General Chemistry 1413Chem-4 PDF

General Chemistry 1413Chem-4 For Engineering Students Prepared by Dr. Atef Korchef Associate Professor Department of Chemistry, College of Science, King Khalid University (KKU) 1  Introduction to the...

General Chemistry 1413Chem-4 For Engineering Students Prepared by Dr. Atef Korchef Associate Professor Department of Chemistry, College of Science, King Khalid University (KKU) 1  Introduction to the course General chemistry course 1413Chem-4 designed for engineering students delivers a general introduction to chemistry course that incorporates both lectures and laboratory experiments in developing and understanding chemical concepts and practices. Theory 3 hours/week Practice 2 hours/week Chapters: Chapter 1. Matter and Measurements Chapter 2. Atoms, Molecules, Ions, and Moles Chapter 3. Calculations in Chemistry Chapter 4. Electron Configuration and Properties of Atoms Chapter 5. Molecular Forces and Properties of Matter. Chapter 6. Gases Assessment details Quiz: 5 marks Book Reference Homework: 5 marks Understanding General Chemistry by Atef Korchef, Mid-exam: 25 marks CRC Press, Taylor and Francis Group, Edition 2022. Practical exam: 25 marks https://www.routledge.com/Understanding-General- Final exam: 40 marks Chemistry/Korchef/p/book/9781032189147 Total: 100 marks 2 Chapter 1 At the end of this chapter, the student will be able to ❑ define chemistry and matter and its states, ❑ differentiate between chemical change (chemical properties) and physical Matter and change (physical properties) Measurements ❑ differentiate between basic and derived quantities ❑ differentiate between intensive and extensive properties ❑ choose the appropriate units in the système internationale (SI) of measurement units ❑ define and use the basic methods and tools used in measurements such as significant figures and rounding off. ❑ define accuracy and precision 3  Matter Chemistry is the study of matter and the changes that matter undergoes Matter is defined as anything that has a mass and occupies space (volume) and is made up of particles. States of Matter – Solid particles close together in an orderly fashion little freedom of motion a solid has a fixed volume and shape – Liquid particles close together but not held rigidly in position particles are free to move past one another a liquid sample has a fixed volume but conforms to the shape of the part of the container it fills – Gas particles randomly spread apart particles have complete freedom of movement a gas sample assumes both the shape and volume of container. 4  Classification of Matter Matter is either classified as a substance or a mixture of substances. Substance can be either an element or a compound. A mixture can be either homogeneous or heterogeneous. Compounds could be separated to their elements by chemical methods such a heat decomposition and electrolysis. Mixtures could be separated to pure substances by physical methods such as ultrafiltration and distillation Such as ultrafiltration and distillation Such as heat decomposition and electrolysis 5  Substances Element is made up of one kind of atom and cannot be separated into simpler substances by chemical means. Examples. Carbon (C), Phosphorus (P), oxygen (O), hydrogen (H), Nitrogen (N), magnesium (Mg), Copper (Cu), Helium (He), iron (Fe),… https://www.teachoo.com/12688/3446/Atom/category/Concepts/ Atom is the building-up unit of the element. 