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28-Aug-16 WHAT IS THE TEXT BOOK? GENERAL CHEMISTRY • Chemistry - The Molecular Nature of Matter and Change (5th edition) (CHEMW2014) • Syllabus for 57NKN • A copy of the text book is on reserve at the TLU Library Thanh M Le CHEMW2014 will be studied from Chapter to Chapter 12 in the text book 28-Aug-16 28-Aug-16 WHAT DO STUDENTS NEED TO COMPLETE? THE COURSE GRADING • Homework = 10 recitation assignments • Exam = chapter exams + final exam Recitations = 10*10 points 100 points You will not be allowed to attend to the final exam, IF: Chapter exams = 4*100 points 400 points • you miss out any chapter exam without the Final exam = 1*200 points 200 points Total possible points: 700 points lecturer’s permission • you are present in the CHEMW2014 class less than 80% of the prescribed time 28-Aug-16 28-Aug-16 www.facebook.com/groups/generalchemistry1 28-Aug-16 28-Aug-16 28-Aug-16 6 28-Aug-16 Lecture PowerPoint Chapter Chemistry The Molecular Nature of Matter and Change Keys to the Study of Chemistry Martin S Silberberg 28-Aug-16 • Physical Properties: properties a substance shows by itself 1.1 Some Fundamental Definitions without interacting with another substance - color, melting point, • Chemistry: is the study of matter, its properties, the boiling point, density changes that matter undergoes, and the energy associated • Chemical Properties: properties a substance shows as it with these changes interacts with, or transforms into, other substances - • Matter: anything that has both mass and volume - the “stuff” flammability, corrosiveness of the universe: books, planets, trees, professors, students • A physical change: occurs when a substance alters its • Composition: the types and amounts of simpler substances physical form, not its composition that make up a sample of matter • A chemical change, also called a chemical reaction: occurs • Properties: the characteristics that give each substance a when a substance (or substances) is converted into a different unique identity substance (or substances) 28-Aug-16 28-Aug-16 Example 1: The scenes below represent an atomic-scale view of substance A undergoing two different changes (B or C) Decide whether Exercise 1: Decide whether each of the following processes is each scene shows a physical or a chemical change? primarily a physical or a chemical change, and explain briefly: (a) Frost forms as the temperature drops on a humid winter night (b) A cornstalk grows from a seed that is watered and fertilized Solution: (c) A match ignites to form ash and a mixture of gases a) A  B: each particle of substance A is composed of one blue and two red spheres Sample B is composed of two different types of particles, some have two red spheres (d) Perspiration evaporates when you relax after jogging while some have one red and one blue As A changes to B, the chemical composition has changed A  B is a chemical change (e) A silver fork tarnishes slowly in air b) A  C: each particle of C is still composed of one blue and two red spheres, but the particles are closer together and are more organized The composition remains unchanged, but the physical form is different A  C is a physical change 28-Aug-16 28-Aug-16 28-Aug-16 The States of Matter Energy in Chemistry • A solid has a fixed shape and volume • Energy is the ability to work Solids may be hard or soft, rigid or flexible • Potential energy is energy due to the position of an object • A liquid has a varying shape that conforms • Kinetic energy is energy due to the movement of an object to the shape of the container, but a fixed Total Energy = Potential Energy + Kinetic Energy • Lower energy states are more stable, and are favored over volume A liquid has an upper surface higher energy states • A gas has no fixed shape or volume and • Energy is neither created nor destroyed, it is conserved, and therefore does not have a surface can be converted from one form to another 28-Aug-16 28-Aug-16 Example 2: A lower energy state is more stable 1.4 Chemical Problem Solving • All measured quantities consist of a number and a unit • Units are used like numbers, in other words, units can be multiplied, divided, and canceled • Try to make a habit of including units in all calculations whenever you practice • Example 3: Area = cm × cm = (3 × 4) (cm × cm) = 12 cm2 28-Aug-16 28-Aug-16 10 Conversion Factors Exercise 2: To wire your stereo equipment, you need 325 • A conversion factor is a ratio of equivalent quantities used to express a quantity in different units centimeters of speaker wire that sells for $0.15/ft What is the price of the wire? The conversion factor: • Example 4: some conversions factors: foot (ft) = 30.48 centimeters (cm)  Between mile and feet is: mi = 5280 ft  Between mile and kilometer is: mi = 1.609 km Exercise 3: A furniture factory needs 31.5 ft2 of fabric to  Between joule and calorie is: J = 0.000239 cal upholster one chair Its Vietnamese supplier sends the fabric in bolts of exactly 200 m2 What is the maximum number of chairs that can be upholstered by bolts of fabric (1 m = 3.281 ft)? 