IB Chemistry HL Study Guide PDF

STUDY GUIDE: HL www.ib.academy  IB Academy Chemistry Study Guide Available on learn.ib.academy Author: Tim van Puffelen Design Typesetting This work may be shared di...

STUDY GUIDE: HL www.ib.academy  IB Academy Chemistry Study Guide Available on learn.ib.academy Author: Tim van Puffelen Design Typesetting This work may be shared digitally and in printed form, but it may not be changed and then redistributed in any form. Copyright © 2020, IB Academy Version: CheHL.3.0.200227 This work is published under the Creative Commons BY-NC-ND 4.0 International License. To view a copy of this license, visit creativecommons.org/licenses/by-nc-nd/4.0 This work may not used for commercial purposes other than by IB Academy, or parties directly licenced by IB Academy. If you acquired this guide by paying for it, or if you have received this guide as part of a paid service or product, directly or indirectly, we kindly ask that you contact us immediately. Laan van Puntenburg 2a ib.academy 3511ER, Utrecht [email protected] The Netherlands +31 (0) 30 4300 430  0 Welcome to the IB.Academy Study Guide for IB Chemistry High Level. We are proud to present our study guides and hope that you will find them helpful. They are the result of a collaborative undertaking between our tutors, students and teachers from schools across the globe. Our mission is to create the most simple yet comprehensive guides accessible to IB students and teachers worldwide. We are firm believers in the open education movement, which advocates for transparency and accessibility of academic material. As a result, we embarked on this journey to create these study guides that will be continuously reviewed and improved. Should you have any comments, feel free to contact us. For this Chemistry HL guide, we incorporated everything you need to know for your final exam. The guide is broken down into manageable chapters based on the syllabus topics. The chapter is then further divided into subtopics. For more information and details on our revision courses, be sure to visit our website at ib.academy. We hope that you will enjoy our guides and best of luck with your studies. IB.Academy Team 3 TABLE OF CONTENTS 1. Quantitative chemistry 7 – Types and states of matter – Chemical reactions – Mole concept and chemical calculations 2. Atomic structure 21 – Types of particles – Notation – Isotopes: abundance and Ar – Atomic shells/subshells/orbitals – Electromagnetic spectrum – Ionization energies 3. Periodicity 31 – The Periodic Table (PT) – Periodic trends – Transition elements 4. Bonding 39 – Metallic Bonding – Ionic bonding – Covalent bonding – Intermolecular forces – Properties of molecular compounds – Molecular orbitals – Hybridization: mixing atomic orbitals – Ozone and oxygen 5. Energetics 61 – Temperature vs heat vs enthalpy – Energy diagrams – Hess’s law – Energy calculations – Energy cycles – Entropy 6. Kinetics 79 – Collision Theory – Rate equation and reaction order 5  TABLE OF CONTENTS 7. Equilibrium 87 – Dynamic equilibrium – Equilibrium law expression – States of matter – Le Chatelier’s principle – Equilibrium calculations – Relation between ∆G and Kc 8. Acids and bases 95 – Acid and base definitions – Strong vs weak – pH scale – Buffers – pH curves – Acid Deposition 9. Redox 105 – Oxidation states – Reactions – Reactivity – Electrochemical cells – The Winkler method and the BOD 10. Organic chemistry 115 – Fundamentals of organic chemistry – Isomers – Reactions – Reaction mechanisms – Reactions overview and retrosynthesis 11. Measurement and data 137 processing – Graphical Techniques – Spectroscopic identification 6 QUANTITATIVE CHEMISTRY 1 1.1 Types and states of matter all matter mixtures substances elements compounds non-metals (semimetals) metals ionic molecular compounds compounds Classification of matter The chemical composition of a Substance has a definite chemical composition and characteristic properties substance is expressed in a chemical formula, Mixture contains multiple substances that retain their individual properties which shows the number of each atom in because they are not chemically bonded, which can be separated using a substance (e.g. H2 O), the difference between the individual properties of each substance. or the ratio of ions in an ionic compound (e.g. MgCl2 ). Homogeneous mixture: Heterogeneous mixture: Components are in the same phase, Components are not all in the same particles are distributed equally over phase, there are physical boundaries the mixture between the components. e.g. solution, alloy e.g. suspension, emulsion Classification of matter: type of element(s): metal metal ionic compound metal + non-metal molecular compound non-metal Elements are atoms that have the same number of protons. Elemental substances contain one type of element (e.g., Na, Fe, H2 , Cl2 , S8 ,... ) Compounds at least two different elements combine to form a compound 7  QUANTITATIVE CHEMISTRY Types and states of matter Note that for ionic molecular formula structural formula empirical formula compounds only the empirical formula is used, because ions H H assemble in a whole number ratio in a example C 2 H4 C C CH2 lattice, but not as H H molecules. showing number of atoms bonding between the simplest number atoms ratio of atoms. Which compound has the empirical formula with the greatest mass? Example molecular divisible empirical formula by formula A. C 2 H6 2 CH3 B. C4 H10 2 C 2 H5 ← greatest mass C. C5 H10 5 CH2 D. C6 H6 6 CH Find the empirical formula given weight percentage composition A compound is found to contain 64.80 % C, 13.62 % H, and 21.58 % O2 by weight. What is the empirical formula for this compound? 