6A, and 7A atoms form 1−, 2−, and 3− ions, respectively  Transition metals do not observe the octet rule: Fe 2+ and Fe I ONIC B ONDING  E LECTRON CONFIGURATIONS OF IONS

15  A chemical bond formed by sharing a pair of electrons  The attractions and repulsions among electrons and nuclei in the H 2 molecule  Quantum mechanical calculation tells us that the concentration of electron density between the nuclei leads to a net attractive force that constitutes the covalent bond holding the molecule together 8.3 C OVALENT B ONDING

16  The formation of covalent bonds can be represented using Lewis symbols 8.3 C OVALENT B ONDING  L EWIS S TRUCTURES  For the nonmetals, the number of valence electrons in a neutral atom is the same as the group number the number of covalent bonds

17  The Lewis structures of CO 2 and N C OVALENT B ONDING  M ULTIPLE B ONDS  Bond length: the distance between the nuclei of atoms involved in a bond

18  Compare the electron-sharing of Cl 2 and H 2 with the sharing of H 2 O and NaCl  Polar covalent bond and nonpolar covalent bond  If the difference of the covalent bond-forming atoms in relative ability to attract electrons is large enough, an ionic bond is formed  Electronegativity is defined as the ability of an atom in a molecule to attract electrons to itself 8.4 B OND P OLARITY AND E LECTRONEGATIVITY

19  Linus Pauling ( ) developed the first and most widely used electronegativity scale  Electronegativity can be used to estimate whether a given bond will be nonpolar or polar covalent, or ionic  The electronegativity of an atom in a molecule is related to its ionization energy and electron affinity  Fluorine, the most electronegative element, has an electronegativity of 4.0  The least electronegative element, cesium, has an electronegativity of B OND P OLARITY AND E LECTRONEGATIVITY  E LECTRONEGATIVITY

20 8.4 B OND P OLARITY AND E LECTRONEGATIVITY  E LECTRONEGATIVITY

21  The greater the difference in electronegativity between two atoms, the more polar their bond 8.4 B OND P OLARITY AND E LECTRONEGATIVITY  E LECTRONEGATIVITY AND BOND POLARITY

22 8.4 B OND P OLARITY AND E LECTRONEGATIVITY  E LECTRONEGATIVITY AND BOND POLARITY

23  We can indicate the polarity of the HF molecule in two ways: 8.4 B OND P OLARITY AND E LECTRONEGATIVITY  D IPOLE M OMENTS  Polarity helps determine many of the properties of substances such as hydrogen bonding and solvation  How can we quantify the polarity of molecule? For the two equal and opposite charges, the dipole moment is: , dipole moment (C·m or debye, D) Q, charge (C) r, distance (m) 1 D = 3.34 × C·m

24 Sample Exercise 8.5 Dipole Moments of Diatomic Molecules The bond length in the HCl molecule is 1.27 Å. (a) Calculate the dipole moment, in debyes, that would result if the charges on the H and Cl atoms were 1+ and 1–, respectively. (b) The experimentally measured dipole moment of HCl(g) is 1.08 D. What magnitude of charge, in units of e, on the H and Cl atoms would lead to this dipole moment? Solution

25  The actual charges on the atoms decrease from 0.41  in HF to  in HI 8.4 B OND P OLARITY AND E LECTRONEGATIVITY  D IPOLE M OMENTS

26  The ability to quickly categorize the predominant bonding interactions in a substance as covalent or ionic imparts considerable insight into the properties of that substance  By considering the interaction between a metal and a nonmetal SnCl 4 : colorless liquid, mp -33 ˚C, bp 114 ˚C, polar covalent  To use the difference in electronegativity SnCl 4 : 1.2; NaCl: 2.1 MnO: 2.0, ionic; Mn 2 O 7 : 2.0, polar covalent The increase in the oxidation state of a metal leads to an increase in the degree of covalent character in the bonding 8.4 B OND P OLARITY AND E LECTRONEGATIVITY  D IFFERENTIATING IONIC AND COVALENT BONDING

27 1.Find the sum of valence electrons of all atoms in the polyatomic ion or molecule. –If it is an anion, add one electron for each negative charge. –If it is a cation, subtract one electron for each positive charge. PCl (7) = D RAWING L EWIS S TRUCTURES  Lewis structures can help us understand the bonding in many compounds and are frequently used when discussing the properties of molecules

28 2.Arrange atoms. The central atom is the least electronegative element that isn’t hydrogen. Connect the outer atoms to it by single bonds. Keep track of the electrons: = D RAWING L EWIS S TRUCTURES

29 3.Complete the octets around all the atoms bonded to the central atom. Hydrogen is an exception Keep track of the electrons: = 20; = D RAWING L EWIS S TRUCTURES

30 4.Place any leftover electrons on the central atom Keep track of the electrons: = 20; = 2; = D RAWING L EWIS S TRUCTURES