28-Aug-16 11 28-Aug-16 12 28-Aug-16 SI Base Units (International System of Units) 1.5 Measurement in Scientific Study Physical Quantity Unit Unit (Dimension) Name Abbreviation kilogram kg meter m second s • Our current system of measurement began in 1790, when the Mass newly formed National Assembly of France Length • In 1960, another international committee met in France to Time establish the International System of Units The units of this Temperature system are called SI units, from the French Système Electric Current Amount of substance International d’Unités Luminous intensity 28-Aug-16 13 kelvin K ampere A mole mol candela cd 14 28-Aug-16 Common Decimal Prefixes used with SI Units Common SI-English equivalent quantities Quantity SI to English Equivalent English to SI Equivalent Length km = 0.6214 mile m = 1.094 yard m = 39.37 inches cm = 0.3937 inch mi = 1.609 km yd = 0.9144 m ft = 0.3048 m in = 2.54 cm Volume cubic meter (m3) = 35.31 ft3 dm3 = 0.2642 gal dm3 = 1.057 qt cm3 = 0.03381 fluid ounce ft3 = 0.02832 m3 gal = 3.785 dm3 qt = 0.9464 dm3 qt = 946.4 cm3 fluid ounce = 29.57 cm3 Mass kg = 2.205 pounds (lb) g = 0.03527 ounce (oz) lb = 0.4536 kg oz = 28.35 g 28-Aug-16 16 Exercise 4: A graduated cylinder contains 19.9 mL of water When a small piece of galena, an Some volume equivalents in SI ore of lead, is added, it sinks and the volume m3 = 1000 dm3 = 1000 L increases to 24.5 mL What is the volume of the piece of galena in cm3 and in L? dm3 = 1000 cm3 = L = 1000 mL Exercise 5: Many international computer communications are cm3 = 1000 mm3 = mL = 100= μL carried out by optical fibers in cables laid along the ocean floor If one strand of optical fiber weighs 1.19 x 10-3 lb/m, what is the mm3 = μL mass (in kg) of a cable made of six strands of optical fiber, each long enough to link New York and Paris (8.94 x 103 km)? 28-Aug-16 17 28-Aug-16 18 28-Aug-16 Density Densities of Some Common Substances • At a given temperature and pressure, the density of a substance is a characteristic physical property and has a specific value *At 28-Aug-16 19 Substance Physical State Density (g/cm3) Hydrogen gas 0.0000899 Oxygen gas 0.00133 Grain alcohol liquid 0.789 Water liquid 0.998 Table salt solid 2.16 Aluminum solid 2.70 Lead solid 11.3 Gold solid 19.3 room temperature (20°C) and normal atmospheric pressure (1atm) 28-Aug-16 20 Temperature • Distinguish between temperature and heat: Exercise 6: Lithium, a soft, gray solid with the lowest density of  Temperature (T) is a measure of how hot or cold a substance is relative to another any metal, is a key component of advanced batteries A slab of  Heat (Q) is the energy that flows between objects that are at different temperatures substance lithium weighs 1.49x103 mg and has sides that are 20.9 mm by 11.1 mm by 11.9 mm Find the density of lithium in • Temperature Scales  Kelvin (K): the “absolute temperature scale” begins at absolute zero and has only g/cm3 positive values Note that the kelvin is not used with the degree sign (°)  Celsius (oC): the Celsius scale is based on the freezing and boiling points of water This is the temperature scale used most commonly around the world The Celsius and Exercise 7: A cube of gold cm on a side weighs 18.3 ounces Kelvin scales use the same size degree although their starting points differ What is the density of gold in g/cm3? The conversion factor:  Fahrenheit (oF): the Fahrenheit scale is commonly used in the US The Fahrenheit scale has a different degree size and different zero points than both the Celsius and ounce (oz) = 28.3495 grams (g) Kelvin scales T (K) = T (oC) + 273.15; 28-Aug-16 T (oF) = T (oC)*9/5 - 32 21 Example 5: the number of significant figures in a measurement 1.6 Uncertainty in Measurement: Significant Figures • Every measurement includes some uncertainty The rightmost digit of any quantity is always estimated • The recorded digits, both certain and uncertain, are called significant figures • The greater the number of significant figures in a quantity, the greater its certainty 28-Aug-16 28-Aug-16 23 24 28-Aug-16 Example 6: Determining which digits are significant Determining Which Digits are Significant • All digits are significant, except zeros that are used only to position the decimal point Make sure the measured quantity has a decimal point  Start at the left and move right until you reach the first nonzero digit  Count that digit and every digit to its right as significant If no decimal point is present, zeros at the end of the number are not significant 28-Aug-16 25 28-Aug-16 26 Example 7: How many digits in these numbers are significant a) 1.030 mL b) 5300 L c) 5300 L d) 0.00004715 m e) 0.0000007160 cm3 Exercise 8: For each of the following quantities, determine the number of significant figures in each quantity Solution: a) 1.030 mL has significant figures (a) 0.0030 L (b) 0.1044 g (c) 53,069 mL (d) 0.00204010 