1. Tabulate and assume 100 g C H O grams 64.80 g 13.62 g 21.58 g 2. Convert the masses to C H O moles (divide by the grams 64.80 g 13.62 g 21.58 g atomic mass) moles 5.396 mol 13.49 mol 1.344 mol 3. Divide by the lowest, C H O seeking the smallest grams 64.80 g 13.62 g 21.58 g whole-number ratio moles 5.396 mol 13.49 mol 1.344 mol simplest 4 10 1 ratio 4. Write the empirical formula C4 H10 O 8 QUANTITATIVE CHEMISTRY Types and states of matter 1 Derive the molecular formula from the empirical formula To determine the molecular formula −1 instead of the empirical From the previous, derive the molecular formula if the molecular mass is 222.4 g mol. formula, the molecular mass must also be 1. molecular mass 222.4 g mol−1 given. =3 mass of the empirical formula 4 · 12.01 + 10 · 1.01 + 16.00 g mol−1 2. write the molecular formula C12 H30 O3 (since the molecule is 3 times the mass of the empirical formula) Phase changes and states of matter gas (g) The state of a substance is indicated as (s), (l) or (g). on dep i rat ion sub osi po Additionally, when a sat ti lim on eva den substance is dissolved in ati con water we can indicate the on phase as (aq). freezing liquid solid (l) (s) melting solid (s) liquid (l) gas (g) movement of particles vibrational free movement free movement distance between particles close close far apart fixed volume yes yes no, compressable fixed shape yes, rigid no no 9  QUANTITATIVE CHEMISTRY Chemical reactions 1.2 Chemical reactions Chemical reaction a process that leads to the transformation of one set of chemical substances to another, thus changing their chemical formulae Combustion reaction a chemical reaction between a fuel and O2 ; which compounds form depends on which elements the fuel contains fuel contains combustion product effect (environmental) C CO2 (complete) greenhouse gas CO (incomplete) toxic to animals H H2 O — S SO2 acid rain (see chapter 8) N NOx acid rain (see chapter 8) Balancing and stoichiometry To balance reactions we use the conservation of mass, which states that the number of atoms before and after a reaction must be equal, and the conservation of charge, which states that the charge before and after a reaction must also be equal. Stoichiometric coefficients the numbers placed in front of substances in order to balance chemical reactions Stoichiometry the quantitative relationships between substances in a chemical reaction (molar ratios). Balance the reaction:... C10 H22 +... O2 −−→... H2 O +... CO2 Example The trick to balancing chemical reactions is to balance elements in order of occurance. Both C and H occur in one substance before and after the reaction arrow, so balance these first: 1 C10 H22 + O2 −−→ 11 H2 O + 10 CO2 31 Next, balance O: 1 C10 H22 + O −−→ 11 H2 O + 10 CO2 (multiply by 2) 2 2 2 C10 H22 + 31 O2 −−→ 22 H2 O + 20 CO2 10 QUANTITATIVE CHEMISTRY Mole concept and chemical calculations 1 1.3 Mole concept and chemical calculations Mole (n) the amount of substance which contains NA = 6.02 × 1023 particles (atoms, molecules, ions, electrons, or other physical particles) Avogadro’s constant (NA) 6.02 × 1023 particles/mol When buying eggs, you can request one dozen eggs from old-lady Mme. Oeuf. It is a convenient expression, since the packaging contains 12 eggs. So a dozen is an alternative way to express “12”. The mole is analogously an alternative way to express the number of entities (6.02 × 1023 ). This number is convenient to represent chemical amounts. A dozen is a grouping of 12, so: A mole is a grouping of 6.0 × 1023 , so: 2 dozen is a grouping of 24. 2 mole is a grouping of 1.2 × 1024. The mole concept is a necessity in chemical calculations. Since we constantly deal with huge numbers of particles in chemistry, expressing the number of particles in moles is more convenient. But more importantly, particles react and form in a particular stoichiometric ratio (molar ratio) in chemical reactions. Take 2H2 + O2 −−→ 2H2 O; two molecules of H2 will react with one molecule of O2. This does not mean that two grams of H2 will react with one gram of O2 , since the masses of H2 and O2 molecules are not equal. This does mean that two moles of H2 will react with one mole of O2 , but also that 1.8 × 10−3 mol H2 will react with 0.9 × 10−3 mol O2. Furthermore, in chemistry we constantly use huge ensembles of molecules. Using moles allows us to use much simpler numbers. 11  QUANTITATIVE CHEMISTRY Mole concept and chemical calculations Relative and average mass The relative masses are all measured relative to the atomic mass unit (u): defined as 1/12 of the atomic mass of a 12C atom. And they are all average: the weighed arithmetic mean of all isotopes and their terrestrial natural abundancies. Relative atomic mass (Ar ) the weighed mean of all the isotopes of one element and their natural abundances, relative to one atomic mass unit, which is 1/12 of the mass of a 12C atom Relative molecular mass (M r ) is the sum of the relative atomic masses of all Molecules have a the atoms in the molecular formula well-defined number of atoms, but ionic Relative formula mass (M r ) applies to ionic compounds, and it is similar to compounds do not consist of a particular the relative molecular mass and also calculated in the same way. number of ions, rather of ions in a particular Molar mass M the mass of a substance per one mole expressed in g mol−1 ratio. 