Figure 8.12: A periodic table illustrating the building-up order.

Mendeleev’s periodic table generally organized elements by increasing atomic mass and with similar properties in columns. In some places, there were missing elements whose properties he predicted. When gallium, scandium, and germanium were isolated and characterized, their properties were almost identical to those predicted by Mendeleev for eka-aluminum, eka-boron, and eka-silicon, respectively.

Figure 8.14: Mendeleev’s periodic table.

Periodic law states that when the elements are arranged by atomic number, their physical and chemical properties vary periodically. We will look in more detail at three periodic properties: atomic radius, ionization energy, and electron affinity.

Effective Nuclear Charge Effective nuclear charge is the positive charge that an electron experiences from the nucleus. It is equal to the nuclear charge, but is reduced by shielding or screening from any intervening electron distribution (inner shell electrons).

Effective nuclear charge increases across a period Effective nuclear charge increases across a period. Because the shell number (n) is the same across a period, each successive atom experiences a stronger nuclear charge. As a result, the atomic size decreases across a period.

Atomic Radius While an atom does not have a definite size, we can define it in terms of covalent radii (the radius in covalent compounds).

Figure 8.17: Representation of atomic radii (covalent radii) of the main-group elements.

Atomic radius is plotted against atomic number in the graph below Atomic radius is plotted against atomic number in the graph below. Note the regular (periodic) variation.

A representation of atomic radii is shown below.

Refer to a periodic table and arrange the following elements in order of increasing atomic radius: Br, Se, Te. 35 Br 34 Se 52 Te Te is larger than Se. Se is larger than Br. Br < Se < Te

First Ionization Energy (first ionization potential) The minimum energy needed to remove the highest-energy (outermost) electron from a neutral atom in the gaseous state, thereby forming a positive ion

Periodicity of First Ionization Energy (IE1) Like Figure 8-18

Fig. 8.15

Left of the line, valence shell electrons are being removed Left of the line, valence shell electrons are being removed. Right of the line, noble-gas core electrons are being removed.

Identifying Elements by Its Successive Ionization Energies Problem: Given the following series of ionization energies (in kJ/mol) for an element in period 3, name the element and write its electron configuration: IE1 IE2 IE3 IE4 580 1,815 2,740 11,600 Plan: Examine the values to find the largest jump in ionization energy, which occurs after all valence electrons have been removed. Use the periodic table! Solution:

Identifying Elements by Its Successive Ionization Energies Problem: Given the following series of ionization energies (in kJ/mol) for an element in period 3, name the element and write its electron configuration: IE1 IE2 IE3 IE4 580 1,815 2,740 11,600 Plan: Examine the values to find the largest jump in ionization energy, which occurs after all valence electrons have been removed. Use the periodic table! Solution: The largest jump in IE occurs after IE3 so the element has 3 valence electrons thus it is Aluminum ( Al, Z=13), its electron configuration is : 1s2 2s2 2p6 3s2 3p1

Fig. 8.16

Ranking Elements by First Ionization Energy Problem: Using the Periodic table only, rank the following elements in each of the following sets in order of increasing IE! a) Ar, Ne, Rn b) At, Bi, Po c) Be, Na, Mg d) Cl, K, Ar Plan: Find their relative positions in the periodic table and apply trends! Solution:

Trends Going down a group, first ionization energy decreases. This trend is explained by understanding that the smaller an atom, the harder it is to remove an electron, so the larger the ionization energy.

Generally, ionization energy increases with atomic number. Ionization energy is proportional to the effective nuclear charge divided by the average distance between the electron and the nucleus. Because the distance between the electron and the nucleus is inversely proportional to the effective nuclear charge, ionization energy is inversely proportional to the square of the effective nuclear charge.

Small deviations occur between Groups IIA and IIIA and between Groups VA and VIA. Examining the valence configurations for these groups helps us to understand these deviations: IIA ns2 IIIA ns2np1 VA ns2np3 VIA ns2np4 It takes less energy to remove the np1 electron than the ns2 electron. It takes less energy to remove the np4 electron than the np3 electron.

Electrons can be successively removed from an atom Electrons can be successively removed from an atom. Each successive ionization energy increases, because the electron is removed from a positive ion of increasing charge. A dramatic increase occurs when the first electron from the noble-gas core is removed.

Refer to a periodic table and arrange the following elements in order of increasing ionization energy: As, Br, Sb. Sb is larger than As. As is larger than Br. 35 Br 33 As 51 Sb Ionization energies: Sb < As < Br

Overall periodic trends Chemistry 140 Fall 2002 Overall periodic trends Note: Electronegativity has similar trend as electron affinity 28

Reactivity of the Alkali Metals 2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g) Lithium video 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) Sodium video 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g) Potassium video Trend?