m (e) 57,600 s (f) 0.0007001 cm3 b) 5300 L has significant figures c) 5300 L has only significant figures Exercise 9: Express the number in 8c), 8d), 8e), 8f) in d) 0.00004715 m has significant figures, = 4.715x10-5 m exponential notation e) 0.0000007160 cm3 has significant figures, = 7.160x10-7 cm3 28-Aug-16 27 28-Aug-16 28 Rules for Significant Figures in Calculations Example 8: Significant Figures in Calculations For multiplication and division The answer contains the a) Multiplication: 9.21 cm x 6.8 cm x 0.3744 cm = 23.4225 cm3 = 23 cm3 same number of significant figures as there are in the b) Addition: measurement with the fewest significant figures 83.5 mL + 23.28 mL = 106.78 mL = 106.8 mL For addition and subtraction The answer has the same c) Subtraction: 865.9 mL - 2.8121 mL = 863.0879 mL = 863.1 mL number of decimal places as there are in the measurement with the fewest decimal places 28-Aug-16 29 28-Aug-16 30 28-Aug-16 Rules for Rounding Off Numbers Example 9: • If the digit removed is more than 5, the preceding number a) 5.379 rounds to 5.38 if significant figures are retained increases by b) 0.2413 rounds to 0.241 if significant figures are retained • If the digit removed is less than 5, the preceding number is c) 17.75 rounds to 17.8; but 17.65 rounds to 17.6 unchanged d) 17.6500 rounds to 17.6; but 17.6513 rounds to 17.7 • If the digit removed is 5, followed by zeros, or with no following digits, the preceding number increases by if it is •Be sure to carry two or more additional significant figures odd and remains unchanged if it is even through a multistep calculation and round off the final answer • If the is followed by other nonzero digits, rule is only followed 28-Aug-16 31 28-Aug-16 32 Example 10: Significant figures of measuring devices Example 11: Perform the following calculations and round each answer to the correct number of significant figures (a) 16.3521 cm2 - 1.448 cm2 4.80x104 mg (b) 1g 1000 mg 11.55 cm3 7.085 cm Solution: (a) 16.3521 cm2 - 1.448 cm2 7.085 cm (b) 4.80x104 mg 1g 1000 mg 11.55 cm3 The mass (6.8605 g) The volume (68.2 mL) = 14.904 cm2 7.085 cm = 48.0 g 11.55 cm3 = 2.104 cm = 4.16 g/ cm3 34 28-Aug-16 Precision, Accuracy, and Error Example 12a: Precision and accuracy in a laboratory calibration • Precision refers to how close the measurements in a series precise and accurate are to each other • Accuracy refers to how close each measurement is to the actual value • Systematic error produces values that are either all higher or all lower than the actual value This error is part of the experimental system precise but not accurate • Random error produces values that are both higher and lower than the actual value 28-Aug-16 35 28-Aug-16 36 28-Aug-16 Example 12b: Precision and accuracy in a laboratory calibration random error Chapter 1: Suggested Practice Problems 1, 4, 8, 26-42(e), 44, 45, 52-68(e), 72, 73 systematic error 37 28-Aug-16 28-Aug-16 38 28-Aug-16 40 39 28-Aug-16 28-Aug-16 39 30-Aug-16 2.1 Elements, Compounds, and Mixtures Chapter • Element - the simplest type of substance with unique physical and chemical properties An element consists of only one type of atom The Components of Matter • Atom - the smallest constituent unit of ordinary matter that has the properties of a chemical element • Molecule - a structure that consists of two or more atoms which are chemically bound together and thus behaves as an independent unit • Compound - a substance composed of two or more elements which are chemically combined • Mixture - a group of two or more elements and/or compounds that Thanh M Le 30-Aug-16 are physically intermingled Example 1: • Elements, • Atoms, • Molecules, • Compounds • Mixtures on the atomic scale 30-Aug-16 Exercise 1: The scenes below represent an atomic-scale view of samples of matter Describe each sample as an element, compound, or mixture 30-Aug-16 30-Aug-16 • Mass fractions (or mass percents) of any element in a 2.2 Mass Conservation & Mass fractions compound AxBy: • Law of Conservation of Mass: “the total mass of substances does not change during a chemical reaction” • Mass of element in sample: = • Example 3: a) Calculate %mN in NH4NO3 • Example 2: CaO + 56.08g (calcium oxide) 30-Aug-16 + CO2 → CaCO3 44.00g = 100.08g (carbon dioxide) b) Calculate mN in 140 kg of NH4NO3 Solution: (calcium carbonate) 30-Aug-16 30-Aug-16 Mass Number, Atomic Number, Atomic Symbol, and Isotopes, 2.5 The Atomic Theory today Nucleus Shell of atom Relative Mass (amu) Relative charge Proton (p+) 1.00727 +1 Neutron (n0) 1.00866 Electron (e-) 0.00054858 -1 • Isotope = atoms of an element with the same number of protons, but a different number of neutrons • Number of neutrons = N • Number of protons = P • We have: Z=P A=Z+N • Each element normally has many isotopes: • Example 4: isotopes of carbon • The atomic mass unit (amu) = 1.66054ì10-24 g; 30-Aug-16 ã The atomic mass (also called atomic weight) of an element is the average mass of the masses of isotopes which are weighted