1.3.1 Unit conversion Roadmap to chemical calculations It is much easier to measure some physical quantity of a sample, such as its mass, than it is to count the number of particles in the sample. Therefore, you will have to convert various quantities to mole and back. The figure below gives an overview of the quantities that can be converted to moles and back, and what other quantity you will need to do so. N◦ of particles no unit NA concentration mass Vsolvent Mr mol dm−3 mol g Vm m3 volume (g) 12  QUANTITATIVE CHEMISTRY Mole concept and chemical calculations 1 Mole ↔ number of particles The number of particles can be expressed in the amount of moles, or in the number of particles. Because 1 mol = 6.02 × 1023 particles. The quantities express the same thing, but use different values in doing so. It’s like saying “a dozen eggs” to express 12 eggs. The relationship between the number of particles and the amount in mol is given by: N◦ of particles = n · NA N◦ of particles N◦ of particles = amount [no units] NA n n = chemical amount [mol] mol−1 mol NA = Avogadro’s constant 6.02 × 1023 mol−1 How many N-atoms are there in 1.0 × 10−2 mol NH3 NO3 ?. Example N◦ of particles = n · NA = 1.0 × 10−2 mol · 6.02 × 1023 mol−1 = 6.02 × 1021 molecules, per molecule there are 2 N-atoms (mole ratio molecules : N-atoms = 1 : 2), so 2 · 6.02 × 1021 = 1.2 × 1022 N-atoms. What is the total number of atoms in 0.20 mol of propanone, CH3 COCH3 ? N◦ of particles = n · NA = 0.20 mol · 6.02 × 1023 mol−1 = 1.2 × 1023 molecules, per molecule there are 10 atoms (mole ratio molecules : atoms = 1 : 10), so 10 · 1.2 × 1023 = 1.2 × 1024 atoms. Mole ↔ gram The molar mass (M ) can be calculated from the formula of the substance, which is the mass of a substance per one mole particles (in g mol−1 ). It allows us to convert between the mass and the amount of particles in moles. The relationship between the amount in mol and mass in g is given by: m =n·M m g m = mass [g] n M n = chemical amount [mol] mol g mol−1 M = molar mass [g mol−1 ]. What is the amount in moles of 4.00 g in NaOH? Example M (NaOH) = 22.99 + 16.00 + 1.01 = 40.0 g mol−1 m 4.00 g n= = = 0.100 mol NaOH M 40.0 g mol−1 13  QUANTITATIVE CHEMISTRY Mole concept and chemical calculations Mole ↔ [concentration] Solution a homogeneous mixture of a substance (the solute) dissolved in another substance (the solvent) (Molar) concentration (C) is the amount of solute (in mol) per unit volume (in dm3 ), often written using [... ], and expressed in mol dm−3 Standard solution a solution with a known concentration of solute The solute concentration is independent of the solvent volume. When we dissolve 4 mol sugar in 2 dm3 2 dm3 water, it has a particular ‘sweetness’. This sweetness is a measure of the 1 dm3 concentration of sugar in water. If we would poor out 1 dm3 from the solution, it would still be equally sweet 4 mol 2 mol = 2 mol dm−3 = 2 mol dm−3 (same concentration) but only contain 2 dm3 1 dm3 half the sugar content (half the amount in moles). The relationship between the amount in mol and concentration in mol dm−3 is given by: n = Vsolvent · C n mol n = chemical amount [mol] C Vsolvent Vsolvent = solvent volume [dm3 ] mol dm−3 dm3 C = concentration [mol dm−3 ]. What amount of NaCl (in moles) is required to prepare 250 cm3 of a Example It’s very useful to remember that 0.200 mol dm−3 solution? cm3 · mol dm−3 = mmol n = Vsolvent · C = 250 cm3 · 0.200 mol dm−3 = 50 mmol Which solution contains the greatest amount (in mol) of solute? Vsolvent ·C = n in mol A. 10.0 cm3 of 0.500 mol dm−3 NaCl 10.0 ·0.500 = 5 mmol B. 20.0 cm3 of 0.400 mol dm−3 NaCl 20.0 ·0.400 = 8 mmol 3 −3 C. 30.0 cm of 0.300 mol dm NaCl 30.0 ·0.300 = 9 mmol ← D. 40.0 cm3 of 0.200 mol dm−3 NaCl 40.0 ·0.200 = 8 mmol 14  QUANTITATIVE CHEMISTRY Mole concept and chemical calculations 1 Mole ↔ volume gas Avogadro’s law equal volumes of all gases, at the same temperature and pressure, have the same number of molecules Assuming the gas is an ideal gas, Avogadro’s law is tested quite Molar volume (Vm ) the volume of one mole gas, expressed in dm3 mol−1 or often: it should be m3 mol−1 , at a particular pressure and temperature. understood that the molar ratio can be applied to gas volumes. The volume of an ideal gas at constant temperature and pressure is proportional to the number of particles (in moles). So when twice the number of particles are placed into a cylinder, then the volume becomes twice as large. The volume of one mole gas (molar volume) at STP is 22.7 dm3 mol−1. 22.7 dm 45.4 dm 273 K 100 kPa The relationship between the amount in mol and gaseous volume in dm3 is given by: Vgas = n · Vm Vgas dm3 n = chemical amount [mol] n Vm Vgas = gas volume [dm3 ] mol dm3 mol−1 Vm = molar volume [dm3 mol−1 ]. Calculate the volume of nitrogen gas produced by the decomposition of Example 2.50 mol of NaN3 (s) at STP in the reaction 2 NaN3 (s) −−→ 2 Na(s) + 3 N2 (g). 