6  A compound is formed when two or more elements are chemically combined in definite ratios. The compound is made up of at least two kinds of atoms. The properties of the compounds are different from the properties of their elements. Compounds can be separated into their elements by chemical means such as electrolysis. Examples: salt (NaCl), water H2O, ethane (C2H6), carbon dioxide CO2, HgO, CaCO3, CaSO4·H2O, Na2CO3, Mg(OH)2 Elements | Atoms | Molecules & Compounds | Tutway Compounds can be separated into their elements by chemical means such as: Heat decomposition of mercuric oxide (HgO) HgO(sd) Hg(lq) + ½ O2(g) Electrolysis of water : H2O(lq) H2(g) + 1/2O2(g) 7  Mixtures Mixture: physical combination of two or more pure substances – Substances retain distinct identities Types of Mixtures – Homogeneous Mixtures : A homogeneous mixture is one in which the components are uniformly distributed. The composition of the mixture is uniform throughout Example: sugar dissolved in water, air, sea water, NaOH solution (and all solutions), … ❑ Solution is a homogenous mixture of two or more substances that are chemically unreacted. Solute ✓ Solution is composed of the solute and the solvent. ✓ Solute: the substance exists in the smallest amount ✓ Solvent: the substance exists in the largest amount. Solvent Example: in NaCl solution, NaCl (sd) is the solute and water H2O is the solvent. 8  Heterogeneous Mixtures : mixture is made of different substances that remain physically separate. Heterogeneous mixtures always have more than one phase (composition is not uniform throughout) Example: sugar mixed with iron filings, sand and chalk powders in water, water and oil Mixtures can be separated to their components by different physical methods such as filtration and distillation: Filtration: used to separate a heterogeneous solid-liquid mixture Example: Mixture of water and sand Distillation: used to resolve a homogenous solid-liquid mixture Example: Salt NaCl in water, seawater 9  Properties of matter ❑ Chemical properties: The ability of a substance to combine with or change into one or more other substances Examples: Heat of combustion, enthalpy of formation, electronegativity ❑ Physical properties: Characteristics that can be observed or measured without changing the composition of the substance Examples: Temperature, color, volume, mass, area, pressure, melting point, boiling point Intensive properties are properties which do not depend on the amount of matter Examples: Temperature, density, boiling point, concentration, solubility, color Extensive properties are properties which depend on the amount of matter Examples: mass, volume, surface (area), amount of substance (number of moles) 10  Chemical Change ❑ A chemical change involves making or breaking chemical bonds to create new substances. Chemical changes include the followings: Oxidation reduction (Redox) reactions: loss and gaining of electrons: The oxidation of iron(II) to iron(III) by hydrogen peroxide in acidic medium Fe2+ Fe3+ + e- H2O2 + 2H+ +2 e- 2H2O Overall equation: 2Fe2+ + H2O2 + 2H+ 2Fe3+ + 2H2O (Redox reaction) Reaction of base and acid (neutralization reaction): HClsol + NaOHsol H2O lq + NaCl aq Heat decomposition: HgO(sd) Hg(lq) + ½ O2 (g) Electrolysis of water: H2O(Lq) H2(g) + 1/2O2(g) Iron rusting (corrosion): 4Fe + 3O2 2Fe2O3 Combustion (of ethane C2H6): 2C2H6 + 7O2 4CO2 + 6H2O Burning of wood 11  Physical Change ❑ A physical change alters a substance without changing its chemical identity. ❑ No new substance is created, no formation of new chemical bonds during a physical change. ❑ Physical changes occur when substances are mixed but don’t chemically react Physical changes include the following: Phase changes such as vaporization, condensation, freezing, sublimation, melting and deposition. Example: H2O(lq) H2O(g) Dissolving sugar and salts in water Mixing sand with water or mixing oil with water crushing 12  PHYSICAL QUANTITIES and the System International (SI) units ❑ Fundamental (basic) quantities cannot be defined in terms of other physical quantities Examples: quantity Length Mass Temperature Time Amount of substance Examples:, SI basic unit m kg K (Kelvin) s (second) mol (meter) ❑ Derived quantities can be defined in terms of the fundamental physical quantities Examples: quantity area Volume density concentration Pressure SI derived m2 m3 kg/m3 mol/m3 Pa (Pascal) unit  = m/V P = F/A 13 m: mass and V: volume F: force and A: area  SI Prefixes Practice: Convert the following to the desired unit: 1 nm = ……. m 1m = ………nm 23 Gg = …………dg 2 ng = ………mg 2.126 mg = ……..