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Chapter 7 Periodicity Chemistry, The Central Science, 10th edition

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Presentation on theme: "Chapter 7 Periodicity Chemistry, The Central Science, 10th edition"— Presentation transcript:

1 Chapter 7 Periodicity Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 7 Periodicity John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc. “Tweaked” by Robert Hernandez Seabreeze Highschool 2015

2 Match the constants with their value. Which does NOT have units?
Avagadro’s # Planck’s Constant Speed of Light Faraday’s Constant ___ x 10-34 ___ 3.00 x 108 ___ 96,500 ___ x 1023 ? B m2 kg / s C m/s D C/mol A

3 Development of Periodic Table
Elements in the same group generally have similar chemical properties. (though not identical)

4 “Father” of the Periodic Table
“The elements, if arranged according to their atomic weights, exhibit an apparent periodicity of properties.” Dimitri Mendeleev Arranged elements by atomic weight. The FIRST periodic table 

5 Development of Periodic Table
Mendeleev predicted the discovery of germanium (which he called eka-silicon) as an element with an atomic weight between that of zinc and arsenic, but with chemical properties similar to those of silicon.

6 Smart uncle who fixed the mistakes of the “Father” of the Periodic Table
..The insides of all the atoms are very much alike, and from these results it will be possible to find out something of what the insides are made up of.  Henry Moseley Rearranged Mendeleev’s table by atomic number. Moseley’s spectrum research 

7 Periodic Trends Recognize the underlying reasons explaining observed trends in… Effective Nuclear Charge (shielding) 2. Atomic and Ionic Size. 3. Ionization energy 4. Electron affinity (& Electronegativity)

8 Periodic Trends Driving Forces
Attract Repel + - - -

9 Periodic Trends Driving Forces
DISTANCE CHARGE Strength - - Strength + + - + +

10 1. Shielding (Effective Nuclear Charge)
shielding effect: inner electrons protect the outer electrons from the nuclear charge. effective nuclear charge, (Zeff): Z is atomic number S is a shielding constant across a period Zeff = Z − S Na atom - Valence electron (-1) Outer electron 10- [Ne] core (10-) Inner electrons 11+ Nucleus (11+) Zeff = Z − S 1 11 10

11 1. Shielding (Effective Nuclear Charge)
Least Most

12 decreases across a period
2. Sizes of Atoms decreases across a period Because Zeff increases (more protons) Because energy levels (n) increase increases down a group

13 2. Sizes of Atoms Smallest Largest
The DOWNWARD increase is much more dramatic than ACROSS.

14 Rank these atoms from SMALLEST to LARGEST
___ ___  LARGEST ? F O Cl S

15 2a. Sizes of Ions (in pm) - Cations are smaller than their neutral atoms, because… outermost electron(s) are removed and the atom loses a shell excess protons (greater Zeff) 156 60 Li Li+ - 191 95 Na Na+ 238 133 - K K+

16 2a. Sizes of Ions Anions are larger than their neutral atoms, because…
(in pm) Anions are larger than their neutral atoms, because… electrons are added and repulsions are increased. excess electrons (less Zeff) F- - F 62 133 Cl- - Cl 102 181 Br - - Br 120 196

17 decreases across a period
2a. Sizes of Ions decreases across a period increases down a group

18 2a. Sizes of Ions Smallest Largest

19 K, K+, Na, Na+ Rank these Atoms and ions from SMALLEST to LARGEST ?
___ ___  LARGEST ? Na+ Na K+ K

20 F, F-, Cl, Cl- Rank these ATOMS and IONS from SMALLEST to LARGEST ?
___ ___  LARGEST ? F F- Cl Cl-

21 3. Ionization Energy (IE)
Amount of energy required to remove an electron from the ground state of a gaseous atom or ion. First ionization energy is that energy required to remove first electron. Second ionization energy is that energy required to remove second electron, etc.

22 3. Ionization Energy (IE)
It requires more energy to remove each successive electron. When all valence electrons have been removed, the IE takes a huge jump. WHY?

23 Trends in FIRST Ionization Energy (IE) increases across a period
WHY? decreases down a group

24 Trends in First IE IE tends to… .. increase across a period
More Zeff, no extra shielding .. decrease down a group more energy levels cause more shielding from inner electrons as valence electrons are farther from the nucleus.

