Ch. 5 - The Periodic Table

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Ch. 5 - The Periodic Table 250 Atomic Radius (pm) 200 150 100 50 0 0 5 10 15 20 Atomic Number III. Periodic Trends (p. 140-154) I II III

A. Periodic Law When elements are arranged in order of increasing atomic #, elements with similar properties appear at regular intervals. 250 Atomic Radius (pm) 200 150 100 50 0 0 5 10 15 20 Atomic Number

B. Reaction Patterns Valence electrons: main group (s & p) electrons in the outermost energy level O: 2s 2 2p 4 = 6 valence electrons S: 3s 2 3p 4 = 6 valence electrons Se: 4s 2 3d 10 4p 4 = 6 valence electrons!! Similar valence e - within a GROUP result in similar chemical properties in the group.

B. Reaction Patterns Only main group electrons are in the valence This means that 8 is the largest number of valence electrons possible. (2 + 6 = 8) This is where the term full octet comes from. A full octet is the most stable type of electron configuration. The noble gases are the only elements that start with full octets.

B. Reaction Patterns # valence electrons increases, left to right: 1 2 3 4 5 6 7 8 Valence #s go by column, which is why properties go by column.

B. Reaction Patterns Ions: charged atoms created when electrons are gained or lost to create a full octet (noble) configuration. 1+ 2+ 3+ 4± 3-2- 1-0 These common ion charges are based on valence numbers.

Periodic Trends Periodic trends are due to: The organization of electrons (in energy levels, sublevels and orbitals due to the laws of quantum mechanics) The fact that opposites attract: +- The nucleus (positive charge) attracts the electron cloud (negative charge)

Periodic Trends Attraction between charged particles can be mathematically calculated with Coulomb s Law: F elec = kq 1 q 2 r 2 We re not going to solve this equation, but it is the basis of most periodic trends, so let s take a look: F elec = attraction (force) k = a constant r = radius (distance from nucleus to electrons) q 1 = positive charge (protons) q 2 = negative charge (electrons)

Periodic Trends Attraction between charged particles can be mathematically calculated with Coulomb s Law: F elec = kq 1 q 2 r 2 The equation is saying two things: More protons & electrons = more attraction Smaller atom = more attraction

Periodic Trends Vertical patterns: Going down a column, energy levels are being added. For a higher energy level, the valence electrons are at greater distance (r) from the nucleus (Higher energy levels = larger orbitals) The added distance decreases attractive force. 1 2 3 4 5 6 7

Periodic Trends Vertical patterns: When comparing elements in the same GROUP (column/family) Elements at the top of a group (lower atomic number) hold onto their valence electrons tightly. Elements at the bottom of a group (greater atomic number) hold onto their valence electrons loosely. 1 2 3 4 5 6 7

Let me explain you a thing This next section, your job is to listen and UNDERSTAND, not to write down.

Periodic Trends It s all about the Valence Electrons Only outermost (valence) electrons are involved in reactions. Reactions happen because an atom doesn t have a stable (full octet) configuration. Mg: [Ne-10] 3s 2 But actually the inner electrons are always stable already. Only the valence electrons need work.

Periodic Trends Inner electrons are already stable, they don t do anything, let s ignore them. Valence electrons behavior (and element properties) are based on their attraction to the protons in the nucleus. Valence electrons don t feel the attraction of all the positive charge in the nucleus. The valence electrons are shielded from the nucleus by the inner electrons. Mg: [Ne-10] 3s 2

Periodic Trends Mg: [Ne-10] 3s 2 The ten inner electrons cancel the charge of ten protons. Mg: [1s 2 2s 2 2p 6 ] 3s 2 Only two protons are effective Z eff = + 2 10p 2p + 1 st energy level shield: 1s 2 2 nd E.L. shield: 2s 2 2p 6 Result: Z eff = valence Z eff = effective nuclear charge (That s important.)

Periodic Trends Conclusion: The shielding effect causes the values of q 1 and q 2 to count from 1 to 8 (based on valence number) for every period. 1 2 3 4 5 6 7 8 Which is why patterns restart every period even though proton number doesn t restart.

Periodic Trends Horizontal patterns: Going across a period, valence electrons and Z eff increase. (q 1 and q 2 ) F elec = kq 1q 2 r 2 1 2 3 4 5 6 7 8 Greater charge (q 1 & q 2 ) increases attractive force. In a period, higher atomic number = greater attraction between nucleus and electron cloud.

Periodic Trends Horizontal patterns: 1 2 3 4 5 6 7 8 When comparing elements in the same PERIOD (row) Elements at the beginning of a period (lower atomic number) hold onto their valence electrons loosely. Elements at the end of a period (greater atomic number) hold onto their valence electrons tightly.

Periodic Trends Rule of thumb: Up and to the right, electrons are held more tightly. Lower and to the left, electrons are held more loosely.

D. Atomic Radius Atomic Radius Atomic Radius (pm) 250 200 150 100 50 Li Na Ne Ar K 0 0 5 10 15 20 C. Johannesson Atomic Number

D. Atomic Radius Atomic Radius: SIZE the distance from nucleus to edge of electron cloud Or: half the distance between two adjacent nuclei

D. Atomic Radius Increases to the LEFT in a period Increases DOWN in a group

D. Atomic Radius Why larger going down? More energy levels Why smaller to the right? Increased nuclear charge (without additional shielding) pulls e - in tighter

D. Ionization Energy Ionization Energy: Energy required to remove one e - from a neutral atom. The more stable an atom is, the more energy is required to ionize it. i.e., the closer to a full octet, the higher the ionization energy.

E. Ionization Energy First Ionization Energy 1st Ionization Energy (kj) 2500 2000 1500 1000 500 0 He Li Ne Na Ar K 0 5 10 15 20 Atomic Number

E. Ionization Energy (First) Ionization Energy Increases to the right in a period Increases to the top in a group

E. Ionization Energy Why is this opposite of atomic radius? In small atoms, e - are close to the nucleus where the attraction is stronger Stronger attraction means more difficult to remove electrons from. Why small jumps within each group? Stable e - configurations (full and half-full sublevels) don t want to lose electrons

E. Ionization Energy First Ionization Energy 2500 1st Ionization Energy (kj) 2000 1500 1000 500 0 Full sublevel Half-full sublevel 0 5 10 15 20 Atomic Number

E. Ionization Energy Successive Ionization Energies Large jump in I.E. occurs when a CORE (non-valence) e - is removed. Mg = [Ne-10] 3s 2 1st I.E. 2nd I.E. 736 kj 1,445 kj Core e - 3rd I.E. 7,730 kj

E. Ionization Energy Successive Ionization Energies Where is the jump for aluminum? Al = [Ne-10] 3s 2 3p 1 3 v.e. means it should jump between 3rd & 4th ionizations. Al 1st I.E. 577 kj 2nd I.E. 1,815 kj 3rd I.E. 2,740 kj Core e - 4th I.E. 11,600 kj

G. Ionic Radius Ionic Radius Cations (+) Created by losing e - smaller than parent atom Anions ( ) Created by gaining e - larger than parent atom

Examples Which atom has the larger radius? Be or Ba Ba Ca or Br Ca

Examples Which atom has the higher 1st I.E.? N or Bi N Ba or Ne Ne

Examples Which particle has the larger radius? S or S 2- S 2- Al or Al 3+ Al