µg Answer: 1 nm = 10-9 m 1m = 109 nm 23 Gg = 23×109 g =23×109×10 dg =23×1010 dg 2 ng = 2×10-9 g = 2×10-9 ×103 mg = 2×10-6 mg 14 2.126 mg = 2.126×10-3 g = 2.126×10-3×106 µg = 2.126×103 µg = 2126 µg 15  Practice on temperature conversions 300 K = (300-273) °C =27 °C =(1.8×27)+32 °F= 80.6 °F 16  Volume is an extensive property. Volume is a derived quantity. The SI derived unit of volume is m3 (meter cubed) but the unit liter (L) is more commonly used in the laboratory setting. 1m3 = 1000 L 1dm3 = 1L 1cm3 = 1mL Practice: Convert the following to the desired unit: a) 1m3 = ………… mL. b) 1 nL= …………. cm3 Answer: a) 1m3 = 1000 L = 103 L = 103 × 103 mL = 106 mL. b) 1 nL= 1 × 10-9 L = 1 × 10-9 ×103 mL = 10-6 mL =10-6 cm3 17  Density: Density is an intensive property (does not depend on the amount of matter). Density is a derived quantity.  = m/V  = density; m = mass and V = volume – The SI unit for density is kg/m3 There are other common units for density such as (g/cm3 or g/mL, g/L) (*gas densities are usually expressed in g/L) ❑ Practice 1: A sample of mercury has a volume of 0.25 L and a mass of 3.4 kg. Calculate the density of mercury in g/cm3? Answer: 1 L = 1000 cm3, Volume = 0.25 x 1000 = 250 cm3 1 kg = 1000 g, Mass = 3.4 x 1000 = 3400 g  = m/v = 3400 g/ 250 cm3 = 13.6 g/cm3 ❑ Practice 2: A cube of metal has a mass of 4.0 g and a length of 2.0 cm on each side. Calculate its density. a) 2.0 g/cm3 b) 0.50 g/cm3 c) 0.20 g/cm3 d) 5.0 g/cm3 18 V =2×2×2 = 8 cm3;  = m/ V = 4/8 = 0.5 g/cm3 ❑ Solubility is the maximum quantity (amount) of the solute that will dissolve in a given amount of the solvent, at a given temperature. Solubility depends on temperature. The solubility may be stated in various units of concentration such as g/L, mol/L and g/100g of solvent. Example: The solubility of sugar in water at 20 o C is 2040 g/L. Solubility is an intensive property Solubility is a derived quantity Practice: The solubility of potassium nitrate (KNO3) in water is 270 g/L at 20 oC. Calculate the mass of KNO3 dissolved in 17.8 mL of water at 20 o C.  Answer: at 20 o C, 270g KNO3 1000 mL water (=1L) Mass KNO3 17.8 mL water Mass KNO3 = (17.8 × 270) / 1000 = 4.8 g 19  Measurements To measure the width of this object, you cannot report the object’s width any more exactly than to the nearest millimeter. All you can do is estimate the next decimal place in the measurement. It makes no sense to report any digits after the first uncertain one. The length of the object: The concept of significant figures takes this limitation into account. The significant 1.38 cm figures of a measured quantity are Certain Uncertain digit defined as all the digits known with digit or Certain certainty and the first uncertain, or digit Estimated digit estimated, digit. 20  Rules for significant figures ❑ Any non-zero digit is significant 3455 km 4 significant figures (4 SF) 28.7 s 3 significant figures (3 SF) ❑ Zeros between non-zero digits are significant 12051 m 5 significant figures 25400032 nL 8 significant figures 208.0025 cm 7 significant figures ❑ Zeros to the left of the first non-zero digit are not significant 0.0058 mol 2 significant figures ❑ Zeros to the right of the last non-zero digit are significant if a decimal is present 25.00 mL 4 significant figures 30.0 g 3 significant figures ❑ Zeros to the right of the last non-zero digit are not significant if the decimal is not present 2500 mL 2 significant figures 21 2.5×10 mL (Scientific notation) 2 significant figures 3  Simplified Rounding rules If the number is less than 5 round “down”. Rounded off to 2 SF 0.242313 0.24 Rounded off to 3 SF 3.674 3.67 If the number is greater than 5 round “up”. Rounded off to 1 SF 0.0679 0.07 1.5396 Rounded off to 4 SF 1.540 When the first digit dropped is 5, make the preceding digit even Rounded off to 4 SF 2.0845 2.084 2.13275 Rounded off to 5 SF 2.1328 3.305 Rounded off to 3 SF 3.30 Note that zero is an even number 22 Practice: Determine the number of significant figures in each measurement. 