More Sodium Reaction Videos 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) http://www.theodoregray.com/PeriodicTable/ Prepping Na 100 g Na in one piece 150 g Na in small pieces

Electronic Configuration Ions Na 1s 2 2s 2 2p 6 3s 1 Na+ 1s 2 2s 2 2p 6 Mg 1s 2 2s 2 2p 6 3s 2 Mg+2 1s 2 2s 2 2p6 Al 1s 2 2s 2 2p 6 3s 2 3p 1 Al+3 1s 2 2s 2 2p 6 O 1s 2 2s 2 2p 4 O- 2 1s 2 2s 2 2p 6 F 1s 2 2s 2 2p 5 F- 1 1s 2 2s 2 2p 6 N 1s 2 2s 2 2p 3 N- 3 1s 2 2s 2 2p 6

Isoelectronic Atoms and Ions H- 1 { He } Li+ Be+2 N- 3 O- 2 F- { Ne } Na+ Mg+2 Al+3 P- 3 S- 2 Cl- { Ar } K+ Ca+2 Sc+3 Ti+4 As- 3 Se- 2 Br- { Kr } Rb+ Sr+2 Y+3 Zr+4 Sb- 3 Te- 2 I- { Xe } Cs+ Ba+2 La+3 Hf+4

Trends when atoms form chemical bonds Chemistry 140 Fall 2002 Trends when atoms form chemical bonds Empirical Observation “when forming ionic compounds, elements tend to lose or gain electrons to be more like the nearest noble gas” Nonmetals tend to gain e-’s Metals tend to lose e-’s 35

Are ions bigger or smaller than atoms? Chemistry 140 Fall 2002 Are ions bigger or smaller than atoms? Representative cation Na → Na+ + e- Representative anion F + e- → F- 36

Cations Contract Anions Add Trends in ion size Cations are always smaller than parent atom - decreased e- repulsion (clouds contract) - if emptying valence shell, “n” decreases Anions are always larger than parent atoms - increased e- repulsion (clouds expand) Cations Contract Anions Add

Trends in atom & ion size Chemistry 140 Fall 2002 39

Chemistry 140 Fall 2002 Trends in ion size 40

Ranking Ions According to Size Problem: Rank each set of Ions in order of increasing size. a) K+, Rb+, Na+ b) Na+, O2-, F - c) Fe+2, Fe+3 Plan: We find the position of each element in the periodic table and apply the ideas of size: i) size increases down a group, ii) size decreases across a period but increases from cation to anion. iii) size decreases with increasing positive (or decreasing negative) charge in an isoelectronic series. iv) cations of the same element decreases in size as the charge increases. Solution: a) since K+, Rb+, and Na+ are from the same group (1A), they increase in size down the group: Na+ < K+ < Rb+ b) the ions Na+, O2-, and F- are isoelectronic. O2- has lower Zeff than F-, so it is larger. Na+ is a cation, and has the highest Zeff, so it is smaller: Na+ < F- < O2- c) Fe+2 has a lower charge than Fe+3, so it is larger: Fe+3 < Fe+2

Chapter #9 - Models of Chemical Bonding 9.1) Atomic Properties and Chemical Bonds 9.2) The Ionic Bonding Model 9.3) The Covalent Bonding Model 9.4) Between the Extremes: Electronegativity and Bond Polarity 9.5) An Introduction to Metallic Bonding

Sodium Chloride

Depicting Ion Formation with Orbital Diagrams and Electron Dot Symbols - I Problem: Use orbital diagrams and Lewis structures to show the formation of magnesium and chloride ions from the atoms, and determine the formula of the compound. Plan: Draw the orbital diagrams for Mg and Cl. To reach filled outer levels Mg loses 2 electrons, and Cl will gain 1 electron. Therefore we need two Cl atoms for every Mg atom. Solution: Mg + Mg+2 + 2 Cl- 2 Cl .. . .. .. .. . Cl . .. .. .. .. Mg + . Mg+2 + 2 Cl .. ..