2.50 mol Since the molar ratio NaN3 : N2 = 2 : 3, 2 × 3 = 3.75 mol N2 forms. At STP Vm = 22.7 dm3 mol−1 , Vgas = n · Vm = 3.75 · 22.7 = 85.1 dm3 15  QUANTITATIVE CHEMISTRY Mole concept and chemical calculations 1.3.2 Gas laws and ideal gases An ideal gas is a theoretical gas that assumes that: the volume of particles is negligible compared to the volume of the surrounding empty space, and no kinetic energy is lost in the collisions between the particles. Whether these assumptions are justified is outside the scope of the IB syllabus, so from now on we will treat all gases as ideal gases. Pressure the force exerted by the collisions of particles on the walls of its container pressure at sea level = 100 kPa = 1.00 × 105 Pa Temperature the average kinetic energy of particles Tin K = Tin ◦C + 273 and Tin ◦C = Tin K − 273 STP standard temperature and pressure: 273 K and 100 kPa SATP standard ambient temperature and pressure: 298 K and 100 kPa The ideal gas law assumes ideal gas behaviour, and it is an equation that relates the pressure, volume, amount in moles and the temperature of a gas. Critically, SI units must Memorize the ideal gas be used in the ideal gas law: law: pV = nRT. For paper 2 you will have it in the databook, but p = pressure [Pa] you will also need it for V = volume [m3 ] paper 1 questions! pV = nRT n = amount of substance [mol] R = ideal gas constant 8.31 J K−1 mol−1 T = temperature [K] Using the ideal gas law, verify that Vm at STP is 22.7 dm3 mol−1.. Example V Molar volume (Vm ) = the volume (V ) per mole (n), or Vm =. n V RT Rearrange the ideal gas law: = = Vm n p RT 8.31 · 273 K Vm = = = 2.27 × 10−2 m3 mol−1 = 22.7 dm3 mol−1 p 1.00 × 105 Pa 16  QUANTITATIVE CHEMISTRY Mole concept and chemical calculations 1 Questions involving the ideal gas law in paper 1 are often presented ‘at constant mass’; this means that the number of moles n is kept constant. R is also a constant. Rearranging the ideal gas law with the variables to one side, and the constants to the other yields: pV = nR. Since only p, V and T are allowed to change, nR will remain constant. T Mathematically, two situations (1 ) and (2 ) can be related by: p1V1 pV nR = = 2 2 Memorize this formula T1 T2 or how to derive it, because it will allow p1V1 pV you to answer all From = 2 2 we can derive Boyle’s law, Charles’ law and Gay-Lussac’s Law (and questions regarding gas T1 T2 laws! Wikipedia knows which is which). In each of the three gas laws, one of the quantities ( p, V or T ) is fixed, as well n (‘at constant mass’). constant pressure constant volume constant temperature pZ 1 V1 Zp2V2 V@ p1@ p2@V@ p1V1 p2V2 derivation of = Z 1 = 2 Z T1 T2 = gas law T1 T2 T@ @1 T@ @2 V1 V2 p1 p2 = = p1V1 = p2V2 T1 T2 T1 T2 relation V ∝T p ∝T 1 p∝ V V P V graph T T P. At constant temperature, sketch a diagram that shows how p changes when V Example changes. p1V1 = p2V2 holds, so put in some numbers to figure out the relation. Assume p1V1 = 1 · 1 = 1: P P × V = 1 = constant 1 1  ‹ 2 × = 1 = constant ,2 2 2 (1, 1) 1 ‹ 1 × 1 = 1 = constant 2, 1 2 × 2 = 1 = constant 2 V 17  QUANTITATIVE CHEMISTRY Mole concept and chemical calculations 1.3.3 Chemical calculations The basics of all chemical calculations can be summarized as follows: Known substance X 1 2 3 Known substance Y mass g × molar ratio g mass concentration mol L−1 mol mol mol L−1 concentration substance substance volume m3 X Y m3 volume Required N◦ of particles molar mass N◦ of particles dm3 solution molar volume Performing chemical calculations 5.0 g of CH4 undergoes complete combustion. Calculate the volume of the resulting gases under STP assuming that water forms as a gas. 1. Note the reaction equation and list CH4 + 2O2 −→ CO2 + 2H2 O 5.0 g CH4 the information given 2. Convert units to mole CH4 M m = 12.0 + 4 · 1.01 = 16.04 g mol−1 5.0 = 0.312 mol CH4 16.04 3. Use the molar ratio to convert to the The volume of all the gases that form is number of moles of the substance(s) required. Per 1 mol CH4 , 3 mol gas forms asked for 3 0.312 mol CH4 · = 0.935 mol gas. 1 4. Convert moles to required units Under STP Vm = 24.5 dm3 mol−1 , −1 3 0.935 mol · 24.5 dm mol = 22.9 dm3 5. Check significant figures and units Looking back to step 1 the amount of CH4 is given in two significant figures, so the answer should also be written using two significant 3 figures −→ 23 dm 18 QUANTITATIVE CHEMISTRY Mole concept and chemical calculations 1 Limiting and excess reactant 12 table legs and 4 table tops are stored in the warehouse of a table factory. Our intuition tells us that we cannot make 4 tables, since it would require 16 table legs. So even though more table legs are available than table tops, due to the ratio in which they are needed the table legs are the limiting reactant and the table tops are in excess. Doing chemistry we do not have the same intuition, but we perform the same math. Theoretical yield the maximum quantity of product that can be obtained from given quantities of reactants, assuming completion Limiting reactant the reactant that determines the theoretical yield of a product, after the reaction is complete there will be none left Excess reactant the reactant that is not used up by the reaction, after the reaction is complete this substance will still be present All calculations must be done using the amount of the limiting reactant, since it determines how much product will be made (i.e. the available 12 table legs determine that the theoretical yield of tables is 3). Identify the limiting reactant 4.22 g Al reacts with 25.0 g Br2 in the following reaction: 2Al + 3Br2 −→ 2AlBr3. Identify the limiting reactant and use it to determine the theoretical yield of AlBr3. 1. Convert units to moles 4.22 g = 0.156 mol Al 26.98 g mol−1 25.0 g = 0.156 mol Br2 2 · 79.90 g mol−1 2. Divide the number of moles of each 0.156 mol Al: = 0.078 reactant by its reaction coefficient. The 2 reactant with the lowest result will be the 0.156 mol Br2 : = 0.052 limiting reactant. 3 So Br2 is the limiting reactant. 