1234586 cm 7 significant figures 131.21 cm 5 significant figures 0.00018 g 2 significant figures 205 m 3 significant figures 100041 m 6 significant figures 30.002 m 5 significant figures 12.00 mL 4 significant figures 3700 mL 2 significant figures 0.0050700 km 5 significant figures 2.145×10-3 m (Scientific notation) 4 significant figures 1.03×104 m (Scientific notation) 3 significant figures 23  Calculations with measured numbers ❑ Addition and subtraction Answer cannot have more digits to the right of the decimal than any of original numbers Example: 102.50 + 0.231 – 12.1 = 90.6 2 digits after 3 digits after 1 digit after 1 digit after decimal point decimal point decimal point decimal point Calculator answer: 90.631 rounded off to 90.6. ❑ Multiplication and division Final answer contains the smallest number of significant figures ▪ Example: 1.4 x 8.011 / 5.12 = 2.2 2SF 4 SF 3 SF 2SF (Limited by 1.4 to 2 significant figures in answer) (Calculator answer: 2.1905078125 round to 2.2) Note that exact numbers do not limit answer because exact numbers have an infinite number of significant figures Example: A penny minted has a mass of 2.5 g. If we have 3 pennies, the total mass is 3 x 2.5 g = 7.5 g In this case, 3 is an exact number and does not limit the number of significant figures in 24 the result. PRACTICE 1: Calculate the following to the correct number of significant figures : (2.20 - 0.0111)/0.1165 = ? (2.20 - 0.0111) = ? (correct answer: 2.19 (2 decimals)) (2 decimals) (4 decimals) (2 decimals) – Calculator answer: 2.1889 2.19 (with 2 decimals) 2.19/0.1165 = ? 3SF 4SF 3SF – Calculator answer: 18.7982832618 18.8 (3 SF) (2.20 - 0.0111)/0.1165 = 18.8 PRACTICE 2: Calculate the following to the correct number of SF: (8.9 × 56) = ………. (2SF) (2SF) (2SF) Calculator answer: 498.4 (4SF) 498.4 = 4.984 ×102 5.0 ×102 (scientific notation) rounded off to 2 SF (8.9 × 56) = 5.0 × 102 (2SF) (2SF) (2SF) 25  Accuracy and precision ❑ Accuracy is how close a measurement is to the accepted value (or true value). ❑ Precision is how closely measurements of the same thing are to one another Practice: The true value of a neutralization volume is 12.370 mL. Describe accuracy and precision for each set of measurements: Set A (12.345 mL, 12.346 mL, 12.344 mL) Precise, not accurate Set B (12.357 mL, 12.337 mL, 12.393 mL) Not precise, not accurate Set C (12.369 mL, 12.370 mL, 12.371 mL) Precise and accurate Explanation: The values in set A are close to one another but are not close to the true value so set A is precise but not accurate. The values in set B are not close to one another and are not close to the true value so set B is neither precise nor accurate The values in set C are close to one another and close to the true value so set C is both precise and accurate. 26  Practice on chapter 1 Q1. CaSO4·1/2H2O is a) an element b) a compound c) a heterogeneous mixture d) a homogeneous mixture Q2. The density of water at 20oC is 0.9982 g/mL. Which of the following sets of density measurements is precise but not accurate? a) (0.9989, 0.9993, 0.9991) b) (0.9981, 0.9980, 0.9982) c) (0.9881, 0.9880, 0.9882) d) (0.8989, 0.8993, 0.8991) Q3. Which of the following is a derived quantity? a) Mass b) Volume c) Temperature d) Amount of substance Q4. The correct answer to express the arithmetic operation (2.00×0.009)/(9.410-8.0) is a) 0.014 b) 0.01 c) 0.0142 d) 0.0143 Q5. Electrolysis of water is a) a physical change b) a chemical change c) a neutralization process d) none of these Q6. The number of significant figures (SF) in 0.0100 is a) 1 SF. b) 3 SF. c) 4 SF. d) 5 SF. Q7. A temperature of 173 K, when converted to °C equals a) 273 °C b) 446 °C c) -100 °C d) 100 °C Q8. Rounding off 3.1055 to four significant figures results in a) 3.105 b) 3.106 c) 3.1057 d) 3.100 Q9. 