The how and why

History Dmitri Mendeleev u Russian scientist Dmitri Mendeleev taught chemistry in terms of properties. u Mid molar masses of elements were known. u Wrote down the elements in order of increasing mass (this is wrong!!!!) u Found a pattern of repeating properties.

Mendeleev’s Table u Grouped elements in columns by similar properties in order of increasing atomic mass. u Found some inconsistencies - felt that the properties were more important than the mass, so switched order. u Found some gaps. u Must be undiscovered elements. u Predicted their properties before they were found.

The modern table u Elements are still grouped by properties. u Similar properties are in the same column. u Order is in increasing atomic number. u Added a column of elements Mendeleev didn’t know about. u The noble gases weren’t found because they didn’t react with anything.

u Horizontal rows are called Periods u There are 7 periods

u Vertical columns are called Groups. u Elements are placed in columns by similar properties. u Also called families

1A 2A3A4A5A6A 7A 8A 0 u The elements in the A groups are called the representative elements

The group B are called the transition elements u These are called the inner transition elements and they belong here

Periodicity Explained u Valence electron cloud u Outside orbitals u The orbitals fill up in a regular pattern u The outside orbital electron configuration repeats u The properties of atoms repeat when placed in order of Atomic Number

1s11s1 1s 2 2s 1 1s 2 2s 2 2p 6 3s 1 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 1 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 6 7s 1 H 1 Li 3 Na 11 K 19 Rb 37 Cs 55 Fr 87

He 2 Ne 10 Ar 18 Kr 36 Xe 54 Rn 86 1s21s2 1s 2 2s 2 2p 6 1s 2 2s 2 2p 6 3s 2 3p 6 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 6

u Alkali metals all end in s 1 u Alkaline earth metals all end in s 2 u really have to include He but it fits better later. u He has the properties of the noble gases. s2s2 s1s1 s- block

Transition Metals -d block d1d1 d2d2 d3d3 d 5 d5d5 d6d6 d7d7 d8d8 d 9 d 10

The p-block p1p1 p2p2 p3p3 p4p4 p5p5 p6p6

f - block u inner transition elements

Atomic Size u First problem where do you start measuring. u The electron cloud doesn’t have a definite edge. u They get around this by measuring more than 1 atom at a time.

Atomic Size u Atomic Radius = half the distance between two nuclei of a diatomic molecule. } Radius

Trends in Atomic Size u Influenced by two factors. u Energy Level u Higher energy level - electrons are further away from nucleus. u Charge on nucleus – Z eff u More charge (# of protons) pulls electrons in closer.

Group trends u As we go down a group u Each atom has another energy level u So the atoms get bigger. H Li Na K Rb

Periodic Trends u As you go across a period the radius gets smaller. u Same energy level. u More nuclear charge = more protons. u Outermost electrons are closer. NaMgAlSiPSClAr

Atomic Number Atomic Radius (nm) H Li Ne Ar 10 Na K Kr Rb

Ionization Energy u The amount of energy required to completely remove an electron from a gaseous atom. u Removing one electron makes a +1 ion. u The energy required is called the first ionization energy.

Ionization Energy u The second ionization energy is the energy required to remove the second electron. u Always greater than first IE. u The third IE is the energy required to remove a third electron.  IE1<IE2<IE3

SymbolFirstSecond Third H He Li Be B C N O F Ne

SymbolFirstSecond Third H He Li Be B C N O F Ne

What determines IE u  nuclear charge =  IE u  distance from nucleus =  IE u Filled and half filled orbitals have lower energy, so achieving them is easier, lower IE. u Shielding

Periodic trends: Ionization Energy (IE) u All the atoms in the same period have the same energy level. u Same energy level = same shielding, but each atom gains a proton, therefore…. u Increasing nuclear charge …helps pull e - in tighter, therefore it is harder to remove. u So IE generally increases from left to right. u Exceptions at full and 1/2 fill orbitals. e-e- e-e-

Periodic trends: Ionization Energy (IE) u As you go down a group first IE decreases because…. u The electron is further away…and u There are energy levels between the nucleus and the e- thus…. u Shielding the e- from + nucleus. e-e- e-e-

Shielding u The electron on the outside energy level has to look through all the other energy levels to see the nucleus

Shielding u The electron on the outside energy level has to look through all the other energy levels to see the nucleus. u A second electron has the same shielding.