3. Use the number of moles of the limiting The molar ratio of AlBr3 : Br2 = 2 : 3, so reactant from step 1 and the molar ratio 2 0.156 mol · = 0.104 mol AlBr3 to calculate the number of moles of the 3 requested substance 19  QUANTITATIVE CHEMISTRY Mole concept and chemical calculations 20 ATOMIC STRUCTURE 2 2.1 Types of particles Particles Atoms Ions Molecules charged cluster of no charge +/−: atoms covalently N◦ p+ = N◦ e− with e – defi- bonded atoms ciency/sufficiency protons neutrons electrons + cation − anion elements have isotopes have a core e – vs p+ > e− p+ < e− the same Z/p+ different N◦ n0 valence e – Atoms contain subatomic particles: protons, neutrons and electrons Nucleus + protons and neutrons form the atom’s nucleus Electron cloud − electrons occupy the space outside the nucleus in shells/subshells/orbitals notation mass relative mass charge relative charge proton p+ 1.67 × 10−24 g 1u 1.60 × 10−19 C +1 neutron n0 1.67 × 10−24 g 1u 0C 0 electron e− 9.11 × 10−28 g ≈0 u −1.60 × 10−19 C −1 21  ATOMIC STRUCTURE Notation Atom charge = 0. the smallest constituent unit of ordinary matter that has the properties of a chemical element anion: a negative ion Ion charge 6= 0, so the number of e − 6= the number of p+ Negative ion/anion − contains more e − than p+ cation: is pawsitive Positive ion/cation + contains less e − than p+ Element all atoms of the same element have the same number of p+ (i.e. the same atomic number Z) Isotopes have the same chemical properties, Isotopes atoms of the same element but with a different number of n0 , but different physical properties resulting in a different mass number AX. 2.2 Notation massnumber A = N◦ p+ + N◦ n0 12 charge: e − deficiency/sufficiency 6 C chemical symbol, determined by: atomic number Z = N◦ p+ This atom has 8 p+ and 10 n0 , what is the chemical notation for this ion?. Example Z = N◦ p + = 8   nuc A = N◦ p+ + N◦ n0 = 8 + 10 = 18 18 2− 8O ◦ + ◦ − charge = N p − N e = 8 − 10 = −2  − ions have extra e − ! Find the symbol, Z, p+ , n0 , e − for: 11 5 X, 199X – and 27 Al3+ Symbol and Z N◦ p+ N◦ n0 N◦ e − 11 5 X Z= 5, so B 5 11 − 5 = 6 5 19 – 9 X Z= 9, so F 9 19 − 9 = 10 9 + 1 = 10 27 3+ Al Al, so Z= 13 13 27 − 13 = 14 13 − 3 = 10 22  ATOMIC STRUCTURE Isotopes: abundance and Ar 2 2.3 Isotopes: abundance and Ar Relative atomic mass (Ar ) the weighed mean of all the isotopes of one 1 u = 1 g mol−1 element and their natural abundances, relative to one atomic mass unit, which is 1/12 of the mass of a 12C atom     fractional fractional mass of  mass of Ar = abundance × + abundance × +... isotope 1 isotope 2 of isotope 1 of isotope 2. 79 81 Example Calculate Ar of bromine, given that the abundancies of Br and Br are 50.69 % and 49.31 %. Ar (Br) = 50.69 % · 79Br + 49.31 % · 81Br = 0.5069 · 79 + 0.4931 · 81 = 79.90 g mol−1 Calculate the abundancies of 69Ga and 71 Ga, given these are the only stable isotopes of Ga and Ar = 69.72 g mol−1 Since 69Ga and 71Ga are the only stable isotopes we can say that: fractional abundance 69Ga + fractional abundance 69Ga = 1(or 100 %) If we let x = fractional abundance 69Ga then fractional abundance 71Ga = 1 − x Ar (Ga) = x · 69Ga + (1 − x) · 71Ga 69.72 g mol−1 = x · 69 + (1 − x) · 71 = 69x + 71 − 71x x = 0.64 So the abundance of 69Ga is 64 % and the abundance of 71Ga is 100 % − 64 % = 36 % 23  ATOMIC STRUCTURE Isotopes: abundance and Ar Mass spectrometer A mass spectrometer is an analytical instrument that can measure the mass of each isotope in a sample. So if a sample of lead, Pb, is injected into the device, the following spectrum and relative abundances will result: 100 100 90 80 Relative abundance 70 Relative 60 Isotope abundance 50 46.2 42.3 204Pb 3.9 40 206Pb 46.2 30 207Pb 42.3 20 208Pb 100 10 3.9 203 204 205 206 207 208 209 Mass/Charge (m/z) The data of the mass spectrum allows us to calculate the average atomic mass by weighing the isotopic mass against its relative abundance. Note that the abundance is not given as a %, so we have to divide by the sum of all the relative abundances. 3.9 · 204 + 46.2 · 206 + 42.3 · 207 + 100 · 208 Ar (Pb) = = 207.2 g mol−1 3.9 + 46.2 + 42.3 + 100 24 ATOMIC STRUCTURE Atomic shells/subshells/orbitals 2 2.4 Atomic shells/subshells/orbitals Electron shell is also Electron shell n = 1, 2, 3... principal energy level which each contains 2n 2 often referred to as the electrons, further divided in a number of subshells main energy level Subshells s, p, d, f each subshell has a particular number of orbitals, and each has its own geometry Atomic orbital region with a specific geometry that can host two electrons of opposite spin shell max. N◦ of e − N◦ of orbitals electron n 2n 2 s p d f total configuration 1 2 · 12 = 2 1 — — — 1 1s x 2 2 · 22 = 8 1 3 — — 4 2s x 2p x 3 2 · 32 = 18 1 3 5 — 9 3s x 3p x 3d x 4 2 · 42 = 32 1 3 5 7 16 4s x 4p x 4d x 4f x The principal electron shells can be imagined as an onion: the first shell (n = 1) is closest to the nucleus and each of the following shells is further away. The shells are numbered: n = 1, n = 2,... + Every shell contains the s subshell, from the second shell (n = 2) forward all shells contain the p subshell, from the third shell (n = 3) all shells contain the d subshell etc. The shape of the subshells Every s subshell consist of one spherical orbital, which is further away from the nucleus the higher the shell number. The p subshell always contains 3 orbitals which are aligned along the xy z-axis. Since the second shell (n = 2) contains the 2s 2p subshells which can host 2 and 6 electrons respectively, the maximum total number of electrons in the second shell is 8. s orbital p x orbital py orbital p z orbital 25  ATOMIC STRUCTURE Atomic shells/subshells/orbitals 2.4.1 Electron configuration E Electron configuration shows the number of e – in each subshell in the groundstate 4p ˜— ˜ ˜ ˜ ˜ (the lowest energy state) ˜— 3d Aufbau principle = 1s 2s 2p 3s 3p 4s 3d 4p 4s ˜— ˜— ˜— electrons are placed into orbitals with ˜— 3p lowest energy first, which are not 3s necessarily orbitals closest to the ˜— ˜— ˜— nucleus. Importantly: electrons occupy ˜— 2p 4s before 3d orbitals. 