according to their abundances • Atomic mass is calculated from its isotopes mass: Isotope Mass (amu) Percent Abundance (%) A1X A2X A3X M1 y1 M2 y2 M3 y3 30-Aug-16 Exercise 2: Silicon(Si) is essential to the computer industry as a major component of semiconductor chips It has three naturally occurring isoltopes: 28Si, 29Si, and 30Si Determine the number of protons, neutrons, and electrons in each silicon isotope, if atomic number of silicon is 14 Exercise 3: Boron (5B) has two naturally occurring isotopes • Example 5: Silver (47Ag) has two occur naturally, 107Ag and 109Ag, with mass 106.90509 (amu) and 108.90476 (amu), with abundance 51.84% and 48.16%, repectively So, the atomic mass of silver is: Find the percent abundances of 10B and 11B given the atomic mass of B = 10.81 amu, the isotopic mass of 10B = 10.0129 amu, and the isotopic mass of 11B = 11.0093 amu 30-Aug-16 10 Mendeleev's 1871 periodic table 2.6 Periodic Table • At the end of the 18th century, Lavoisier compiled a list of the 23 elements known at that time • By 1870, 65 were known • In 1871, the Russian chemist Dmitri Mendeleev published the most successful of these Dmitri Mendeleev (1836–1907) organizing schemes as a table of the elements • By 1925, 88 were known Today, there are 116 and still counting! 30-Aug-16 11 30-Aug-16 12 29-Sep-16 • Example 10: From electronegativity (EN) values and their Predicting the Polarity of Molecules periodic trends, predict the overall molecular dipole: • Step Draw and name the molecular shape • Step Use relative EN values to decide on the direction of (a) Ammonia, NH3; (b) Boron trifluoride, BF3 Solution each bond dipole • Step Combine all the bond dipoles in the molecule as a whole • Step If the combine vector is zero, then the molecule to be nonpolar Vice versa, the molecule to be polar 25 29-Sep-16 26 29-Sep-16 Chapter 10: Suggested Practice Problems 1-21(odd), 26, 28, 30-33, 34-48(even), 50-52, 63-65, 68, 70, 72, 81, 91, 55, 57, 59, 67, 94 29-Sep-163 27 29-Sep-16 28 06-Oct-16 11.1 Valence Bond (VB) Theory and Orbital Hybridization Chapter 11 • The basic principle of VB theory:  A covalent bond forms when the orbitals of two atoms Theories of overlap and a pair of electrons occupy the overlap region  The space formed by the overlapping orbitals can accommodate a maximum of two electrons and these Covalent Bonding electrons must have opposite (paired) spins  The greater the orbital overlap, the stronger the bond  Extent of orbital overlap depends on orbital shape and direction 06-Oct-16 06-Oct-16 • Example 1: Use the VB theory to depict the orbital Orbital Hybridization overlap and spin pairing in diatomic molecules: • Hybridisation (or hybridization) is the concept of mixing H2, HF, F2 atomic orbitals into new hybrid orbitals suitable for the pairing of Solution electrons to form chemical bonds in VB theory • Features of Hybrid Orbitals  The number of hybrid orbitals formed equals the number of atomic orbitals mixed Hydrogen, H2  The type of hybrid orbitals formed varies with the types of atomic orbitals mixed  The shape and orientation of a hybrid orbital maximizes overlap with the other atom in the bond Hydrogen fluoride, HF 06-Oct-16 Fluorine, F2 06-Oct-16 Types of Hybridization Type of hybrid Atomic orbitals used Geometry of hybrid orbitals Number of hybrid orbitals formed sp s, p linear sp2 s, p, p trigonal planar sp3 s, p, p, p tetrahedral sp3d s, p, p, p, d trigonal pyramidal sp3d2 s, p, p, p, d, d octahedral sp2 hybridization sp sp3 hybridization hybridization 06-Oct-16 06-Oct-16 06-Oct-16 sp hybridization sp3d hybridization AO-sp sp3d2 hybridization 06-Oct-16 06-Oct-16 sp3 hybridization sp2 hybridization AO-sp2 06-Oct-16 AO-sp3 06-Oct-16 10 • Example 2: Describe how mixing of the atomic orbitals Determine the hybridisation in a molecule of the central atom(s) leads to hybrid orbitals in CH3OH • Step 1: From molecular formula, write Lewis structure Solution • Step 2: Determine molecular shape (molecular geometry), and The electron-pair geometry of both C and O is tetrahedral, so C & O has sp3 hybridization electron group arrangement (electron-pair geometry) • Step 3: Choose the most suitable hybridisation which has geometry of hybrid orbitals as same as the molecule’s electron-pair geometry 06-Oct-16 11 06-Oct-16 12 06-Oct-16 • Example 3: Describe how mixing of the atomic orbitals • Exercise 1: Use partial orbital diagrams to show of the central atom(s) leads to hybrid orbitals in SF4 how the atomic orbitals of the central atom mix to Solution form hybrid orbitals in: (a) beryllium fluoride, BeF2; The electron-pair geometry of S is trigonal (b) silicon tetrachloride, SiCl4; bipyramidal, so the central S atom is sp3d hybridized (c) xenon tetrafluoride, XeF4 3d ↑↓ ↑ ↑ 3p ↑↓ 06-Oct-16 • Exercise 2: Determine the hybridization form of the 3d ↑ ↑ ↑ ↑ center atom in the following molecules: ↑ sp3d hybridized S atom 3s isolated S atom 13 Limitations of the Hybridization Model a) CH4 b) SO2 c) SO42- d) NO3- e) C6H6 