. .. . . .. .. .. .. .. . Depicting Ion Formation from Orbital Diagrams and Electron Dot Symbols - II Problem: Use Lewis structures and orbital diagrams to show the formation of potassium and sulfide ions from the atoms, and determine the formula of the compound. Plan: Draw orbital diagrams for K and S. To reach filled outer orbitals, sulfur must gain two electrons, and potassium must lose one electron. Solution: 2 K + 2 K+ + S - 2 S . .. . . 2 - K .. .. .. .. .. . + S 2 K+ + S K

Three Ways of Showing the Formation of Li+ and F - through Electron Transfer

Lewis Electron-Dot Symbols for Elements in Periods 2 & 3

The Reaction between Na and Br to Form NaBr The Elements The Reaction!

Melting and Boiling Points of Some Ionic Compounds Compound mp( oC) bp( oC) CsBr 636 1300 NaI 661 1304 MgCl2 714 1412 KBr 734 1435 CaCl2 782 >1600 NaCl 801 1413 LiF 845 1676 KF 858 1505 MgO 2852 3600

Figure 9.11: Potential-energy curve for H2.

Covalent Bonding in Hydrogen, H2

Figure 9.10: The electron probability distribution for the H2 molecule.

Covalent bonds animation http://www.chem1.com/acad/webtext/chembond/cb03.html animation http://www.chem.ox.ac.uk/vrchemistry/electronsandbonds/intro1.htm

For elements larger than Boron, atoms usually react to develop octets by sharing electrons. H, Li and Be strive to “look” like He. B is an exception to the noble gas paradigm. It’s happy surrounded by 6 electrons so the compound BH3 is stable. Try drawing a Lewis structure for methane.

Draw Lewis dot structures for the halogens. Notice that these all follow the octet rule! Try oxygen and nitrogen. These also follow the octet rule!

Bond Lengths and Covalent Radius

Figure 9.14: The HCl molecule.

Figure 9.12: Molecular model of nitro-glycerin. What is the formula for this compound?

Rules for drawing Lewis structures 1. Count up all the valence electrons 2. Arrange the atoms in a skeleton 3. Have all atoms develop octets (except those around He)

Make some Lewis Dot Structures with other elements: SiH4 H2O NH3 CH2O C2H6 C2H6O

Figure 9.9: Model of CHI3 Courtesy of Frank Cox.

Make some Lewis Dot Structures with other elements: CH4 H2O NH3 CH2O C2H6 C2H6O

Look at all these structures and make some bonding rules:

Rules for drawing Lewis structures 1. Count up all the valence electrons 2. Arrange the atoms in a skeleton 3. Have all atoms develop octets (except those around He) 4. Satisfy bonding preferences!

A model of ethylene.

A model of acetylene.

A model of COCl2.

The Relation of Bond Order, Bond Length and Bond Energy Bond Bond Order Average Bond Average Bond Length (pm) Energy (kJ/mol) C O 1 143 358 C O 2 123 745 C O 3 113 1070 C C 1 154 347 C C 2 134 614 C C 3 121 839 N N 1 146 160 N N 2 122 418 N N 3 110 945 Table 9.4

Conceptual Problem 9.103

Fig. 9.14

Figure 9.15: Electronegatives of the elements.

The Periodic Table of the Elements 2.1 He 0.9 1.5 2.0 2.5 3.0 3.5 4.0 Ne Electronegativity 0.9 1.2 1.5 1.8 2.1 2.5 3.0 Ar 0.8 1.0 1.3 1.5 1.6 1.6 1.5 1.8 1.8 1.8 1.9 1.6 1.6 1.8 2.0 2.4 2.8 Kr 0.8 1.0 1.2 1.4 1.6 1.8 1.9 2.2 2.2 2.2 1.9 1.7 1.7 1.8 1.9 2.1 2.5 Xe 0.7 0.9 1.1 1.3 1.5 1.7 1.9 2.2 2.2 2.2 2.4 1.9 1.8 1.8 1.9 2.0 2.2 Rn 0.7 0.9 1.1 Ce Pr Nd Pm Yb Lu 1.1 1.1 1.1 1.2 1.2 1.1 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.3 1.3 1.5 1.7 1.3 1.3 1.3 1.3 1.3 1.3 1.3 1.3 1.3 1.5 Th Pa U Np No Lr

Fig. 9.16

Fig. 9.17

Determining Bond Polarity from Electronegativity Values Problem: (a)Indicate the polarity of the following bonds with a polarity arrow: O - H, O - Cl, C - N, P - N, N - S, C - Br, As - S (b) rank those bonds in order of increasing polarity. Plan: (a) We use Fig. 9.16 to find the EN values, and point the arrow toward the negative end. (b) Use the EN values. Solution: a) the EN of O = 3.5 and of H = 2.1: O - H the EN of O = 3.5 and of Cl = 3.0: O - Cl the EN of C = 2.5 and of P = 2.1: C - P the EN of P = 2.1 and of N = 3.0: P - N the EN of N = 3.0 and of S = 2.1: N - S the EN of C = 2.5 and of Br = 2.8: C - Br the EN of As = 2.0 and of O = 3.5: As - O b) C - Br < C - P < O - Cl < P - N < N - S < O - H < As - O 0.3 < 0.4 < 0.5 < 0.9 < 0.9 < 1.4 < 1.5