600 nm equals a) 6 x 10-3 m b) 0.6 µm c) 6 x 10-2 µm d) 6 x 103 m Q10. Which of the following is an intensive property? a) Volume b) Mass c) Amount of substance d) Density Q11. Which of the following is the smallest measure of the mass of a substance? a) 2 Mg b) 2 mg c) 2 ng d) 2 µg 27 Chapter 2 At the end of this chapter, the student will be able to ❑ describe the atomic structure ❑ differentiate between the properties of the atom’s components Atoms, Molecules, ❑ differentiate between the atom, the Ions and Moles molecule, the ion, and the mole ❑ Calculate the number of protons, neutrons, and electrons in an atom or an ion ❑ recognize the isotopes of an element ❑ differentiate between the chemical formula, the empirical formula, and the structural formula ❑ define the ionic compounds 28  The atomic structure Atom consists of three subatomic particles: Protons (p) : have a positive charge (+) Neutrons (n): are neutral and have a mass slightly greater than that of the protons Electrons (e): have a negative charge (-) and a tiny mass Most of the atom is empty Mass of proton ≈ mass of neutron >> mass of electron Atom is the smallest particle (building up unit) of an element, having the same chemical properties as the bulk element. 29  M ass Num ber A X Elem ent Atom ic Num ber Z Atomic Number (Z) is the number of protons in the nucleus Mass Number (A) is the number of protons and neutrons (P + n) Number of neutrons (n) = Mass Number (A) – Atomic Number (Z) For neutral atom: Number of protons = Number of electrons Examples: M a ss N u m b e r 12 n o. o f (P ) = 6 C n o. o f (e ) = 6 A to m ic N u m b e r 6 n o. o f (n ) = 12 - 6 = 6 197 A u 79 no. of (P) = 79 no. of (e) = 79 no. of (n) = 197 - 79 = 118 30  Isotopes Isotopes are atoms of a given element that differ in the number of neutrons (same atomic number) Examples: 1 H 2 H 3 H Hydrogen Isotopes of hydrogen 1 1 1 H D T Deuterium Tritium Isotopes of carbon Isotopes of oxygen 31  Ions: Anions and Cations A neutral atom becomes an ion either by losing electrons (cation) or by gaining electrons (anion) Na Na+ + 1e- Cl + 1e- Cl- Mg Mg2+ + 2e- O + 2e- O2- Fe Fe3+ + 3e- 32  Molecules A molecule is a neutral group of two or more atoms bonded together that may or may not be the same element. Homonuclear molecules formed by one type of atoms Examples: O2 , O3 , H2 , N2 , Cl2 , Br2 , I2 , F2, … Heteronuclear molecules formed by more than one type of atoms. Examples: H2O, NH3 , NaCl , NaOH, CaSO4·2H2O , Na2CO3 , …. Heteronuclear molecule is the smallest particle (building up unit) of a compound. 33 ❑ Molecular Formula represents the actual number and types of atoms in a molecule Examples: H2 , Cl2 , H2O2 , C6H6 , C6H12O6 , CaCO3, Fe2O3 , CaSO4·2H2O , …. ❑ Empirical or Simplest Formula is the formula that gives only the relative number of atoms of each type in a molecule (gives rise to the smallest set of whole numbers of atoms) Molecular formula = D × (simplest formula) D means how many times simplest formula repeated to give molecular formula. D = 1,2,3… Examples: H2O2 D=2 Empirical formula: HO C6H12O6 D= 6 Empirical formula: CH2O H2 O D=1 Empirical formula: H2O C 3 H6 O3 D= 3 Empirical formula: CH2O ❑ Structural Formula is the formula that indicates the attachment of atoms Examples: 34  PRACTICE 1. Which of the following is the simplest formula of C6H14? a) C6H14 b) CH c) C3H7 d) CHO 2. Which of the following is the structural formula of methane CH4? a) CH4 b) CH c) C2H8 d) 3. CH2O is the simplest formula of a) C5H1OO b) H-O-H c) CHO d) C3H6O3 4. The molecular formula is the formula that a) gives only the relative number of atoms of each type in a molecule b) represents the actual number and types of atoms in a molecule c) indicates the attachment of atoms in a molecule d) represents a group of atoms of the same element containing different number of protons 5. Which of the following is the building up unit of a compound? a) Atom b) Homonuclear molecule c) Heteronuclear