f) C2H2 06-Oct-16 14 11.2 Orbital Overlap in Covalent Bonds • Hybridization is not always consistent with observed molecular • A sigma (σ) bond = end-to-end overlap of orbitals • A pi (π) bond = sideways overlap of orbitals shapes This is particularly true for the bonding of larger atoms For example: the bond angle in H2S (92o) is closer to the angle • A pi bond is weaker than a σ bond because sideways overlap is less effective than end-to-end overlap • All single bonds = σ bonds between unhybridized p orbitals • A double bond = one σ bond + one π bond • d-orbitals not hybridize effectively with s and p orbitals, which are much lower in energy and more stable 06-Oct-16 15 Example 4: Single bond, double bond in overlap forms 06-Oct-16 16 • Example 5: Describe the types of bonds and orbitals in acetone, (CH3)2CO Solution sp2 sp3 06-Oct-16 17 06-Oct-16 18 06-Oct-16 11.3 Molecular Orbital (MO) Theory • Exercise 3: Describe the types of bonds and orbitals in: • The basic principle of MO theory: a) hydrogen cyanide, HCN b) carbon dioxide, CO2 c) methane, CH4 d) methanol, CH3OH e) Acetaldehyde, CH3CHO f) Acetic acid, CH3COOH  The combination of orbitals to form bonds is viewed as the combination of wave functions  Atomic wave functions (AOs) combine to form molecular wave functions (MOs) • Exercise 4: How many sigma (σ) and pi (π) bonds in each of  Addition of AOs forms a bonding MO, which has a the structure of urea & caffeine (below) region of high electron density between the nuclei  Subtraction of AOs forms an anti-bonding MO, which has a node, or region of zero electron density, between 06-Oct-16 19 MO diagram 06-Oct-16 the nuclei 20 MO configuration & magnetic property • A MO electron configuration shows the type of MO and the • An MO diagram shows the relative energy and number of electrons in each MO, also shows the AOs from which each MO number of e- each contains (σ1s)(σ*1s) (σ2s)(σ*2s) (p2p)(p2p)(σ2p) (p*2p)(p*2p)(σ*2p) is formed (σ1s)(σ*1s) (σ2s)(σ*2s) (σ2p)(p2p)(p2p) (p*2p)(p*2p)(σ*2p)  ∑electrons in MOs = ∑electrons in original AOs • Magnetic property of molecule depends on the lone electrons  MOs are filled in order of increasing energy in the MO configuration:  An MO can hold a maximum of e- with opposite spins  Orbitals of equal energy are half-filled, with spins parallel,  Diamagnetic: the MO configuration has no lone electrons before pairing spins 06-Oct-16  Paramagnetic: the MO configuration has lone electrons 21 06-Oct-16 22 Bonding MO & antibonding MO of AO-1s MO bond order ã Bond order = ẵ(e- in bonding MO – ∑e- in antibonding MO):  The higher the bond order → the stronger the bond  The higher the bond order → the higher the bond energy  The higher the bond order → the shorter the bond length • The bonding MO is lower in energy than the original AOs 06-Oct-16 23 • The antibonding MO is higher in energy than the original AOs 06-Oct-16 24 06-Oct-16 Molecular Orbital Diagrams of AO-2s & AO-2p (homonuclear diatomic molecules) Bonding MO & antibonding MO of AO-2p without 2s-2p mixing with 2s-2p mixing MO energy levels for B2, C2, and N2 MO energy levels for O2, F2, and Ne2 06-Oct-16 25 06-Oct-16 26 • Example 7: Draw the MO diagram, write molecular • Example 6: Draw the MO diagram, write molecular electron configuration & determine bond order for He2+ electron configuration & determine bond order for H2 and He2 Solution • The molecular electron configuration (σ1s)2 • The bond order: = ẵ (2-0) = ã Conclusion: molecule H2 has σ bond • • Electron configuration: Electron configuration: (σ1s)2(σ*1s )1 • 06-Oct-16 27 E 2p 2s 1s MO X2 s*2p p*2p p*2p s2p p2p p2p s*2s s2s s*1s s1s Bond order for He2 = 28 MO diagram for O2, F2, Ne2 Atom X 2p s*2p p*2p p*2p 2p p2p p2p s2p s*2s 2s 2s 1s 1s 29 MO X2 Atom X E MO configuration: (σ1s)(σ*1s) (σ2s)(σ*2s) (p2p)(p2p) (σ2p)(p*2p)(p*2p) 06-Oct-16 ã Bond order for He2 = ẵ 06-Oct-16 MO diagram for X2 (except O2, F2) Atom X (σ1s)2(σ*1s )2 + s2s s*1s s1s Atom X 2p 2s 1s MO configuration: (σ1s)(σ*1s) (σ2s)(σ*2s) (σ2p)(p2p)(p2p)(p*2p)(p*2p) 06-Oct-16 30 06-Oct-16 Example 8: MO diagram, MO configuration and bond order for N2 + Electron configuration for atom N: 1s2 2s2 2p3 Atom N MO for N2 p*2p E 2p p2p 2s s*2p p*2p s2p p2p s*2s s2s • Example 9: Draw the MO diagram, & determine bond order for O2, and O2+ Atom N Solution 2p σ *2p 2s s1s Bond orders: p *2p ↑ O2 = ½(8 – 4) = ↑↓ ↑↓ p2p ↑↓ ↑↓ O2+ = ½(8 – 3) = 2.5 ↑↓ σ2p ↑↓ ↑↓ σ *2s ↑↓ ↑↓ σ2s ↑↓ ↑ s*1s 1s O2 + O2 ↑ O2+ has a shorter, stronger bond than O2 1s + MO configuration N2: (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (p2p)2(p2p)2 (σ2p)2 + Bond order = ½(10 - 4) = 06-Oct-16 32 Molecular Orbital Diagrams (heteronuclear diatomic molecules) The MO diagram for HF The MO diagram for NO Chapter 11: Suggested Practice Problems 1, 5-17(odd), 20, 21, 23, 25 06-Oct-16 06-Oct-16 33 35 06-Oct-16 06-Oct-16 35 34 13-Oct-16 12.1 Physical States and Phase Changes Chapter 12 • There are two types of electrostatic forces:  Intramolecular force: bonding forces within a molecule; Intermolecular Forces: influence chemical properties  Intermolecular force: forces between molecules; influence Liquids, Solids and physical properties Phase Changes 13-Oct-16 13-Oct-16 13-Oct-16 • There are three basic states of matter:  Gas: conforms to shape and volume of container  Liquid: conforms to shape of container; volume limited by surface  Solid: maintains its own shape and volume • Types of Phase Changes: 13-Oct-16 • Enthalpy of phase changes (∆Hphase change): • Heats of vaporization and fusion for some common substances  Condensing and freezing are exothermic changes (∆H 0) • Example 1: Heat of vaporization & heat of fusion 13-Oct-16 H2O(l) → H2O(g) ∆Hovap = 40.7 kJ/mol (at 100°C) H2O(s) → H2O(l) ∆Hofus = 6.02 kJ/mol (at 0°C) H2O(g) → H2O(l) ∆Hocon= -40.7 kJ/mol H2O(l) → H2O(s) ∆Hofre = -6.02 kJ/mol 13-Oct-16 13-Oct-16 • Example 2: Find the heat (in kJ) need to provide to a system 12.2 Heat & Pressure in Phase Changes when 24.3 g of H2O liquid from 85oC to H2O gas at 117oC: • Within a phase, a change in heat is accompanied by a change in temperature: Qheating/cooling = C×m×∆T (or Q = C×n×∆T)  C = the heat capacity of water (J/g.°C) or (J/mol.°C)  m = the amount (mass or mole) of water  ∆T = the temperature change • During a phase change,a change in heat occurs at a constant temperature: Qphase change = m×∆Hophase change (or Q = n×∆Hophase change) CH2O = 75.4 J/mol.°C ∆Hovap = 40.7 kJ/mol Solution 24.3 g H2O = 1.35 mol H2O ã Q1 = nH2OìCH2OìT = (1.35 mol) (75.4 J/mol.C) (100-85.0)C= ã Q2 = nH2OìHovap = (1.35 mol) (40.7 kJ/mol) = 54.9 kJ • Q3 = nH2OìCH2OìT = (1.35 mol) (75.4 J/mol.C) (117-100)C= 13-Oct-16 ã Qtotal heat = 1.53 kJ + 54.9 kJ + 0.760 kJ = 57.2 kJ • A Phase Diagram: A graph that describes phase changes of a substance The Effects of Temperature and Intermolecular Forces on Vapor Pressure under various combinations of temperature and pressure: • Regions (bounded areas) = one phase • The higher the temperature is, the higher the vapor pressure • Interfaces (lines) = between different regions (equilibria between phases) • The weaker the intermolecular forces are, the higher the vapor • Isolated points (point) = critical point, triple point (unique T/P pressure combinations) • The Clausius - Clapeyron equation shows the nonlinear relationship between vapor pressure and temperature: CO2 or: 13-Oct-16 13-Oct-16 • Exercise 1: The molecular scenes below represent a phase change of water CH2O = 75.4 J/mol.°C, ∆Hovap H 2O 10 12.3 Types of Intermolecular Forces = 40.7 kJ/mol, • Types of intermolecular (van der Waals) forces: ∆Hofus = 6.02 kJ/mol Find the heat when 3.25 mol of H2O  ion-dipole: when ions are nearby polar molecules (dipole) undergoes this change:  dipole-dipole: when polar molecules lie near one another  hydrogen bond: a special type of dipole-dipole force arises between molecules that have one of H-O, H-N, H-F bonds  ion-induced dipole: when ions are nearby nonpolar molecules  dipole-induced dipole: when polar molecules are nearby nonpolar molecules  dispersion (London): when nonpolar molecules is nearby together 13-Oct-16 11 13-Oct-16 12 13-Oct-16 • The interactions is illustrated in the following diagram • Compare the streng of the intermolecular forces: ion-dipole > hydrogen bond > dipole-dipole > > ion-induced dipole > dipole-induced dipole > dispersion • Boiling point: the stronger the intermolecular forces → the higher the boiling point of the substances 13-Oct-16 13 13-Oct-16 14 13-Oct-16 15 13-Oct-16 16 • Example 3: Which of the following substances exhibits • Example 4: For each pair of inorganic substances, identify the hydrogen bonding? For those that do, draw two molecules of intermolecular forces in each substance, and select the the substance with the H-bonds between them a) C2H6 b) CH3OH substance with the higher boiling point c) CH3-CO-NH2 a) MgCl2 or PCl3 Solution b) CH3NH2 or CH3F Solution • C2H6 has no H-bonding sites (a non-polar molecule) a) Mg2+ and Cl- are held together by ionic bonds (a salt), while • CH3OH and CH3CONH2 have H-bonds between their molecules PCl3 is covalently bonded and the molecules are held together by dipole-dipole interactions Ionic attractions are much stronger than dipole interactions and so MgCl2 has the higher boiling point b) CH3NH2 and CH3F are both covalent compounds and have polar bonds The dipole in CH3NH2 can H-bond while that in CH3F cannot Therefore, CH3NH2 has the stronger interactions and the higher boiling point 17 18 13-Oct-16 • Example 5: For each pair of organic substances, identify the • Exercise 2: In each group following, identify all the intermolecular forces in each substance, and select the substance with the higher boiling point intermolecular forces present for each substance, and select a) CH3OH or CH3CH2OH b) hexane (CH3CH2CH2CH2CH2CH3) or 2,2-dimethylbutane the substance with the higher boiling point: (CH3C(CH3)2CH2CH3) a) CH3Br or CH3F Solution a) Both CH3OH and CH3CH2OH can H-bond, but