Fig. 9.18

Percent Ionic Character as a Function of Electronegativity Difference (En) Fig. 9.19

Lewis Structures of Simple Molecules H H H .. CH4 H C C .. O H Methane H H Ethyl Alcohol (Ethanol) .. .. .. .. C F .. O K+ .. .. Cl .. .. .. .. .. O .. O CF4 .. KClO3 Potassium Chlorate Carbon Tetrafluoride

Resonance: Delocalized Electron-Pair Bonding - I Ozone : O3 .. .. .. .. .. O O .. .. .. .. .. .. O O .. O O II I Resonance Hybrid Structure .. O .. .. .. .. O O One pair of electron’s resonances between the two locations!!

Delocalized Electron-Pair Bonding - II Resonance: Delocalized Electron-Pair Bonding - II H H C C C C H C C H H H H C C H C H C H H C C C H H C H C H C H H C C Benzene Resonance Structure H

Lewis Structures of Simple Molecules Resonance Structures -III Nitrate .. .. .. O .. N .. .. O .. .. O .. .. .. O .. .. O N .. .. N .. .. .. O .. .. O .. .. .. .. O O

Lewis Structures for Octet Rule Exceptions .. .. .. B Cl .. .. .. .. F .. .. .. .. F Cl .. .. F Each chlorine atom has 8 electrons associated. Boron has only 6! Each fluorine atom has 8 electrons associated. Chlorine has 10 electrons! . .. .. .. .. .. .. N .. .. Cl Be Cl .. .. .. O O Each chlorine atom has 8 electrons associated. The beryllium has only 4 electrons. NO2 is an odd electron atom. The nitrogen has 7 electrons.

Resonance Structures - Expanded Valence Shells .. .. .. .. .. .. .. .. .. .. .. .. .. .. F F F F .. .. .. .. .. .. F S F .. P .. F .. .. .. .. .. .. F .. .. .. .. F F .. F p = 10e- S = 12e- Sulfur hexafluoride .. Phosphorous pentafluoride .. .. O S H .. O S H .. .. Resonance Structures .. .. .. .. .. Sulfuric acid S = 12e-

Lewis Structures of Simple Molecules . . . . -2 Sulfate O . . . . Resonance Structures-V . . . . O S O . . . . -2 . . . . O o * o o O o o Plus 4 others for a total of 6 x o o o o o x . . o * O x o S O o o x o o -2 . . O . . o o x x . . . . o o O o o O S O . . . . o o . . O . . . . x = Sulfur electrons o = Oxygen electrons

VSEPR: Valence Shell Electron Pair Repulsion: A way to predict the shapes of molecules Pairs of valence electrons want to get as far away from each other as possible in 3-dimensional space.

Balloon Analogy for the Mutual Repulsion of Electron Groups Two Three Four Five Six Number of Electron Groups

AX2 Geometry - Linear .. .. .. .. .. .. Molecular Geometry = Linear Arrangement Cl Be Cl BeCl2 1800 Gaseous beryllium chloride is an example of a molecule in which the central atom - Be does not have an octet of electrons, and is electron deficient. Other alkaline earth elements also have the same valence electron configuration, and the same geometry for molecules of this type. Therefore this geometry is common to group II elements. .. .. .. .. O C O CO2 1800 Carbon dioxide also has the same geometry, and is a linear molecule, but in this case, the bonds between the carbon and oxygens are double bonds.