31 5.If there are not enough electrons to give the central atom an octet, try multiple bonds 8.5 D RAWING L EWIS S TRUCTURES

32 Sample Exercise 8.8 Lewis Structure for a Polyatomic Ion Practice Exercise Draw the Lewis structure for (a) ClO 2 –, (b) PO 4 3– Draw the Lewis structure for the BrO 3 – ion. Solution The total number of valence electrons is, therefore, 7 + (3  6) + 1 = 26. For oxyanions— BrO 3 –, SO 4 2–, NO 3 –, CO 3 2–, and so forth—the oxygen atoms surround the central nonmetal atom. Answers: (a) (b)

33  The charge the atom would have if all the atoms in the molecule had the same electronegativity  How to assign formal charges. For each atom, count the electrons in lone pairs and half the electrons it shares with other atoms. Subtract that from the number of valence electrons for that atom: the difference is its formal charge. 8.5 D RAWING L EWIS S TRUCTURE  F ORMAL C HARGE

34  The concept of formal charge helps us choose a preferred Lewis structure  How to choose the correct structure. Choose the Lewis structure in which the atoms bear formal charges closest to zero Choose the Lewis structure in which any negative charges reside on the more electronegative atoms  F ORMAL C HARGE 8.5 D RAWING L EWIS S TRUCTURE preferred

35  F ORMAL C HARGE 8.5 D RAWING L EWIS S TRUCTURE preferred

36 Oxidation numberFormal chargeActual partial charge +1 − − −0.178 Dipole moment 8.5 D RAWING L EWIS S TRUCTURE

37  Consider the Lewis structure of ozone, O R ESONANCE S TRUCTURES  Both Lewis structures are not corresponding to the structure of ozone  Resonance structure: An alternate way of drawing a Lewis dot structure for a compound

38  How can we understand the resonance structure?  The rules for drawing Lewis structures do not allow us to have a single structure that adequately represents the ozone molecule 8.6 R ESONANCE S TRUCTURES

39 Draw the resonance structures for HCO 2 − and NO 3 −.

40 8.6 R ESONANCE S TRUCTURES  R ESONANCE IN B ENZENE All six C-C bonds are of equal length 1.40 Å C-C single bond 1.54 Å; C=C double bond 1.34 Å

41 8.7 E XCEPTIONS TO THE O CTET R ULE  O DD NUMBER OF ELECTIONS  In a few molecules and polyatomic ions, such as ClO 2, NO, NO 2, and O 2 −, the number of valence electrons is odd  Complete pairing of these electrons is impossible  An octet around each atom cannot be achieved  NO contains =11 valence electrons

42 8.7 E XCEPTIONS TO THE O CTET R ULE  L ESS THAN AN OCTET OF VALENCE ELECTRONS  Consider boron trifluoride, BF 3  In the Lewis structure of BF 3, there are only six electrons around the boron atom and the formal charges are zero  If you try to complete the octet around boron: Most important

43 8.7 E XCEPTIONS TO THE O CTET R ULE  M ORE THAN AN OCTET OF VALENCE ELECTRONS  Consider phosphorus pentachloride, PCl 5  Elements from the third period and beyond have unfilled nd orbitals that can be used in bonding  The larger the central atom is, the larger the number of atoms that can surround it Draw the Lewis structure of SF 4, I Cl 4 −, and PO 4 3− The orbital diagram for the valence shell of a phosphorus atom

44 8.8 S TRENGTH OF C OVALENT B ONDS  The stability of a molecule is related to the strengths of the covalent bonds it contains  The strength of a bond is measured by determining how much energy is required to break the bond  The bond enthalpy for a Cl-Cl bond, D(Cl-Cl), is measured to be 242 kJ/mol

45 8.8 S TRENGTH OF C OVALENT B ONDS

46  The bond enthalpy is always positive as energy is always required to break chemical bonds  A molecule with strong chemical bonds generally has less tendency to undergo chemical change than does with weak bonds

47 8.8 S TRENGTH OF C OVALENT B ONDS  B OND ENTHALPIES AND THE ENTHALPIES OF REACTION  We can use average bond enthalpies to estimate the enthalpies of reactions in which bonds are broken and new bonds are formed  Consider a gas-phase reaction: CH 4 (g) + Cl 2 (g) → CH 3 Cl(g) + HCl(g)

48 8.8 S TRENGTH OF C OVALENT B ONDS  B OND ENTHALPIES AND THE ENTHALPIES OF REACTION

49 8.8 S TRENGTH OF C OVALENT B ONDS  B OND ENTHALPY AND BOND LENGTH

50  Enormous amounts of energy can be stored in chemical bonds → can be used as an explosives  Characteristics of an explosive Very exothermic decomposition Gaseous products Rapid decomposition Controllably stable

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52 −2−−++++

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