Chapter 8 Periodicity 2010

EM radiation Announcements is a wave and a --Exam 3 Oct 3...... Includes chapters 7/8/9/10 units called particle The excluded items include: 1. Classical distinction between energy and matter (p. 217) are Quant having 2. Numerical problems involving the Rydberg equation (equations a 7.3 and 7.4) absorbe emitted amplitude energy frequency wavelength 3. Spectral analysis in the laboratory (pp. 226-227) d 4. Numerical problems involving the Heisenberg uncertainty involve energy changes in related by related by principle (p. 231) 5. Trends among the transition elements (p. 261) atomsatoms c = !v 6. Trends in electron affinity (pp. 265-266) electron E = hv 7. Pseudo-noble gas configuration (p. 269) s molecules 8. Lattice energy (pp. 283-285) described by 9. IR spectroscopy (p. 292) e- ﬁlling e- conﬁguration comprising 10.Numerical problems involving electronegativity (p. 296) Wave functions gives 11.Electronegativity and oxidation number (p. 297) 12.Section 11.3: MO theory and electron delocalization having quantum determined by 13.All sections in chapter 12 except 12.3 uunmbers Aufbau Rules

Quantum described by Core Numbers Electrons Wave Function e- filling spdf electronic configuration comprising Orbital size (Orbital) Valence Principal define & energy determined by Electrons n = 1,2,3,.. s

described by Aufbau basis for defines Rules Periodic Quantum which involve Table Angular Numbers momentum, l Orbital which summarizes defines Oribital Pauli Hund’s shape which are Energy Exclusion Rule Periodic Properties Magnetic Orbital defines ml orientation

Electron Spin, ms deﬁnes spin

I. The Periodic Law and the Periodic Table

• 1864 Newland “Law of Octaves

• 1869 Dimitri Mendeleev and Lother Meyer When the elements are arranged in order of increasing atomic mass, certain sets of chemical and physical properties recur periodically. Electronic Configuration and Periodicity • 1913 Henry Mosely relates X-ray frequency to Chapter 8 atomic number When the Elements Were Discovered Quantum numbers (n,l,ml,ms) specify “allowed states” or “orbitals” which are regions of space where electrons are likely to be found around the nucleus.

1. Principal Quantum Number (n): Defines the size and energy level of the orbital. n = {1,2,3,4,.....}. Also called a shell (K = 1, L = 2, M = 3, N = 4, .....).

2. Angular Momemtum Quantum Number (l): Defines the “shape” of the orbital which is a volume in space where the electron is likely to be found. Also called a subshell. l = {0,1,2,3...up to n-1} where (0=s, 1=p, 2=d, 3=f)

3. Magnetic Quantum Number (ml): Defines the spatial orientation of an orbital of the same energy. ml = {-l, 0, +l}

4. Magnetic Spin Quantum Number (ms): Defines the orientation of “electron spin”. ms = {+1/2 or -1/2}.

Electronic configuration of the elements: The lowest energy (ground state) four quantum numbers describe an electron electronic configuration of all in a ground state atom. elements are constructed by filling lowest energy orbitals sequentially in what is called the “Aufbau Process”. 4s Name Symbol Permitted Values Property 3p 1. Lower energy (n-quantum number) principal n positive integers (1,2,3, orbital energy (size) orbitals fill first. 3s !) 2. Hund’s Rule-orbitals fill one electron at angular l 2p momentum integers from 0 to n-1 orbital shape a time before electrons are paired. 3. Pauli Exclusion Principle: No two orbital orientation in 2s magnetic ml integers from -l to 0 to electrons can have same 4-quantum +l space numbers)

- spin ms +1/2 or -1/2 direction of e spin Electrons ﬁll the lowest energy orbitals ﬁrst, 2 at a time! 1s

The order of filling of the orbitals can be Chemists use spdf notation and orbital box remembered using a mnemonic device. diagrams to denote or show the “ground state Memorize this to help you! electronic configuration” of elements.

For an Hydrogen atom spdf orbital box Spin quantum orbital energy only depends Element Notation diagram number. An on the n quantum number. arrow denotes 1 an electron For many electrons atoms H 1s with “spin the energy of an orbital or up” (+1/2) or electron depends on both “spin- n and l (3s < 3p < 3d) He 1s2 down” (-1/2).

n principal quantum #

l quantum number # of electrons in orbital The Pauli Exclusion principle states: “No two The order of filling of the orbitals can be remembered electrons can have the same 4-quantum numbers”. using a mnemonic device. Memorize how to write it The spin numbers can not be the same (spin up out as it determines electronic structure. and spin down allowed only).

