Noble Gases and Valence e - Ionization Energy and Bonding

1. Noble Gases and Valence e- Ionization Energy and Bonding

2. A look back at the Periodic Table • Halogens, group 7A or 17: F, Cl, Br, I • Nonmetals • Element form is diatomic F2 Cl2 Br2 I2 • At room temperature, F2 and Cl2 are gases, Br2 is liquid, I2 is solid • Reactivity decreases down group • F2 is the most reactive of all elements • I2 is the least reactive halogen

3. Trends in Element Families • Alkali metals, group 1A or 1: Li, Na, K, Rb, Cs • Metals • stored under oil to keep them from reacting with O2 • Impart bright, characteristic colors to flames • Li = red Na = orange K, Rb, Cs = lavender • All react with water to produce H2 • Reactivity increases down group • Cs is most reactive alkali metal • Li is least reactive alkali metal

4. Trends in Element Families • Noble gases, group 8A or 18: He, Ne, Ar, Kr, Xe • Nonmetals • Element form is monatomic • At room temperature, all are gases • Most significant property is that they are almost completely unreactive • Compounds of Xe, Kr, and recently Ar have been made with F and O, but they decompose very easily • No compounds of He have ever been prepared

5. How do Atoms Join? • The behavior of the noble gas, alkali metal, and halogen families is key to understanding bonding • Noble gases generally do not form compounds • Alkali metals form compounds by losing one electron to form a +1 ion (cation) • Halogens form compounds by gaining one electron to form a –1 ion (anion)

6. Ion sizes • A cation is always smaller than its parent atom • An anion is always larger than its parent atom

7. Let’s Practice… • Which ion is larger? • Se or Se-2 • Which ion is smaller? • Al or Al+3

8. Ionization energy • Ionization energy (IE) is the energy needed to remove the outermost electron from an atom • Low IE electron is easier to remove • High IE electron is harder to remove

9. Ionization energy • IE decreases down a group (easier to remove e– as you go down a group) • IE generally increases across a period from left to right (harder to remove e– as you go across a period)

10. Ionization Energy

11. Let’s Practice… • Which has a larger ionization energy? • Cs or F • Which has a smaller ionization energy? • Be or Mg

12. 2 He 3 Li 9 F 10 Ne 11 Na 18 Ar 19 K 17 Cl Group 8A Noble gases “happy to be me” Group 1A Alkali metals “just one less electron . . .” Group 7A Halogens “just one more electron . . .”

13. Valence electrons • This behavior suggests that • The number of electrons in a noble gas is especially stable • That number of electrons forms a noble gas core of stable electrons unavailable for bonding • Only electrons outside the noble gas core participate in the formation of chemical bonds between atoms • Those electrons are called VALENCE ELECTRONS

14. Valence electrons 1 H 2 He 1A 7A 2A 3A 5A 8A 4A 6A 5 B 3 Li 6 C 7 N 4 Be 8 O 9 F 10 Ne 11 Na 12 Mg 13 Al 14 Si 15 P 16 S 17 Cl 18 Ar 20 Ca 19 K 31 Ga 32 Ge 33 As 34 Se 35 Br 36 Kr FOR MAIN-GROUP ELEMENTS: number of valence electrons = group number

15. •• • • •• • O O • C • C Li Li • • Carbon Group 4A 4 valence electrons Oxygen Group 6A 6 valence electrons Lithium Group 1A 1 valence electron Depicting Valence Electrons • Show the valence electrons for a main-group element using a LEWIS DOT STRUCTURE

16. • • •• • •• •• • • • N N N • N • • • • • •• Lewis Dot Structures • Dot number is important, positions are not Nitrogen Group 5A 5 valence electrons

17. •• •• •• •• •• •• •• F • I • Cl •• • Br • •• •• •• •• Lewis Dot Structures • All the atoms in the same family have the same dot structure Halogens Group 7A 7 valence electrons

18. Let’s Practice… • Draw the Lewis Dot Structure for: • Si • Rb • As • Br

19. Chemical Bonding • The stability of the noble gases suggests that for main group elements, Atoms form chemical bonds by losing, gaining, or sharing valence electrons in order to achieve the same number of valence electrons as the nearest noble gas

20. Octet Rule • Noble gases have 8 valence shell electrons (except He) • Atoms lose, gain, or share valence electrons in order to achieve 8 valence shell electrons • This is the OCTET RULE • Atoms near He (H and Li) will try to achieve just 2 valence shell electrons

21. Let’s Practice… • Octet rule predicts that an atom of Flourine will try to _____(gain/lose) _____(1,2,3,4) electron(s) when forming a compound. • Octet rule predicts that an atom of Calcium will try to _____(gain/lose) _____ (1,2,3,4) electron(s)when forming a compound.