2s Pauli’s exclusion principle per orbital a maximum of 2 e – with opposite spin are allowed. Hund’s rule instead of forming pairs in the ˜— same orbital, electrons rather occupy 1s empty orbitals to minimize repulsion. Write the electron configuration of atoms Note that when asked to write the full electron Write the full and condensed electron configuration for Fe. configuration, the condensed form is incorrect! 1. Determine the total number of e− 26Fe, is an atom so: number of p+ = number of e – = 26 2. Allocate each electron to each subshell 2 2 6 2 6 2 6 1s 2s 2p 3s 3p 4s 3d according to the Aufbau principle. Place max. 2 e – in the s subshells, max. 6 e – in Abbreviated form: the p subshells, max. 10 e – in d subshell. 2 6 [Ar] 4s 3d To write the electron configuration of ions, we have to add or remove electrons from the electron configuration of the atom equal to the charge of the ion. As the 3d sublevel becomes populated with electrons, the relative energies of the 4s and 3d fluctuate relative to one another and the 4s ends up higher in energy as the 3d sublevel fills. This means that 4 s electrons are removed prior to 3 d electrons. 26 ATOMIC STRUCTURE Electromagnetic spectrum 2 Write the electron configuration of ions Write the condensed electron configuration for Fe3+ 1. Write the electron configuration of the [Ar]4 s2 3 d6 atom 2. Add electrons / remove electrons from Fe3+ has 3 fewer e – than the atom. First the outermost shell (4 s before 3 d) remove two 4 s electrons, and then 5 remove one 3 d electron: [Ar] 3d 4 9 d and d exceptions Memorize these two The situation when subshells are completely filled, or incorrect correct exceptions because half-filled, is energetically favoured. An electron from 2 3d4 Cr [Ar]  4s [Ar] 4s1 3d5 they have been tested the 4 s subshell can be promoted to attain a half-filled 2 3d9 often! Cu [Ar]  4s [Ar] 4s1 3d10 d-subshell (d5 ) or full d-subshell (d10 ). 2.5 Electromagnetic spectrum Electromagnetic radiation a form of energy that propagates through space at the speed of light as electromagnetic waves, or photons Ephoton = energy of a photon [J] h = Planck’s constant 6.63 × 10−34 J s hc Ephoton = hν = ν or f = frequency [s−1 ] λ λ = wavelength [m] γ -ray X-ray Ultra violet Visible light 10−5 10−4 10−3 10−2 10−1 100 101 102 103 104 105 106 107 108 109 Infrared Wavelength λ (µm) Micro wave Radio wave Visible light (Vis) is an example of electromagnetic (EM) radiation. The colour of light is tied to the amount of energy of a photon. But visible light is only a small part of the EM spectrum; at the higher energy end of the spectrum we find ultraviolet (UV), X-ray and γ -ray, at the lower energy end of the spectrum we find infrared (IR), microwaves and radio waves. 27  ATOMIC STRUCTURE Electromagnetic spectrum Electron energy levels The energy level of an electron depends on which atomic orbital it occupies. The lowest energy level is called the groundstate; an electron can move to a higher energy level (excited state) by absorption of a photon. And similarly, an electron can move from an excited state to a lower energy level by emitting a photon. The transition between electron energy levels is only possible when the electron absorbs or emits a photon with exactly the same amount of energy as the difference between the energy levels. Energy transitions are discrete: of a particular amount of energy. photon emitted e− + proton lower energy state higher energy state When excited electrons ‘fall’ from a higher to a lower energy state, photons with a discrete amount of energy are emitted. The emission spectrum of atoms is a line spectrum: only light of a particular colour (discrete energy) is emitted. Continuous spectrum 380 400 420 440 460 480 500 520 540 560 580 600 620 640 660 680 700 720 740 760 780 Hydrogen 380 400 420 440 460 480 500 520 540 560 580 600 620 640 660 680 700 720 740 760 780 28 ATOMIC STRUCTURE Ionization energies 2 Hydrogen energy levels E n=6 ionised atom n=5 n=∞ n=4 n=6 n=3 n=5 n=4 n=2 excited n=3 states infrared UV Vis n = ··· → n = 3 n=1 n=2 visible light n = ··· → n = 2 ground n=1 ultraviolet state n = ··· → n = 1 Electrons that ‘fall’ to the groundstate (n = 1) emit photons with the greatest amount of energy (UV radiation). The length of the arrows is proportional to the amount of energy. Electrons that ‘fall’ to n = 2 emit visible light and to n = 3 emit infrared radiation. Also note that the energy levels converge at higher energy: the difference between the energy levels becomes smaller up to the point where the difference is 0. The energy difference between the more energetic photons is increasingly smaller. Therefore line spectra converge at higher energy. 2.6 Ionization energies First ionization energy the energy required to remove one mole electrons from one mole of gaseous atoms, to produce one mole of gaseous 1+ ions. He first ionization energy (kJ mol−1 ) Evidence for shells Ne From He over Ne to Ar: down the group it becomes easier to F Ar remove an electron since it is N Cl further away from the nucleus H O P (in a higher shell). Be C Mg S From He to Li: the first electron in B Si Ca a new shell is easily removed Al Li Na K because the effective nuclear charge is low (+1), so it is weakly attracted. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 atomic number 29  ATOMIC STRUCTURE Ionization energies From Li to Ne: across a period it becomes harder to remove an electron since the effective nuclear charge increases, so the attraction increases. Evidence for subshells From Be [He]s2 to B [He]s2 p1 : an electron is added to the p-subshell, which has a slightly higher energy than the s-subshell, meaning it can be removed more easily. From N [He]s2 p3 to O [He]s2 p4 : an electron is paired in a p-orbital causing increased repulsion, meaning it can be removed more easily. Successive ionisation energies Note that y -axis is plotted logarithmically, log(IE) so abrupt changes in the y direction are compressed but are still easy to identify. 1 2 3 4 5 6 7 8 9 number of electrons removed In the figure above we see the abrupt change occurring after the 3rd electron, which means that it is much harder to remove electron 4 than electron 3. All elements in group 13 (B, Al,Ga... ) will have this characteristic, since after removing 3 electrons, removing the 4th will break the noble gas configuration. In the figure above we see the abrubt change occurring after the 3rd electron, which means that it is much harder to remove electron 4 than electron 3. All elements in group 13 (B, Al,Ga... ) will have this characteristic, since after removing 3 electrons, removing the 4th will break the noble gas configuration. 30 PERIODICITY 3 3.1 The Periodic Table (PT) Period is a row in the periodic table, and it represents the principal electron shells (n = 1, 2,... ) Valence electrons are just the outermost Group is a column in the periodic table, and it groups elements with similar electrons chemical properties due to having the same number of valance e – 1 IA 18 VIIIA 1 1.0079 2 4.0025 1 H Z mass He Hydrogen 2 IIA 13 IIIA 14 IVA 15 VA 16 VIA 17 VIIA Helium 3 6.941 4 9.0122 X 5 10.811 6 12.011 7 14.007 8 15.999 9 18.998 10 20.180 Name 2 Li Be B C N O F Ne Lithium Beryllium Boron Carbon Nitrogen Oxygen Flourine Neon 11 22.990 12 24.305 13 26.982 14 28.086 15 30.974 16 32.065 17 35.453 18 39.948 3 Na Mg Al Si P S Cl Ar Sodium Magnesium 3 IIIA 4 IVB 5 VB 6 VIB 7 VIIB 8 VIIIB 9 VIIIB 10 VIIIB 11 IB 12 IIB Aluminium Silicon Phosphorus Sulphur Chlorine Argon 19 39.098 20 40.078 21 44.956 22 47.867 23 50.942 24 51.996 25 54.938 26 55.845 27 58.933 28 58.693 29 63.546 30 65.39 31 69.723 32 72.64 33 74.922 34 78.96 35 79.904 36 83.8 4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Potassium Calcium Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc Gallium Germanium Arsenic Selenium Bromine Krypton 37 85.468 38 87.62 39 88.906 40 91.224 41 92.906 42 95.94 43 96 44 101.07 45 102.91 46 106.42 47 107.87 48 112.41 49 114.82 50 118.71 51 121.76 52 127.6 53 126.9 54 131.29 5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Rubidium Strontium Yttrium Zirconium Niobium Molybdenum Technetium Ruthenium Rhodium Palladium Silver Cadmium Indium Tin Antimony Tellurium Iodine Xenon 55 132.91 56 137.33 57 138.91 72 178.49 73 180.95 74 183.84 75 186.21 76 190.23 77 192.22 78 195.08 79 196.97 80 200.59 81 204.38 82 207.2 83 208.98 84 209 85 210 86 222 6 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Caesium Barium Lanthanide Halfnium Tantalum Tungsten Rhenium Osmium Iridium Platinum Gold Mercury Thallium Lead Bismuth Polonium Astatine Radon 87 223 88 226 89 227 104 261 105 262 106 266 107 264 108 277 109 268 110 281 111 280 112 285 113 284 114 289 115 288 116 293 117 292 118 294 7 Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Uub Uut Uuq Uup Uuh Uus Uuo Francium Radium Actinide Rutherfordium Dubnium Seaborgium Bohrium Hassium Meitnerium Darmstadtium Roentgenium Ununbium Ununtrium Ununquadium Ununpentium Ununhexium Ununseptium Ununoctium Alkali Metal 57–71 58 140.12 59 140.91 60 144.24 61 145 62 150.36 63 151.96 64 157.25 65 158.93 66 162.50 67 164.93 68 167.26 69 168.93 70 173.04 71 174.97 Transition Metal Halogen La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Lanthanum Cerium Praseodymium Neodymium Promethium Samarium Europium Gadolinium Terbium Dysprosium Holmium Erbium Thulium Ytterbium Lutetium Noble Gas 89–103 90 232.04 91 231.04 92 238.03 93 237 94 244 95 243 96 247 97 247 98 251 99 252 100 257 101 258 102 259 103 262 Lanthanide Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr Actinide Actinium Thorium Protactinium Uranium Neptunium Plutonium Americium Curium Berkelium Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium Element type Metals the majority of elements (in the figure: from blue to green), found on the left in the PT. Non-metals form mostly molecules, (in the figure: from orange to red), found on the right in the PT. Metalloids have intermediate properties The division between elements that are metals and non-metals starts between Al (which is a metal) and B, and staircases down to the right. Later on it will be assumed knowledge and it is crucial to quickly tell if an element is metallic or not. If you have never heard of the element, chances are that it’s a metal. 