(n, l, ml and ms)

Example:

Atomic Orbital Box Full-electronic Condensed-electronic Number/Element Diagram configuration configuration

Li 1s22s1 [He]2s1

Electronic configuration using Aufbau Process s-block Atomic Orbital Box Full-electronic Condensed-electronic main group p-block Number/Element Diagram configuration configuration main group d-block H 1s1 1s1 transition metals He 1s2 1s2 f-block written with noble inner transition metals gas configuration Li 1s22s1 [He]2s1

Be 1s22s2 [He]2s2

Atomic Orbital Box Full-electronic Condensed-electronic Number/Element Diagram configuration configuration

B 1s22s22p1 [He]2s22p1

2 2 2 2 2 C 1s 2s 2p [He]2s 2p

1s22s22p3 [He]2s22p3

1s22s22p4 [He]2s22p4 1s22s22p5 [He]2s22p5 1s22s22p6 [He]2s22p6 Unpaired electrons in orbitals gives rise to Odd-filling behavior here! paramagnetism and is attracted to a magnetic field. 4th and 9th position. Diamagnetic species contain all paired electrons and is “repelled” by the magnetic field.

• Diamagnetic atoms or ions: Diamagnetic – All e- are paired. all electrons paired – Weakly repelled in a magnetic field. 2p • Paramagnetic atoms or ions: Paramagnetic – Unpaired e- exist in an orbital unpaired electrons – Attracted to an external magnetic field. 2p

Unpaired electrons in orbitals gives rise to When a cation is formed from an atom of a paramagnetism and is attracted to a magnetic field. transition metal, electrons are removed first from Diamagnetic species contain all paired electrons and the ns orbital, then from the (n-1)d orbital. is “repelled” by the magnetic field.

Magnetic field off Magnetic field on Magnetic field on Fe: [Ar]4s23d6 Fe2+: [Ar]4s03d6 or [Ar]3d6 Fe: [Ar]4s23d6 Fe3+: [Ar]4s03d5 or [Ar]3d5 Mn: [Ar]4s23d5 Mn2+: [Ar]4s03d5 or [Ar]3d5 Paramagentic Diamagentic

Paramagnetic Diamagnetic Metals loose electrons (oxidized) to become Isoelectronic species are two different elements cations. Non-metals gain electrons to become with the same electronic configuration--but not the anions. The electronic configuration of each same nuclear configuration. reflects this change in the number of electrons. oxidation Na: [1s22s22p63s1] =====> Na+: [1s22s22p6] = [Ne] Na [Ne]3s1 Na+ [Ne] Metals lose electrons so oxidation 2 2 6 2 p1 3+ 2 2 6 Ca [Ar]4s2 Ca2+ [Ar] that cation has a noble-gas Al: [1s 2s 2p 3s 3 ] =====> Al : [1s 2s 2p ] = [Ne] 2 1 3+ outer electron configuration. reduced Al [Ne]3s 3p Al [Ne] N: [1s22s22p3] =====> N3-: [1s22s22p6] = [Ne] reduced O: [1s22s22p4] =====> O2-: [1s22s22p6] = [Ne] H 1s1 H- 1s2 or [He] reduced Non-metals gain F: [1s22s22p5] =====> F-: [1s22s22p6] = [Ne] F 1s22s22p5 F- 1s22s22p6 or [Ne] electrons so that anion has a noble-gas outer O 1s22s22p4 O2- 1s22s22p6 or [Ne] Na+, Al3+, F-, O2-, and N3- are all said to be “isoelectronic electron configuration. with Ne” as they have the same electronic N 1s22s22p3 3- 2 2 6 N 1s 2s 2p or [Ne] configuration....all subshells are filled.

Metals and non-metals form ions with electronic Metals and non-metal ions tend to form electronic configurations closest to their nearest noble gas states closest to their nearest noble gas configuration. configuration. 1A 2A 3A 4A 5A 6A 7A 8A

What is the spdf and condensed electron Using the periodic table on the inside cover of the configuration of Mg and Mg2+ ? Mg 12 electrons text and give the full and condensed electrons configurations, partial orbital diagrams showing 2 2 6 2 2 Mg 1s 2s 2p 3s [Ne]3s valence electrons, and number of inner electrons for 2+ 2 2 6 0 0 the following elements: Mg 1s 2s 2p 3s [Ne]3s = [Ne] (a) potassium (K: (b) molybdenum (c) lead (Pb: Z What are the possible quantum numbers for the last Z = 19) (Mo: Z = 42) = 82) (outermost) electron in Cl? Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s23p5 2 + 2 + 6 + 2 + 5 = 17 electrons Last electron added to 3p orbital

n = 3 l = 1 ml = -1, 0, or +1 ms = ! or -!

