Chemical Bonding

1. Chemical Bonding Chapter 7

2. Chemical Bonding • The __________ are generally involved in chemical bonding. • Ionic bonding – bonding resulting from the _________ among ions. Electrons are transferred from one atom to another. • Covalent bonding – bonding resulting from _______ electron pairs between two atoms.

3. Ionic Compounds Covalent Compounds Melting Point High (>400C) Low (<300C) Solubility in polar solvents Most are soluble Most are insoluble Solubility in nonpolar solvents Most are insoluble Most are soluble Conductivity (molten) High Low Conductivity (aqueous) High Low Properties of Ionic and Covalent Compounds

4. Lewis Dot Formulas • Keeps track of electrons in the __________. • Electrons in the __ and __ orbitals. • Mg and Br Lewis structures • Elements in the _____ group have the similar Lewis structures. • Illustrates electron distribution upon bonding.

5. Ionic Bonding • Cation – an ion that has ______ electrons than protons and is _____ charged. • Na+ and Mg2+ • Anion – an ion that has ______ electrons than protons and is ______ charged. • Cl-, O2- • Polyatomic ions – an ion that contains more than one atom. • NH4+ and SO42-

6. Ionic Bonding • Octet rule – upon the formation of ionic compounds, most representative elements achieve _________ electron configurations. • When the EN difference is _____ between two elements the elements are likely to form an ionic compound. Table 6-3 • Nonmetals and metals • Lithium and Fluorine

7. Ionic Bonding – Group IA with Group VIIA • 2Li(s) + F2(g)  2LiF(s) • Characteristics of elements versus ionic compound. • Electronegativity difference and electron configurations. How could these element be isoelectronic with a noble gas? • Li  Li + + e- (cation, ionization) • F + e-  F- (anion, electron affinity)

8. Ionic Bonding – Group IA with Group VIIA • Draw Lewis structures for the reaction of lithium with fluorine. • Why is there a 1:1 ratio of Li to F? • The formula (e.g. LiF) represents the ____ of ions in the compound. • Individual units of LiF do no exist. • Structure of LiF (NaCl) and the arrangement of ions and CD demonstration.

9. Ionic Bonding – Group IA with Group VIIA • All Group IA metals react with Group VIIA nonmetals to form ionic compounds of the same formula. • 2M(s) + X2(g)  2M+X-(s) • M = ______, X = ________ • How many possible combinations? The _____ apart two elements are on the periodic table the more ionic their bonding will be. Why? Table 6-3

10. Ionic Bonding – Group IA with Group VIA • Lithium (Li) and oxygen (O) • What is the electronegativity difference and electron configurations? • How will they be isoelectronic with a noble gas? • Li  Li+ + e- (cation, ionization energy) • O + 2e-  O2- (anion, electron affinity) What will be the structure of this ionic compound? Why?

11. Ionic Bonding – Group IA with Group VIA • 4Li(s) + O2(g)  2Li2O(s) • For the formation of O2-(gain of two electrons), two lithium atoms are required to be ionized (loss of one electron each). • All Group IA metals react with Group VIA to from ionic compounds with formula M2X. • Demonstration of structure (notice the ratio of atoms and difference with the MX structure).

12. Ionic Bonding – Group IIA with Group VIA • Calcium (Ca) and Oxygen (O) • Electron configurations • What will the ratio of calcium to oxygen be in the ionic compound? • Ca  Ca2+ + 2e- (cation, 1st and 2nd ionization) • O + 2e-  O2- (electron affinity) 2Ca(s) + O2(g)  2CaO(s) (MX formula)

13. Summary of Binary Ionic Compounds • Examine table 7-2 in book. The melting and boiling points for ionic compounds are large due to strong attractive forces that are present between ions.

14. Summary of Binary Ionic Compounds • Possible ionic compounds • IA, IIA, and IIIA can be cations (metals) • VA, VIA, and VIIA can be anions (nonmetals) • Generally, monatomic ions possess charges that are between +3 and –3. • Subscripts reflect the ratio to make the overall charge equal to zero. • d- and f-transition elements also form many ionic compounds. • Most will not be isoelectronic with a noble gas.