25 notice Boron (5) and Oxygen (8)

26 IE and Boron’s “dip” Be B 1s 2s 2p 1s 2s 2p
4 Be 9.0 1s 2s 2p 5 B 10.8 1s 2s 2p Removing an electron from a “p” orbital is easier than from the tighter “s” orbital. 26

27 IE and Oxygen’s “dip” N O 1s 2s 2p 1s 2s 2p
7 N 14.0 1s 2s 2p 8 O 16.0 1s 2s 2p Removing an electron from a filled “p” orbital is easier because of the other electron’s repulsion. 27

28 3. Ionization Energy (IE)
Fluorine has the HIGHEST IE Largest F Smallest

29 4. Electron Affinity Energy change when; 1 mol of gaseous atoms…
- - Energy change when; 1 mol of gaseous atoms… gains 1 mol of electrons… to form 1 mol of gaseous ions. Cl(g) + e−  Cl−(g) ∆E = -349 kJ/mol - - - - -

30 4. Electron Affinity Cl(g) + e−  Cl−(g) ∆E = -349 kJ/mol
∆E > 0 (+) endothermic (unstable, doesn’t form) ∆E < 0 (–) exothermic (more –, more stable) DE = Ef − Ei 30

31 Trends in Electron Affinity (EA)
Generally, electron affinity becomes more exothermic across a period. (many exceptions caused by orbital issues) more neg. across a period less neg. down a group

32 Fluorine has the HIGHEST EA too.
4. Electron Affinity .. MOST NEGATIVE Fluorine has the HIGHEST EA too. F LEAST Negative…

33 Electron Affinity Ionization Energy
IE measures the ease with which an atom loses an electron. E always +, lower is easier. EA measures the ease with which an atom gains an electron. E > 0 or E < 0 (–), more – is easier. (more exothermic)

34 Trends in Electronegativity (EN)
ability of an atom to attract electrons when bonded with another atom. increases across a period decreases down a group

35 Fluorine has the HIGHEST EN .
Electronegativity MOST Fluorine has the HIGHEST EN . F LEAST

36 Group Trends

37 Group 1: Alkali Metals Soft, metallic solids. Found only as compounds.
Low densities and melting points. Lowest ionization energies. Reactivity tends to increase as go down a group. Why low IE’s?

38 Group 1: Alkali Metals Their reactions with water are famously exothermic. Produce bright colors when placed in flame.

39 Group 2: Alkaline Earth Metals
Higher densities and melting points than alkali metals. Low ionization energies, but not as low as alkali metals.

40 Group 2: Alkaline Earth Metals
Be does not react with water, Mg reacts only with steam, but others react readily with water. Reactivity tends to increase as go down a group.

41 Group 17 (VIIA): Halogens
Large, negative EA Large EN Therefore, tend to oxidize (take e-) other elements easily

42 Group 18 (VIIIA): Noble Gases
Huge ionization energies Positive EA (zero EN) Therefore, relatively unreactive Monatomic gases

43 Group 18 (VIIIA): Noble Gases
Xe forms three compounds: XeF2 XeF4 (at right) XeF6 Kr forms only one stable compound: KrF2 The unstable HArF was synthesized in 2000.

44 Properties of Metal, Nonmetals, and Metalloids

45 Metals vs. Nonmetals Differences between metals and nonmetals tend to revolve around these properties.

46 Metals Compounds formed between metals and nonmetals tend to be ionic.
Metal oxides tend to be basic in water.

47 Nonmetals Compounds formed between nonmetals only are molecular (covalent) compounds. Most nonmetal oxides are acidic in water.

48 Metalloids Have some characteristics of metals, some of nonmetals.
For instance, silicon looks shiny, but is brittle and is a semi-conductor.

49 And, Finally… PES Z Dispense, I do.

50 Photoelectron Spectroscopy (PES)
1. Which peak is H and which is He? higher peak = more e–’s 1s2 Relative # of e–’s He 1s1 H Binding Energy (eV or MJ)  further left = more energy required (stronger attraction due to more protons)

51 Photoelectron Spectroscopy (PES)
2. What element is this? 2p6 Ne 1s2 Relative # of e–’s He 1s1 1s2 2s2 H Binding Energy Draw this spike line on Graph . Watch the proportions. (eV or MJ)

52 PES (A) PES (B) 3d10 2p6 3p6 1s2 2s2 3s2 4s2 4p2 Ge n = 1 n = 2 n = 3
Identify element (A) 3d10 2p6 3p6 1s2 2s2 3s2 4s2 4p2 Ge n = 1 n = 2 n = 3 n = 4 PES (B) Identify element (B) 4s1 ? K

53 4. Write the complete electron configuration of element (X), and identify the element.
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1 Ga 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1 PES worksheet PES (X) 3d10 2p6 3p6 4s2 1s2 2s2 4p1 3s2


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