molecule d) Electron 35  The mole Mole - the amount of a substance that contains the same number of entities as there are atoms in exactly 12g of carbon-12. This amount is 6.022x1023. The number is called Avogadro’s number and is abbreviated as NA. (NA = 6.022x1023 ) One mole (1 mol) contains NA = 6.022x1023 entities entities: atoms (ex: H, C, O, Cl, Ca, Fe, …); molecules (ex: H2, O2, CaCl2, NaOH, …); ions (ex: H+, Cl-, OH-, SO42-, NH4+, …) Examples: 1 mole of C = 6.022×1023 of C atoms = 12 g of C 1 mole of H2O = 6.022×1023 of H2O molecules = 18 g H2O 1 mole of NO3- = 6.022×1023 of NO3- ions = 62 g NO3- 36  Introduction to the periodic table of elements The arrangement of elements is in order of increasing atomic number (Z) The vertical columns of the table (numbered from 1 to 18 ) are called groups or families. Element in the same group have similar but not identical characteristics. The horizontal rows of the table (numbered from 1 to 7) are called periods. 37  Ionic compounds Most elements are metals. Metals easily lose electrons. Nonmetals tend to gain electrons Metals and nonmetals form compounds with each other. These compounds are called ionic compounds. metal ion + nonmetal ion → ionic compounds ❑ Formula of ionic compounds If the charge of cation = charge of anion, the ionic compound will be 1:1 [cation : anion] Examples: The formula of the ionic compound formed between Na+ and Cl- NaCl Mg2+ and S2- MgS Al3+ and N3- AlN If the charge of the cation and the charge of the anion are different, use the cross over method to determine the formula of the ionic compound. Examples: 38 Q1. 12C, 13C and 14C are called Practice on chapter 2 a) molecules b) compounds c) ions d) isotopes Q2. The formula of the ionic compound formed between Fe2+ and SO42- ions is a) FeSO4 b) Fe3(SO4)2 c) Fe2(SO4)3 d) Fe3(SO4) Q3. The simplest (empirical) formula of glucose C6H12O6 is a) C6H12 O6 b) C3H4O3 c) CHO d) CH2O Q4. Cations form when a) neutrons are added to or removed from an atom b) electrons are removed from an atom c) electrons are added to an atom d) all of these Q5. The formula that indicates the attachment of atoms in a molecule is the a) structural formula b) ionic formula c) molecular formula d) simplest formula Q6. Which of the following is a homonuclear molecule? a) H2O b) CaF2 c) KNO3 d) Cl2 Q7. The atomic number of an element is the number of a) protons and electrons b) neutrons and electrons c) protons d) protons and neutrons Q8. The numbers of neutrons (n) and electrons (e-) in radium ion 2+ are a) 226 n and 88e- b) 138 n and 88 e- c) 138 n and 86 e- d) 226 n and 86 e- Q9. The horizontal rows in the periodic table are called a) groups or families b) isotopes c) periods d) ionic compounds Q10. The atomic mass of iron (26Fe) is 56. The atomic mass of the ferric ion (Fe3+) is a) 26 b) 56 c) 29 d) 59 3+ Q11. When the cation Ti gains an electron, it becomes a) Ti+ b) Ti4+ c) Ti2+ d) Ti- Q12. The ionic compound is formed between a) two metals b) two nonmetals c) a metal and a nonmetal d) two cations Q13. One mole of H2O contains a) NA molecules of H2O b) NA atoms of hydrogen c) 3 molecules of H2O d) 2×NA molecules of oxygen 39 Chapter 3 At the end of this chapter, the student will be able to ❑ explain the relationship between the molar mass and the atomic mass or Calculations in molecular mass ❑ calculate the number of moles, and the Chemistry mass percentage of an element in a compound ❑ Calculate the number of atoms or molecules knowing the number of moles ❑ Determine the molecular formula of a compound using chemical calculations ❑ recognize a balanced chemical equation ❑ show the relationship between the number of moles of the reactants and the number of moles of the products in a balanced chemical equation 40  Atomic mass Atomic mass (the mass of one atom) is the average mass of all the isotopes of an atom. It is commonly expressed in unified atomic mass units (amu) 1 atomic mass unit is defined as 1/12 of the mass of a single carbon-12 atom: 1amu = (1/12) × 12/(6.022 ×1023) = 1/(6.022 ×1023)=1.660x10-24 g 1amu = (1/NA) g = 1.660x10-24 g Hydrogen (H) atom has a mass of 1 amu Carbon (C) atom has a mass of 12 amu Oxygen (O) atom has a mass of 16 amu The atomic masses of the elements are given in the periodic table of elements. 