15. Why Do Ionic Compounds Form? • KBr formation from the elements • The reaction is extremely exothermic (i.e. gives off energy). Why? • Takes energy for ionization and, in many cases, to add electron(s) to atoms. • Examination of the formation of Li2O. • This compound forms readily from the elements.

16. Formation of Li2O(s) from the Gaseous Atoms • Required energy to form the ions • Li(g)  Li+(g) + e- IE1 = _______ (Table 6-1) • Two Li atoms need to be ionized per O atom (Li2O). • O(g) + e- O-(g) EA1 = ______ (Table 6-2) • This actually gives off energy • O-(g) + e- O2-(g) EA2 = 844 kJ • How much energy does it require to form the ions for the solid to form? Why does it form?

17. Why Does Li2O Form? • There are attractive forces between the created ions. where q- and q+ are the charge magnitudes and d is the distance between charges. As the charges come closer together the, d decreases and the attractive force increases. The energy associated with the attraction of gaseous ions to form an ionic solid is the __________.

18. Energy Changes for 2Li(g) + O(g)  2Li2O(s) 2Li+(g) +O2-(g) EA2 of O-(g) = 844 kJ/mol 2Li+(g) + 2e- +O(g) EA of O(g) = -141 kJ/mol 2Li+(g) + 1e- +O-(g) Crystal lattice energy of Li2O = -2799 kJ/mol 1st IE of Li = 520 kJ/mol (x 2 = 1040) 2Li(g) + O(g) Net energy change for 2Li(g) + O(g) Li2O(s) = -1056 kJ/mol Li2O(s)

19. Your Turn to Determine Net Energy Change • Mg(g) + O(g)  MgO(s) • IE2 for Mg = _____ (in your notes) • Crystal lattice energy MgO(s) = -3850 kJ You have enough information to determine the net energy change for the reaction shown above. • Use the tables and previous information. Note: A negative value indicates that the reaction is favored (i.e. gives off energy).

20. Melting Point of an Ionic Solid is Influenced by the Attractive Forces • Attractive force increases with charge density of the ions • Charge density is the ratio of charge to size • MgO versus NaCl • Mg2+ versus Na+ Which has the greater charge density? • O2- versus Cl- The greater charge density brings the ions closer together increasing the attractive force. More energy is given off (crystal lattice energy).

21. Melting Point of an Ionic Solid is Influenced by the Attractive Forces • How is the attractive force changed with increasing charge density? • Melting points • MgO 2852 C • NaCl 801 C

22. Covalent Bonding • Covalent bonding occurs when two atoms _________. Generally the EN difference between the two atoms is _______. • Atoms bond covalently to achieve __________ that are full. This is similar to ionic bonding.

23. Covalent Bond Formation • Formation of covalent bonds involves repulsive and attractive forces. As the atoms approach each other different forces are experienced. • Repulsion between electrons in each atom • Repulsion between nuclei • Attraction between electron(s) of one atom and the nucleus of another. Atoms will approach until the attractive forces minus the repulsive forces is the greatest (show this).

24. Potential Energy Diagram The energy of the bonded atoms is lower than the separated atoms. This difference is called the __________.

25. Covalent Bonds • Covalent bonds usually consists of shared pairs of electrons (draw). • Single covalent – shares two electrons • H2 • Double covalent – shares four electrons • C2H4 • Triple covalent – shares six electrons • C2H2 Few quadrupole bonds do exist but only with transition or inner-transition elements.

26. Drawing Lewis Structures for Molecules and Polyatomic Ions • The Lewis formula illustrates the distribution of ________ over the molecule or ion. • Lewis formula is based on the octet rule • The representative elements achieve ___________ configuration in most compounds. • ________ electrons in the outermost shell (except H and He).