The Two Molecular Shapes of the Trigonal Planar Electron-Group Arrangement

AX3 Geometry - Trigonal Planar .. .. .. .. .. .. All of the boron Family(IIIA) elements have the same geometry. Trigonal Planar ! F F BF3 B Boron Trifluoride 1200 .. .. .. F AX2E SO2 .. - .. .. .. .. O S O .. .. NO3- .. 1200 .. N .. .. .. .. O O The AX2E molecules have a pair of Electrons where the third atom would appear in the space around the central atom, in the trigonal planar geometry. 1200 Nitrate Anion

The Three Molecular Shapes of the Tetrahedral Electron-Group Arrangement

.. AX4 Geometry - Tetrahedral H Methane H 109.50 CH4 H C H C H H H H All molecules or ions with four electron groups around a central atom adopt the tetrahedral arrangement H H 109.50 N H .. 107.30 109.50 + H+ N H H Ammonia is in a tetrahedral shape, but it has only an electron pair in one location, so the smaller angle! all angles are the same! Ammonium Ion H

The Four Molecular Shapes of the Trigonal Bipyramidal Electron- Group Arrangement

AX5 Geometry - Trigonal Bipyramidal .. .. .. .. .. .. F I .. .. .. 86.20 .. .. .. I 1800 Br .. F .. .. .. .. .. I .. F .. AX3E2 - BrF3 .. AX2E3 - I3- .. .. .. Cl .. .. Cl .. .. P Cl .. AX5 - PCl5 .. .. Cl .. .. .. .. Cl

The Three Molecular Shapes of the Octahedral Electron-Group Arrangement

AX6 Geometry - Octahedral .. .. .. .. .. .. .. .. .. .. F F .. .. .. .. F .. F .. .. F .. F .. .. .. .. S I .. .. .. .. F F .. .. .. .. F .. F .. .. .. F AX5E Iodine Pentafluoride AX6 Sulfur Hexafluoride .. .. .. .. .. .. F .. F .. Xe .. .. .. .. .. F .. F Xenon Tetrafluoride Square planar shape

Using VSEPR Theory to Determine Molecular Shape 1) Write the Lewis structure from the molecular formula to see the relative placement of atoms and the number of electron groups. 2) Assign an electron-group arrangement by counting all electron groups around the central atom, bonding plus nonbonding. 3) Predict the ideal bond angle from the electron-group arrangement and the direction of any deviation caused by the lone pairs or double bonds. 4) Draw and name the molecular shape by counting bonding groups and non-bonding groups separately.

Hybrid Orbital Model

The sp Hybrid Orbitals in Gaseous BeCl2

The sp3 Hybrid Orbitals in NH3 and H2O

The sp3d Hybrid Orbitals in PCl5

The sp3d2 Hybrid Orbitals in SF6 Sulfur Hexafluoride -- SF6

Figure 10.26: Sigma and pi bonds.

Figure 10.27: Bonding in ethylene.

Figure 10.28: Bonding in acetylene.

Restricted Rotation of -Bonded Molecules A) Cis - 1,2 dichloroethylene B) trans - 1,2 dichloroethylene

Postulating the Hybrid Orbitals in a Molecule Problem: Describe how mixing of atomic orbitals on the central atoms leads to the hybrid orbitals in the following: a) Methyl amine, CH3NH2 b) Xenon tetrafluoride, XeF4 Plan: From the Lewis structure and molecular shape, we know the number and arrangement of electron groups around the central atoms, from which we postulate the type of hybrid orbitals involved. Then we write the partial orbital diagram for each central atom before and after the orbitals are hybridized.

Postulating the Hybrid Orbitals in a Molecule Problem: Describe how mixing of atomic orbitals on the central atoms leads to the hybrid orbitals in the following: a) Methyl amine, CH3NH2 b) Xenon tetrafluoride, XeF4 Plan: From the Lewis structure and molecular shape, we know the number and arrangement of electron groups around the central atoms, from which we postulate the type of hybrid orbitals involved. Then we write the partial orbital diagram for each central atom before and after the orbitals are hybridized. Solution: a) For CH3NH2: The shape is tetrahedral around the C and N atoms. Therefore, each central atom is sp3 hybridized. The carbon atom has four half-filled sp3 orbitals: 2s 2p sp3 Isolated Carbon Atom Hybridized Carbon Atom

.. The N atom has three half-filled sp3 orbitals and one filled with a lone pair. 2s sp3 2p .. H C N H H H H

b) The Xenon atom has filled 5 s and 5 p orbitals with the 5 d orbitals empty. Isolated Xe atom 5 d 5 s 5 p Hybridized Xe atom: sp3d2 5 d

b) continued:For XeF4. for Xenon, normally it has a full octet of electrons,which would mean an octahedral geometry, so to make the compound, two pairs must be broken up, and bonds made to the four fluorine atoms. If the two lone pairs are on the equatorial positions, they will be at 900 to each other, whereas if the two polar positions are chosen, the two electron groups will be 1800 from each other. Thereby minimizing the repulsion between the two electron groups. .. F F F F 1800 Xe Xe .. F F F F Square planar

That's all, folks!!