41  Molecular mass Molecular Mass (mass of one molecule) = Sum of atomic masses of all the atoms in the molecule Practice: Determine the molecular mass of H2SO4 and C6H12O6 Atomic masses: H=1, C=12, O= 16, S= 32 One molecule of H2SO4 contains 2 atoms of H, 1atom of S and 4 atoms of O: Molecular mass of H2SO4 = 2×(atomic mass of H)+1×(atomic mass of S) + 4×(atomic mass of O) = (2 × 1) + (1 × 32) + (4 × 16) = 98 amu One molecule of C6H12O6 contains 6 atoms of C, 12 atoms of H and 6 atoms of O: Molecular mass of glucose C6H12O6 = 6 ×12 + 12×1 + 6×16 = 180 amu 42 Practice: Calculate in grams: a) the mass of one atom of selenium (Se) b) the mass of one molecule of H2O Atomic masses: (H = 1, O = 16, Se = 78.96) NA= 6.022× 1023 Answer: a) The atomic mass of Se equals 78.96 amu The mass of one atom of Se in grams is 78.96/NA = 78.96/(6.022× 1023) = 1.311×10-22 g b) The molecular mass of H2O is (2×1 + 1×16) = 18 amu The mass of one molecule of H2O in grams is 18/NA = 18/(6.022× 1023) = 2.989×10-23 g 43  Molar Mass ❑ For monatomic elements, the molar mass is the numerical value on the periodic table expressed in g/mol EXAMPLES: Element Atomic mass Molar mass (M) H 1 amu 1 g/mol Na 23 amu 23 g/mol Fe 56 amu 56 g/mol ❑ For molecules, the molar mass is the sum of the molar masses of each of the atoms in the molecular formula. EXAMPLES: (Atomic masses: H=1, C= 12, 0=16 and Ca=40) One molecule of H2O contains 2 atoms H and 1 atom O: The molecular mass of H2O is equal to 2×1+1×16 = 18 amu The molar mass of H2O is 18 g/mol One molecule of CaCO3 contains 1 atom Ca, 1 atom C and 3 atoms O: The molecular mass of CaCO3 is equal to 1×40+ 1×12 + 3×16 = 100 amu The molar mass of CaCO3 is 100 g/mol 44  m Number of moles: n= n : number of moles of the substance M m : mass of the substance in grams M: molar mass of the substance (g/mol) Practice 1: Calculate the number of moles of calcium carbonate CaCO3 in 14.8 g of calcium carbonate. (atomic masses: C = 12, O = 16, Ca = 40) Answer: Molar mass of CaCO3 M = (1 x 40) + (1 x 12) + (3×16) = 100 g/mol Number of moles: n = m / M = 14.8 / 100 = 0.148 mol Practice 2: What is the mass in grams of 0.287 mol of Acetylsalicylic acid C9H8O4? (Atomic masses: H = 1, C = 12, O = 16) Answer: Molar mass of C9H8O4: M = (9×12) + (8×1) + (4×16) = 180 g/mol n= m/M m = n × M = 0.287 × 180 = 51.7 g 45 The number of atoms or molecules: N = n × NA Practice 1: Calculate the number of Se atoms in 1 g of Se. (atomic mass: Se = 78.96) Answer: The molar mass of Se = 78.96 g/mol Number of Se moles n = m / M = 1 / 78.96 = 1.266×10-2 mol Number of Se atoms N = n× NA = 1.266×10-2 ×6.022×1023 = 7.624×1021 atoms Practice 2: Calculate the number of molecules of calcium carbonate CaCO3 in 14.8 g of calcium carbonate. (atomic masses: C = 12, O = 16, Ca = 40) Answer: Molar mass of CaCO3 = (1 × 40) + (1 × 12) + (3×16) = 100 g/mol Number of moles n = m / M = 14.8 / 100 = 0.148 mol Number of molecules N = n× NA = 0.148×6.022×1023 = 8.91×1022 molecules 46 Mass percent of an element in a compound: n x molar mass of element mass % (element) = ( ) x 100% molar mass of compound n is the number of moles of the element in 1 mole of the compound. n is determined from the molecular formula of the compound. Practice : Calculate the mass percent of iron Fe, nitrogen N and oxygen O in Fe(NO3)3. (Atomic masses: N = 14, O = 16, Fe = 56) Answer Molar mass M of Fe(NO3)3 M = M(Fe) + 3 × M(NO3) = M(Fe) + 3 × [M(N) + 3 × M(O)] = 56 + 3 × [14 + 3 ×16] = 242 g/mol mass % (Fe) = [ (1 × M(Fe)) / M(Fe(NO3)3 ] × 100 = [(1 ×56)/242]× 100 = 23.14 % mass % (N) = [ (3 × M(N)) / M(Fe(NO3)3 ] × 100 = [(3 ×14)/242]× 