C) Is ground state F paramagenetic or diamagnetic? (a) for K (Z = 19) There are 18 inner electrons. Use condensed electron configurations to write the reaction 2 2 6 2 6 1 full configuration 1s 2s 2p 3s 3p 4s for the formation of each transition metal ion, and predict 1 condensed [Ar] 4s whether the ion is paramagnetic. orbital diagram

4s1 3d 4p (a) Mn2+(Z = 25) (b) Cr3+(Z = 24) (c) Hg2+(Z = 80) (b) for Mo (Z = 42) 36 inner electrons and 6 valence electrons full configuration 1s22s22p63s23p64s23d104p65s14d5 condensed [Kr] 5s14d5 partial orbital diagram Write the electron configuration and remove electrons starting with ns to match the charge on the ion. If the remaining 5s1 4d5 5p configuration has unpaired electrons, it is paramagnetic. (c) for Pb (Z = 82) 78 inner electrons and 4 valence electrons. 1s22s22p63s23p64s23d104p65s24d10 full configuration 5p66s24f145d106p2 condensed [Xe] 6s24f145d106p2

partial orbital diagram 6s2 6p2

Use condensed electron configurations to write the reaction Identify n and l quantum numbers for each of the for the formation of each transition metal ion, and predict following. whether the ion is paramagnetic.

(a) Mn2+(Z = 25) (b) Cr3+(Z = 24) (c) Hg2+(Z = 80) Write the electron configuration and remove electrons starting with ns to match the charge on the ion. If the remaining configuration has unpaired electrons, it is paramagnetic.

SOLUTION: third shell fourth shell (a) Mn2+(Z = 25) Mn([Ar]4s23d5) Mn2+ ([Ar] 3d5) + 2e- paramagnetic What neutral element has the following orbital-filling (b) Cr3+(Z = 24) Cr([Ar]4s13d5) Cr3+ ([Ar] 3d3) + 3e- paramagnetic diagram?

Identify n and l quantum numbers for each of the Many atomic properties show periodicity and trends. following.

2 4dz Amount of energy to remove 1 3p mole e- from 1 mole of gaseous atoms or element

third shell fourth shell

What neutral element has the following orbital-filling diagram? Gallium = Ga Amount of energy to add 1 mole e- to 1 mole of gaseous atoms or element Many atomic properties show periodicity and trends. Electrons in elements are categorized either as inner core electrons or valence electrons.

1) Inner core electrons : electrons filling the lower n shells of an element. They are located closer to the nucleus.

2. Outer core or VALENCE e- : those e- in the highest energy level (highest n-value). The number of valence e- is given by the Group Number in the periodic table for Group A.

--Responsible for chemistry and bonding of elements forming compounds or ions (true for representative but not transition metals--more complex).

Periodicity in the chemical reactivity of elements Inner core electrons “shield” outer electrons from the occurs because of periodicity in the electronic positive charge of the nucleus. structure of valence electrons!

1-electron 2-electrons 5-electrons outer s-orbital outer d-orbital outer p-orbital

Effective nuclear charge (Zeff) is the electrostatic Z = Z – core e- eff **** force felt by the outer valence electrons taking into Core Valence Radius + “shielding” by core electrons. Configuration Element Z (p ) Zeffective Electrons Electrons (pm) [Ne]3s1 Na 11 10 1 1 186 To a good approximation: effective nuclear 2 charge, Zeff is given by: [Ne]3s Mg 12 10 2 2 160 [Ne]3s23p1 Al 13 10 3 3 143 [Ne]3s23p2 Si 14 10 4 4 132 Zeff = Z – core e- [Ne]3s23p3 P 15 10 5 5 128 [Ne]3s23p4 S 16 10 6 6 127 Effective # of inner non- [Ne]3s23p5 Cl 17 10 7 7 99 Nuclear # protons valence electrons 2 6 charge [Ne]3s 3p Ar 18 10 8 8 98 [Ar]4s1 K 19 18 1 1 227 2 Bigger Zeff means more “pull” or electrostatic force [Ar]4s Ca 20 18 2 2 197 between nucleus and electrons. [Ar]4s23d1 Sc 21 18 3 3 135 Because of increasing effective nuclear charge Many atomic properties show periodicity and trends. across a period, atomic radii decrease across a Period. As n increases down a group so does the radius. Decreasing Atomic Radius

Amount of energy to remove 1 mole e- from 1 mole of gaseous

atoms or element Increasing Increasing Radius Atomic

Amount of energy to add 1 mole n increases e- to 1 mole of gaseous atoms or element

Periodicity of Atomic Radius The radii of cations are smaller than their parent neutral atoms, while anions are larger than its parent.