CH3CH2OH has b) CH3CH2CH2OH or CH3CH2OCH3 more C-H bonds for greater dispersion force interactions c) C2H6 or C3H8 Therefore, CH3CH2OH has the higher boiling point b) Hexane and 2,2-dimethylbutane are both non-polar with only d) NH3, PH3, AsH3 dispersion forces to hold the molecules together Hexane has a e) HF, HCl, HBr, HI larger surface area, and therefore the greater dispersion forces and higher boiling point 19 12.4 Properties of the Liquid State 13-Oct-16 20 • Capillarity: the tendency of a liquid in a capillary tube or • Surface tension: the tension of the surface film of a liquid absorbent material to rise or fall as a result of surface tension caused by the attraction of the particles in the surface layer by  The rising of a liquid through a narrow space against the pull the bulk of the liquid, which tends to minimize surface area of gravity is called capillary action (capillarity)  Unit of surface tension = J/m2  Capillarity results from a competition between the  The stronger the forces are between the particles in a liquid, intermolecular forces within the liquid (cohesive forces, H the greater the surface tension bonding for water) & those between the liquid and the tube walls (adhesive forces) 13-Oct-16 21 13-Oct-16 22 12.6 The Solid State: Structure, Properties, and Bonding • Viscosity: the viscosity of a fluid is a measure of its resistance to gradual deformation by shear stress or tensile stress  Both gases and liquids flow, liquid viscosities are much higher because the intermolecular forces operate over much shorter distances  Viscosity decreases with heating  Molecular shape plays a key role in a liquid’s viscosity • Solids are divided in two broad categories based on the orderliness of their shapes:  Crystalline solids generally have a well-defined shape, because their particles—atoms, molecules, or ions—occur in an orderly arrangement For example: galena, quartz  Amorphous solids have poorly defined shapes because their particles lack long range ordering throughout the sample  Stronger intermolecular forces create higher viscosity 13-Oct-16 23 13-Oct-16 24 13-Oct-16 The crystal lattice and the unit cell • Crystal structure: consists of the same group of atoms, the Crystal lattice basis, positioned around each and every lattice point • Unit cell: the smallest group of atoms of a substance that has the overall symmetry of a crystal of that substance, and from which the entire lattice can be built up by repetition in three dimensions • There are crystal systems and 14 types of unit cells that occur in nature, but we will be concerned primarily with the cubic system, which gives rise to the cubic lattice 13-Oct-16 25 13-Oct-16 26 13-Oct-16 28 •There are three types of cubic unit cells within the cubic system: Simple cubic unit cell (SC) Body-centered cubic unit cell (BCC) Face-centered cubic unit cell (FCC) 13-Oct-16  27 SC The contribution of particular spheres in a unit cell:  Spheres inside of the unit cell → contribute sphere  Spheres at the surface of unit cell → contribute ½ sphere  Spheres at the edge of unit cell → contributed ¼ sphere  Spheres in the corner of unit cell → contributed ⅛ sphere 1/8 atom at corners Atoms in a unit cell = 1/8 x = 13-Oct-16 29 BCC 1/8 atom at corners atom at center FCC 1/8 atom at corners 1/2 atom at faces Atoms in a unit cell Atoms in a unit cell = (1/8 x 8) + = = (1/8 x 8) + (1/2 x 6) = 13-Oct-16 • Coordination number (called Ligancy) of a central atom in a Packing Efficiency molecule (or crystal) is the number of its near neighbors • Packing efficiency (%P) is the fraction of volume in a crystal (atoms, ions, or molecules) structure that is occupied by constituent particles  The SC unit cell %P = 52%  The BCC unit cell: %P = 68% Coordination number = Coordination number =  The hexagonal, FCC unit cells %P = 74% Coordination number + + = 12 13-Oct-16 31 13-Oct-16 32 33 13-Oct-16 34 Relationship between edge length (A) & atomic, ionic radius (r) in the three cubic unit cells 13-Oct-16 • Example 6: Barium is the largest non-radioactive alkaline earth • Example 7: Copper has the most essential industrial metal It has a body-centered cubic unit cell and a density of 3.62 applications of the three coinage metals Its crystal structure g/cm3 What is the atomic radius of Ba? (Vphere=4/3pr3) adopts cubic closest packing, and the edge length of the unit Solution • cell is 361.5 pm What is the atomic radius of copper? mol of Ba = 137.3 g Ba (MBa) = 6.022x1023 Ba atoms (NA) Solution • • Body-centered cubic unit cell has 68% packing efficiency Copper crystallizes in cubic closest packing → its structure has a face-centered cubic unit cell 35 13-Oct-16 36 13-Oct-16 Ionic Solids: Types and Properties of Crystalline Solids • The five most important types of crystalline solids  Atomic Solids: atoms are held together by dispersion forces  Molecular Solids: molecules occupy the lattice points, various combinations of dipole-dipole, dispersion, and Hbonding forces are at work in