31  PERIODICITY Periodic trends Subshells & blocks The elements in the periodic table can be divided into four blocks, based on their electronic configuration. Since elements down a group have the same number of valence electrons, they will also have the same outermost subshell configuration. Take the alkali metals for example: each has a single electron in the outermost shell, but each in a shell further away. The electron configurations of Li, Na and K are [He] 2s1 , [Ne] 3s1 and [Ar] 4s1 respectively. The shell number that contains those electrons can be read off from the period number. s-block d-block p-block Note that H has the s1 1 IA 18 VIIIA configuration but is not 1 H He 1 Hydrogen 2 IIA 13 IIIA 14 IVA 15 VA 16 VIA 17 VIIA Helium a metal, and He has s2 configuration so we 2 Li Be B C N O F Ne 2 Lithium Beryllium Boron Carbon Nitrogen Oxygen Flourine Neon could consider placing it in above Be. But 3 Na Mg Al Si P S Cl Ar 3 Sodium Magnesium 3 IIIA 4 IVB 5 VB 6 VIB 7 VIIB 8 VIIIB 9 VIIIB 10 VIIIB 11 IB 12 IIB Aluminium Silicon Phosphorus Sulphur Chlorine Argon since the valence shell 4 K Ca 4 Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 4 of He is completely full Potassium Calcium Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc Gallium Germanium Arsenic Selenium Bromine Krypton its properties 5 Rb Sr 5 Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 5 correspond much closer Rubidium Strontium Yttrium Zirconium Niobium Molybdenum Technetium Ruthenium Rhodium Palladium Silver Cadmium Indium Tin Antimony Tellurium Iodine Xenon to the noble gases. 6 Cs Ba 6 La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 6 Caesium Barium Lanthanide Halfnium Tantalum Tungsten Rhenium Osmium Iridium Platinum Gold Mercury Thallium Lead Bismuth Polonium Astatine Radon 7 Fr Ra 7 Ac Francium Radium Actinide The nucleus of fluorine s1 s2 d1 d2 d3 d4 d5 d6 d7 d8 d9 d10 p1 p2 p3 p4 p5 p6 has a charge of 9+, the 1st shell is full, reducing f-block the effective attractive La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Lanthanum Cerium Praseodymium Neodymium Promethium Samarium Europium Gadolinium Terbium Dysprosium Holmium Erbium Thulium Ytterbium Lutetium power of the nucleus to the valence electrons to Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr Actinium Thorium Protactinium Uranium Neptunium Plutonium Americium Curium Berkelium Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium 7+. The elements in the first row are called lanthanides, and in the second row actinides. 3.2 Periodic trends Periodic trends are governed by the balance between attractive and repulsive electrostatic forces between the nucleus and the valence electrons. non-shielding helium filled shell Shielding electrons in lower full electron outer shell (7e ) − (2e − , shielding) shells reduce the + -charge that the valence electrons experience Effective nuclear charge (Zeff ) the net + charge that valence electrons neon shell nucleus (9+) experience. Zeff = Z − N◦ shielding electrons Zeff = 9 − 2 = +7 32. PERIODICITY Periodic trends 3 Determine the effective nuclear charge (Zeff ) for O, F, and Cl Example 8O 1s2 2s2 2p4 , it has 2 e − in the filled 1st shell (shielding). Notice how the number of shielding e − does Zeff = Z − N◦ shielding electrons = 8 − 2 = +6 not change across the period! 9F 1 s2 2 s2 2 p5 , so 2 e − in the filled 1st shell (shielding). Zeff = Z − N◦ shielding electrons = 9 − 2 = +7 Notice how Zeff does not change down a 17Cl 1 s2 2 s2 2 p6 3 s2 3 p5 , so 10 shielding electrons (2 + 8e − in the 1st and 2nd shells). group! Zeff = Z − N◦ shielding electrons = 17 − 10 = +7 The valence electrons experience an attractive force to the nucleus proportional to: The effective nuclear charge (Zeff ) The higher the effective nuclear charge, the stronger the valence electrons are attracted to the nucleus. The effective nuclear charge increases → a period. The distance The further away the valence electron is from the nucleus, the weaker it is attracted. The distace between the valence shell and the nucleus increases with increasing shell number, so ↓ a group. The valence electrons mutually repel each other. This repelling force increases when there are more electrons in the valence shell. Comparatively, the effect of the attractive forces is stronger than the repelling forces, which means that only when the effective nuclear charge and the shell stays the same do we use arguments based on the repelling forces between valence electrons (ions). To summarize, attraction between the + nucleus and the − valence electrons increases: → the period Zeff increases, causing the valence electrons to experience stronger attraction to the nucleus. The valence shell number is the same, so the electrons are at the same (approximate) distance. ↑ the group e − are closer to the nucleus, causing the valence electrons to experience stronger attraction to the nucleus. The Zeff stays the same in the same group. H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg 33  PERIODICITY Periodic trends Attraction ↑ (between the nucleus and the valence e – ) → a period, because Zeff increases (larger charge difference) ↑ a group, because distance decreases (larger distance difference) Atomic radius the distance from the nucleus to the valence electron(s). – When attraction ↑, the atomic radius ↓. The effect of adding or removing one electron Ionic radius the distance from the nucleus to the valence electron(s). while the attraction remains equal is Zeff and the shell number stay the same, so the attraction stays the same. Only significant. It’s safe to the repelling forces between the valence e – changes: assume that + ions are – In + ions a number e − are removed. All things equal, the mutually always smaller than repelling forces between valence electrons decreases. The larger the atoms, and - ions always larger. + charge, the smaller the ion. – In − ions a number of e − are added. All things equal, the mutually repelling forces between valence electrons increases. The larger the + charge, the la