22. Three Types of Chemical Bonding • Chemical Bonds can be classified into 3 broad categories, based on how valence electrons are arranged:

23. NaCl, salt Ionic bonds • Ionic bonds form between metals and nonmetals • Metal gives up electrons • Nonmetal accepts electrons • Ions form • Because of opposite charges, ions stick together to form a crystal • Resulting compound is a SALT

24. • •• •• 1– 1+ Cl •• •• •• Na Na  •• Cl • •• Ionic bonds: electron transfer • Metals transfer electrons to nonmetals to form ions: • Na has 1 valence e–, wants to lose it to attain a Ne core • Cl has 7 valence e–, wants to gain 1 to attain an octet, like Ar • Solution is to transfer 1 electron from Na to Cl • Electron transfer forms ions • Ions stick together because of opposite charges

25. Dot structures for ionic compounds • Draw dot structures for these ionic compounds • Show the atoms before and ions after e– transfer • Use an arrow to show e– transfer • Show charges on ions after e– transfer • Write ions near each other but not together (they are not sharing the electrons) • MgO Na2O CaCl2

26. Electronegativity • Electronegativity is the ability to attract bond e– • The higher the EN, the “greedier” the atom

27. Electronegativity • Electronegativity increases across a period and decreases down a group

28. •• •• •• •• Cl •• Cl • •• •• •• Cl •• •• •• •• Cl • •• Covalent bonds: electron sharing • Nonmetals share electrons to form covalent bonds: • Each Cl has 7 valence e–, wants to gain 1 e– to get an octet like Ar • Neither atom is willing to give up an electron • Solution is to share their unpaired electrons • The shared pair of e– is a single covalent bond 

29. •• •• •• •• •• •• Cl Cl •• Cl •• •• •• •• •• Cl •• Covalent bonds • Covalent bonds form between nonmetals • Nonmetals share unpaired valence electrons • Each atom “owns” all of its bond electrons • Each atom achieves an octet (or 2, for hydrogen) • One shared pair of electrons is shown with a single line or ––

30. •• •• • O • Bonding in O2 • Oxygen is a diatomic element • Each unpaired electron must get into a bond • The atoms cannot achieve a filled valence level with a single bond, so a double bond forms: they share two pairs of electrons •• •• •• •• •• •• = •• •• •• •• •• •• O O O O •  O or •

31. How to draw a dot structure • Count valence e– • Put atom with lowest EN in center • Arrange other atoms around it symmetrically • Form single bonds between atoms (1 line = 2 e–) • Put lone pairs around terminal atoms to give each an octet (2 for H), then finish central atom octet • If central atom does not get octet, move in lone pairs to make double or triple bonds

32. Lewis dot structures • Draw a Lewis dot structure for each species • NH3 HBr CO2 • OH1– NH41+ HCN • H2CO NO31– PF3 • SO2 C2H4 Cl2O

33. How equally do atoms share e– ? • Hydrogen and fluorine share one pair of e– in a single covalent bond • In the Lewis dot structure, they appear to share the electrons equally, but do they?

34. How equally do atoms share e– ? • When you put HF in an electric field, the molecules line up, as if the F end were negative and the H end positive. • The electrons are shared unequally.

35. How equally do atoms share e– ? • When bond e– are shared unequally, the bond is said to be a polar covalent bond. • A polar covalent bond has a dipole: one end is more negative than the other end

36. Electronegativity and bond polarity • Bond polarity depends on the difference in EN

37. Evaluating bond type • H–Br • H = 2.1, Br = 2.8 • ∆EN = 0.7 • Bond is polar covalent with Br end more negative H–Br–

38. Evaluating bond type • C = O • C = 2.5, O = 3.5 • ∆EN = 1.0 • Bond is polar covalent with O end more negative • Doesn’t matter whether bond is single or double C=O–

39. Evaluating bond type • KCl • K = 0.8, Cl = 3.0 • ∆EN = 2.2 • Bond is ionic; e– transferred from K to Cl K1+ Cl1–