molecular solids Ionic Solids:  Ionic Solids: the unit cell contains particles (cations & anions), the unit cell has the same cation/anion ratio as the empirical formula  Metallic Solids:  Network Covalent Solids 13-Oct-16 37 Ionic Crystalline Solids 38 • Example 9: Show the empirical formula for unit cell of Calcium • The interparticle forces (ionic bonds) are much stronger than the van der Waals forces in atomic or molecular solids • In the unit cells, cations are surrounded by as many anions as possible, and vice versa, with the smaller of the two ions lying in the spaces (holes) formed by the packing of the larger • The unit cell has the same cation/anion ratio as the empirical formula • Example 8: Show the empirical formula for unit cell of NaCl:  The number of ions Na+ = = 12ãẳ + 1ã1 =  The number of ions Cl- = = 8ã + 6ãẵ = The formula for the unit cell will be: Na4Cl4 = NaCl 13-Oct-16 13-Oct-16 fluorite structure? Solution • The number of ions Ca2+ = 8• + 6ãẵ = ã The number of ions F- = 8•1 = • The empirical formula for the unit cell will be: Ca4F8 = CaF2 39 13-Oct-16 40 12.7 Advanced Materials • Exercise 3: Zinc oxide adopts the • Electronic Materials: are materials studied and used mainly for zinc blende crystal structure (beside) How many Zn2+ & O2- their electrical properties The electric response of materials largely stems from the dynamics of electrons, and their interplay ions are in the with atoms and molecules ZnO unit cell?  Crystal defects in crystals  Doped semiconductors: n-type semiconductor, p-type • Exercise 4: Iron crystallizes in a body-centered cubic structure semiconductor  Placing a p-type semiconductor adjacent to an n-type If the atomic radius of Fe is 126 pm, find the edge length (in creates a p-n junction nm) of the unit cell  A modern computer chip the size of a nickel may incorporate millions of p-n junctions in the form of transistors 13-Oct-16 41 13-Oct-16 42 13-Oct-16 • Liquid Crystals: a substance that flows like a liquid but has • Ceramic Materials: A ceramic material is an inorganic, non- some degree of ordering in the arrangement of its molecules metallic, often crystalline oxide, nitride or carbide material  These materials flow like liquids but, like crystalline solids, Some elements, such as carbon or silicon, may be considered pack at the molecular level with a high degree of order ceramics Ceramic materials are brittle, hard, strong in  Gases have no order, and liquids have little more Both are compression, weak in shearing and tension considered isotropic The properties of a crystal depend on direction, so a crystal is anisotropic  Ceramics are defined as nonmetallic, nonpolymeric solids  Liquid crystal phases can arise in two general ways, and, that are hardened by heating to high temperatures sometimes, either way can occur in the same substance:  High-tech ceramics incorporate these traditional thermotropic phase, lyotropic phase, nematic phase, characteristics in addition to superior electrical and magnetic cholestericphase, smectic phase properties  The liquid crystal displays (LCDs) used in watches, 13-Oct-16 calculators, cell phones, and computers 43 • Polymeric Materials: a polymer (Greek, “many parts”) is an 13-Oct-16 44 • Nano Materials: a material having particles or constituents of extremely large molecule, or macro-molecule, consisting of a nanoscale dimensions, or one that is produced by nanotechnology covalently linked chain of smaller molecules, called monomers Nanomaterials describe, in principle, materials of which a single (Greek, “one part”) unit is sized (in at least one dimension) between and 1000 nanometres (10−9 meter) but is usually 1—100 nm (the usual  Synthetic polymers such as plastics, rubbers, and specialized glasses have revolutionized everyday life definition of nanoscale)  Polymer mass, Polymer size and shape, Polymer  Nanoscale Optical Materials, Nanostructured Materials, crystallinity High-Surface-Area Materials, Nanomachines  Flow Behavior of Polymers, Molecular Architecture of  Nanoscale materials can be made one atom or molecule at Polymers: Dendrimers, Crosslinks, elastomers, copolymer, a time through construction processesinvolving selfassembly and controlled orientation of molecules 13-Oct-16 45 13-Oct-16 46 Chapter 12: Suggested Practice Problems 1, 5-17(odd), 20, 21, 23, 25, 31,32,34,37,39,41,43,45,49,51,53 48 13-Oct-16 13-Oct-16 47 13-Oct-16 48 ... of significant figures in a quantity, the greater its certainty 28-Aug-16 28-Aug-16 23 24 28-Aug-16 Example 6: Determining which digits are significant Determining Which Digits are Significant... 28-Aug-16 28 Rules for Significant Figures in Calculations Example 8: Significant Figures in Calculations For multiplication and division The answer contains the a) Multiplication: 9.21 cm x 6.8 cm... OH- hydroxide hydronium cobalt(II) cobalt (III) manganese(II) CH3COOCNClO3NO2- acetate cyanide chlorate nitrite manganese(III) tin(II)/ stannous tin(IV)/ stannic - NO3 MnO4CO32SO42PO43- nitrate