27. Lewis Formulas for Simple Molecules • Homonuclear • H2 and Cl2 • Heteronuclear • HCl • Other simple molecules • NH3 and H2O

28. Determination of Lewis Formula • S = N – A • S = number of ______ in the molecule or ion. • N = total number of valence shell electrons needed by all the atoms • Each atom, except for __ and __ need 8 electrons. • A = number of ___________ in the valence shells • Number of valence electrons for each atom equals the ______. • Add or subtract if a polyatomic ion. Available electrons are either present as bonding electrons or unshared electrons.

29. Determination of Lewis Formula • CO2 • NH4+ • C2H2 S = N - A

30. Drawing Lewis Structures • Select a reasonable arrangement for the atoms • Most symmetrical arrangement • Select the center atom • Usually requires the most electrons • Usually the least electronegative • Hydrogen and halogens (group VIIA) are generally terminal atoms.

31. Drawing Lewis Structures • Calculate N and A in order to determine S • Create the correct number of bond between atoms. Single bonds first. • Carbon always forms four bonds • In most species, oxygen forms two bonds and nitrogen forms three • Can change for charged species

32. Drawing Lewis Structures • HCN • SO32- • SO3 • Draw arrangement of molecules containing C, N, O, H, and halogens (including cyclics)

33. Resonance • Exhibited by a molecules (or polyatomic ion) that has more than one Lewis formula with the same arrangement of atoms. • SO3 has three possible structures depending on the placement of the double bond. • Three structures • A ‘hybrid’ of the three structures exists. • Bonds are between double and single with the electrons being delocalized

34. Limitations to the Octet Rule • Most covalent compounds of Be Why? • Commonly, Be is involved in the formation of ionic compounds. • BeF2 • Covalent compounds of Group IIIA elements. • How would this violate the rule? • BCl3

35. Limitations to the Octet Rule • Compounds or ions containing an odd number of electrons • NO2 • Compounds or ions in which the central atom must take more than eight valence electrons • AsF5 Many compounds that don’t follow the octet rule are very reactive. • DEMO: Benzyl peroxide and analine

36. Polar and Nonpolar Covalent Bonds • Nonpolar covalent bonds • Electrons are ________ between atoms. • Symmetric charge distribution • Must be the same element • H2 and Cl2 (homonuclear diatomics) • Electron density maps

37. Polar Covalent Bonds • Electrons are _________ between atoms. • Electronegativity differences exist between atoms • Polarity increases with EN difference. • __________ diatomic molecules possess polar covalent bonds. • Higher electron density is over the atom with the greater ___________.

38. Polar Covalent Bonds • Electron density maps for HF and HI Red – high electron density Blue – low electron density What are the electronegativity differences? Identify. High polarity versus low polarity.

39. Indicating Polarity • Due to EN differences partial positive and negative charges exist on the molecule. • Where are the partial positive and negative charges on HF? • Dipoles • An arrow points from the _______ side of the bond to the ________ side of the bond. • The length of the dipole indicates the EN difference or electron density separation. • HF and HI

40. Dipole Moments • The bond polarity is measured by its dipole moment,  = d  q. • d is equal to the distance separating the charge • q is equal to the magnitude of charge • Measurement of dipoles in simple diatomics is made by placing a sample between charged plates.

41. Dipole Moment Measurement Demonstration: bending of a stream of water by a charged balloon.

42. Continuous Range of Bonding Types • Two extremes for bonding • Equal sharing (nonpolar covalent) • H2 and Cl2 (homonuclear) • Complete transfer of electrons (ionic) • Doesn’t exist • Almost all bonds have some covalent and ionic character. The general type of bonding is determined by the _________.

43. Continuous Range of Bonding Types • All polar bonds have partial ionic character • HCl has about 17% ionic character • The amount of ionic character increasing with increases ________. • All ionic compounds have some partial covalent character • Distortion of electron densities produce covalent character • Positively-charged species attract electron density from the negatively-charged species.

44. Polyatomic Ions • Ionic compounds containing polyatomic ions possess polar and ionic bonds. How? • NH4Cl and CaSO4 • Handout and discussion • These compounds are classified as ionic due to its properties. The entire substance is held together by ionic bonds (page 299).

45. Polyatomic Ions