Effective Nuclear Charge, Z eff u AtomZ eff Experienced by Electrons in Valence Orbitals u Li+1.28 u Be u B+2.58 u C+3.22 u N+3.85 u O+4.49 u F+5.13 Increase in Z* across a period

First Ionization energy Atomic number He u He has a greater IE than H. u same shielding u greater nuclear charge H

First Ionization energy Atomic number H He l Li has lower IE than H l more shielding l further away l outweighs greater nuclear charge Li

First Ionization energy Atomic number H He l Be has higher IE than Li l same shielding l greater nuclear charge Li Be

First Ionization energy Atomic number H He l B has lower IE than Be l same shielding l greater nuclear charge l By removing an electron we make s orbital half filled Li Be B

First Ionization energy Atomic number H He Li Be B C

First Ionization energy Atomic number H He Li Be B C N

First Ionization energy Atomic number H He Li Be B C N O u Breaks the pattern because removing an electron gets to 1/2 filled p orbital

First Ionization energy Atomic number H He Li Be B C N O F

First Ionization energy Atomic number H He Li Be B C N O F Ne u Ne has a lower IE than He u Both are full, u Ne has more shielding u Greater distance

First Ionization energy Atomic number H He Li Be B C N O F Ne l Na has a lower IE than Li l Both are s 1 l Na has more shielding l Greater distance Na

First Ionization energy Atomic number

Driving Force u Full Energy Levels are very low energy. u Noble Gases have full orbitals. u Atoms behave in ways to achieve noble gas configuration.

Electron Affinity u The energy change associated with adding an electron to a gaseous atom. u Easiest to add to group 7A. u Gets them to full energy level. u Increase from left to right atoms become smaller, with greater nuclear charge. u Decrease as we go down a group.

Periodic trends: Electron Affinity(EA) e-e- e-e- From left to right atoms become smaller, with greater nuclear charge. e - are attracted by the + charged nucleus. therefore EA will increase from left to right

Periodic trends: Electron Affinity(EA) e-e- e-e- Down a group atoms become larger, and have greater nuclear charge. e - are attracted by the increased +charge but shielding also increases …which has a greater influence!! therefore EA will decrease as you go down a group or family

The energy change associated with adding an electron to a gaseous atom. e - are attracted by a + charge. –Therefore more protons more attraction –But remember “shielding” Easiest to add to group 7A. Gets them to full energy level. Increase from left to right atoms become smaller, with greater nuclear charge. Decrease as we go down a group. Periodic trends: Electron Affinity(EA)

Ionization Energy & Electron Affinity Effective Nuclear Charge INCREASE

Atomic Radius & Shielding INCREASES

Size of Isoelectronic ions u Iso - same u Iso electronic ions have the same # of electrons u Al +3, Mg +2, Na +1, Ne, F -1, O -2 and N -3 u all have 10 electrons u all have the configuration 1s 1 2s 2 2p 6

Size of Isoelectronic ions u Positvie ions have more protons so they are smaller. Al +3 Mg +2 Na +1 Ne F -1 O -2 N -3

Na 2 O (s) + H 2 O (l)  2 NaOH (aq) CaO (s) + H 2 O (l)  Ca(OH) 2 (aq) MgO (s) + 2HCl (aq)  MgCl 2(aq) + H 2 O (l) NiO (s) + H 2 SO 4 (aq)  NiSO 4 (aq) + H 2 O (l) Metal Oxides are BASIC

CO 2(g) + H 2 O (l)  H 2 CO 3(aq) P 4 O 10(s) + 6 H 2 O (l)  4 H 3 PO 4 (aq) CO 2(g) + 2 NaOH (aq)  Na 2 CO 3(aq) + H 2 O (l) SO 3(g) + 2 KOH (aq)  K 2 SO 4 (aq) + H 2 O (l) Non-Metal Oxides are ACIDIC