Periodic Relationships 1
Tabulation of Elements Mendeleev (1869) Arranged by mass Tabulation by chem.& physical properties Predicted missing elements and properties 2
Modern Periodic Table Argon vs. potassium problem. Now ordered by atomic number, not mass. 3
Periodic Table Representative elements (1,2,13-17) -incomplete s or p subshells Noble gases (18) -filled p subshells (except He) -ns 2 np 6 4
Periodic Table Transition elements (groups 3-11) -incompletely filled d subshell -d block Group 12 Lanthanides & actinides -incompletely filled f subshells -f block 5
Groups Similar properties of a group result from similar electron configuration. e.g. Group 1: ns 1 Valence e - Group 2 : ns 2 Group 17: ns 2 np 5 (e - in highest n ) Valence e - : key to bonding; determines chemical properties. 6
Writing Free Elements Metals: use empirical formula (Fe) H 2, N 2, O 2, F 2, Cl 2, I 2 : diatomic Phosphorus : use P 4 Carbon: use empirical formula, C Sulfur (S 8 ) use S Metalloids: use empirical formula 7
e - Configuration of Ions Representative Elements: ions have noble gas e - configuration Na: [Ne]3s 1 Na + [Ne] Ca: [Ar]4s 2 Ca 2+ [Ar] F: 1s 2 2s 2 2p 5 F - [Ne] Note: F - and Na + are isoelectronic 8
Transition Element Cations Aufbau: e - fill ns prior to (n-1)d but Ion: ns removed before (n-1)d Mn: [Ar]4s 2 3d 5 Mn 2+ : [Ar]3d 5. WEIRD. valence e - removed first 9
Transition Metal Ions ns removed before (n- 1)d is one reason that transition elements are similar. Fe 2+, Mn 2+, Co 2+ 10
Periodic Trends 1. Atomic radius 2. Ionic radius 3. Ionization energy 4. Electronegativity Caution: Atoms are not anthropomorphic. 11
Atomic Radius: Trends 12
Atomic Radius Down a group: electrons are placed in higher energy levels (farther from the nucleus). Radius increases down a group. 13
Remember Shielding? Li atom: 1s 2 2s 1 - +3 - - shield outer e - from protons 14
Effective Nuclear Charge Due to shielding, nuclear charge (Z) felt by valence e - is reduced. Z eff = Z s where s is the shielding constant, roughly the number of inner shell e - For Li: Z eff = Z s = 3 ~2 = ~1 15
Shielding: Coulomb s Law e - Z eff F = q 1 q 2 r 2 Since Z eff (q 2 ) is less than Z, the force attracting the e - is less. 16
Shielding by Inner e - Li atom - +3 - - has Z eff less than +3 So Li is larger than expected. 17
Atomic Radius: Trends Across row: decreasing size Z increases & number of shielding e - constant (only inner shell e - shield) Li - vs. - - +3 Be +4 - - (>Z eff ) - - 18
Atomic Radius Model 19
Atomic Radius Try It: Arrange these atoms in order of increasing size. N, O, P, S 20
Ionic Radius: Trends Cation is much smaller than its atom. (loss of an electron shell) Anion is much larger than its atom. (gain of e - with same Z eff ) Li + F Li + F - 1s 2 2s 1 1s 2 2s 2 2p 5 1s 2 1s 2 2s 2 2p 6 21
Ionic Radii (model) Compare isoelectronic ions Na + & F - 22
Ionic Radii: Comparisons For isoelectronic ions: Al 3+ Mg 2+ Na + = [Ne] Larger (WHY?) N 3- O 2- F - Smaller WHY? = [Ne] 23
Ionic Radii Place in order of increasing size? Fe, Fe 2+, and Fe 3+ 24
Ionic Radii Try It: Arrange these ions in order of increasing size. O -2, F -, Na +, Mg 2 + 25
Ionization Energy Chemical properties are determine by the valence electrons. Ionization energy: energy (kj/mol) to remove e - from gaseous atom in ground state. If IE high, e - held tightly. 26
Ionization Energy I 1 energy + X(g) X + (g) + e - endo- or exothermic? I 2 energy + X + (g) X 2+ (g) + e - I 3 energy + X 2+ (g) X 3+ (g) + e - 27
Ionization Energy IE always endothermic (+) (energy put in to remove e - ) I 1 < I 2 < I 3 Why? Coulomb s law I explains Na +, Mg 2+, etc. (after Na loses 1 e -, the number of shielding e - drops from 8 to 2) 28
Ionization Energies explain the electron shell model of the atom 29
I 1 vs. Z I 1 I 1 across Period down Group I 1 30
Ionization Energy Trends Across a row (I 1 increases): Explain I in terms of shielding and Z. 1 eff Na vs. Mg Down a group (I 1 decreases): Explain I in terms of energy levels (n). 1 Na vs. K Explain both using Coulomb s law. 31
I 1 Irregularities: Be vs. B Be: 899 kj/mol 2s 2 B: 801 kj/mol 2s 2 2p 1 Partial Shielding: 2s (lower energy) slightly shields 2p. 32
I 1 Irregularities: N vs. O N: 1400 kj/mol 2s 2 2p 3 O: 1314 kj/mol 2s 2 2p 4 e - repulsion: O is lower than N due to e - repulsion in same orbital. 33
Ionization Energy Which has smaller I 1 and why? O or S Al or Si P or S Which has smaller I 2 and why? Li + or Be + 34
Electronegativity A relative scale to measure the ability of an atom to attract bonding electrons. Trend parallels ionization energy. 35
Electronegativity High EN: F (4.0) O (3.5) Low EN: Na (0.9) Ba (0.9) 36
Metallic Character Metals lose e - to become cations. Which element is the most metallic? (smallest ionization energy) Nonmetals gain e - to become anions. Which element is the least metallic? (largest ionization energy) 37
Especially note: trends and why shielding characteristic reactions Section 8.6 38