Group I

Cations get smaller (greater Zeff) Anions get larger (lower Zeff)

Group VIII

Using only the periodic table rank each set Using only the periodic table rank each set of of main group elements in order of main group elements in order of decreasing atomic size: decreasing atomic size: (a) Ca, Mg, Sr (b) K, Ga, Ca (c) Br, Rb, Kr (a) Ca, Mg, Sr (d) Sr, Ca, Rb (b) K, Ga, Ca SOLUTION: These elements are in Group 2A(2). (c) Br, Rb, Kr (a) Sr > Ca > Mg These elements are in Period 4. (b) K > Ca > Ga (d) Sr, Ca, Rb Rb has a higher n engery level and (c) Rb > Br > Kr is far to the left. Br is to the left of Kr. (d) Rb > Sr > Ca Ca is one energy level smaller than Rb and Sr. Rb is to the left of Sr. First ionization energies of the main-group elements. First Ionization Energy

Ionization energy is the minimum energy (kJ/mol) Ranking Elements by First Ionization Energy required to remove an 1 mole of electrons from one mole of a gaseous atom in its ground state (!H > 0). PROBLEM: Using the periodic table only, rank the elements in each of the following sets in order of decreasing IE1: (a) Kr, He, Ar (b) Sb, Te, Sn (c) K, Ca, Rb (d) I, Xe, Cs + - I1 + X (g) X (g) + e I1 first ionization energy PLAN: IE decreases as you proceed down in a group; IE increases as you go across a period. 2+ - I second ionization energy I2 + X (g) X (g) + e 2

3+ - I3 + X (g) X (g) + e I3 third ionization energy

I1 < I2 < I3

Ranking Elements by First Ionization Energy The ionization energy increases dramatically when an core electron is removed from a non- valence shell. PROBLEM: Using the periodic table only, rank the elements in each of the

following sets in order of decreasing IE1: (a) Kr, He, Ar (b) Sb, Te, Sn (c) K, Ca, Rb (d) I, Xe, Cs

PLAN: IE decreases as you proceed down in a group; IE increases as you go across a period.

SOLUTION: 1s2 2s1 (a) He > Ar > Kr Group 8A(18) - IE decreases down a group. 1s2 2s2

2 2 1 (b) Te > Sb > Sn Period 5 elements - IE increases across a period. 1s 2s 2p 1s2 2s2 2p2 (c) Ca > K > Rb Ca is to the right of K; Rb is below K. 1s2 2s2 2p3 1s2 2s2 2p4 (d) Xe > I > Cs I is to the left of Xe; Cs is furtther to the left and down one period. Identifying an Element from Successive Identifying an Element from Successive Ionization Energies Ionization Energies Name the Period 3 element with the following ionization Name the Period 3 element with the following ionization energies (in kJ/mol) and write its electron configuration: energies (in kJ/mol) and write its electron configuration:

IE1 IE2 IE3 IE4 IE5 IE6 IE1 IE2 IE3 IE4 IE5 IE6 1012 1903 2910 4956 6278 22,230 1012 1903 2910 4956 6278 22,230

SOLUTION: PLAN: Look for a large increase in energy which indicates that all of the valence electrons have been The largest increase occurs after IE , that is, after the removed. 5 5th valence electron has been removed. Five electrons The number valence electrons is reflected in the would mean that the valence configuration is 3s23p3 periodic table for Group A elements....find the and the element must be phosphorous, P (Z = 15). group with that number of valence electrons. The complete electron configuration is 1s22s22p63s23p3.

Main Group (or representative) metals form ionic Properties of Oxides Across a Period basic oxides when reacted with oxygen while non- metals form covalent acidic oxides with oxygen. Covalent Increasing Acidity Oxides basic acidic

1A 2A 3A 4A 6A 7A Ionic (14) (16) (17) Oxides 2 Li2O BeO B2O3 CO2 OF2 Increasing Na2O Basicity

4 K2O CaO Ga2O3 GeO2 SeO3 Br2O7

5 Rb2O SrO In2O3 SnO2 TeO3 I2O7

6 Cs2O BaO Tl2O3 PbO2