100 = 17.36 % mass % (O) = [ (9 × M(O)) / M(Fe(NO3)3 ] × 100 = [(9 ×16)/242]× 100 = 59.50 % 23.14 % + 17.36% + 59.50 % = 100.00% 47  Empirical and Molecular Formulas Calculation Molecular Formula = D× (Empirical Formula); D means how many times simplest formula repeated to give molecular formula. Molar mass of molecular formula D= Molar mass of empirical formula Practice 1: The simplest formula of vitamin C is C3H4O3 and its molar mass is 176 g/mol. What is the molecular formula of vitamin C? (Atomic masses: H = 1, C = 12, O = 16) Answer: Molar mass of empirical formula = 3 xM(C) + 4 x M(H) + 3 x M(O) = (3 x 12) + (4 x 1) + (3 x 16) = 88 g/mol Molar mass of the molecular formula = 176 g/mol D = 176 / 88 = 2 The molecular formula of vitamin C is C6H8O6 48 Practice 2: Elemental analysis of lactic acid (M = 90 g/mol) shows that 100 g of this compound contains 40.0 g (C), 6.7 g (H) and 53.3 g (O). Find the empirical formula and the molecular formula of lactic acid. Atomic masses: (H = 1, C = 12, O = 16) Answer: The number of moles of C: n(C) = m(C)/M(C) = 40/12 = 3.33 mol The number of moles of H: n(H) = m(H)/M(H) = 6.7/1 = 6.7 mol C3.33H6.7O3.33 The number of moles of O: n(O) = m(O)/M(O) = 53.3/16 = 3.33 mol Divide each mole number by the smallest number of moles (3.33): Relative mole ratio of (C) = 3.33 / 3.33 = 1 Relative mole ratio of (H) = 6.7 / 3.33 = 2 The empirical formula is CH2O Relative mole ratio of (O) = 3.33 / 3.33 = 1 The molar mass of the empirical formula CH2O is M(CH2O) = 12 + 2×1 + 16 = 30 g/mol Or the molar mass of the molecular formula is M(lactic acid) = 90 g/mol So, D = M(lactic acid)/M(CH2O) = 90/30 = 3. The molecular formula is obtained by multiplying the subscripts of the empirical formula by 3. The molecular formula of lactic acid is C3H6O3. 49 Balancing Chemical Equations: Determining the coefficients that provide equal numbers of each type of atom on each side of the equation: 1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water C2H6 + O2 CO2 + H2O 2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 2C2H6 NOT C4H12 3. Start by balancing those elements that appear in only one reactant and one product. C2H6 + O2 CO2 + H2O start with C or H but not with O C2H6 + O2 2CO2 + H2O C2H6 + O2 2CO2 + 3H2O 50  4. Balance those elements that appear in two or more reactants or products. 7 C2H6 + O2 2CO2 + 3H2O multiply O 2 by 2 C2H6 + 7 O2 2CO2 + 3H2O remove fraction 2 multiply both sides by 2 2C2H6 + 7O2 4CO2 + 6H2O 5. Check to make sure that you have the same number of each type of atom on both sides of the equation. 2C2H6 + 7O2 4CO2 + 6H2O Reactants Products 4C 4C 12 H 12 H 14 O 14 O Practice: Which of the following is a balanced chemical equation? a) C2H6 + 7 O2 → 2CO2 + 3 H2O b) 2HgO → 2Hg + 2O2 c) 3H2 + N2 → 2 NH3 d) 2H2O → 2H2 + 1/2O2 51  Mass relations in balanced chemical equations Reactants products Practice: From the following balanced chemical equation: N2 (g) + 3 H2 (g) → 2 NH3 (g) Calculate a) the mass in grams of NH3 formed when 1.8 mol H2 reacts. b) the mass in grams of N2 required to form 1 kg NH3. c) the mass in grams of H2 required to react with 6 g N2. Atomic masses (H= 1, N=14) Answer: N2 (g) + 3 H2 (g) → 2 NH3 (g) a) The mass in grams of NH3 formed when 1.8 mol H2 reacts: 3 mol H2 2mol NH3 1.8 mol n(NH3) = ? n(NH3) = (1.8×2)/3 = 1.2 mol NH3 mass of NH3 : m(NH3) = n(NH3) × M(NH3) = 1.2 × (14+ 3×1) = 20.4 g 52  N2 (g) + 3 H2 (g) → 2 NH3 (g) b) The mass in grams of N2 required to form 1 kg (= 1000 g) NH3: 1 mol N2 2 mol NH3 2×14 = 28g N2 2×(14+3×1) =34 g NH3 m(N2) = ? 1000 g m(N2) = (1000/34) ×28 = 823.5 g c) The mass in grams of H2 required to react with 6 g N2: 1 mol N2 3 mol H2 2×14 = 28g N2 3×(2×1) = 6 g H2 6 g N2 m (H2) = ? m(H2) = (6/28) ×6 = 1.28 g Book Reference https://www.routledge.com/Understanding-General- Chemistry/Korchef/p/book/9781032189147 53  Practice on chapter 3 Atomic masses: H = 1, C = 12, N = 14, O = 16, Na = 23, Mg = 24, S = 32, Cl = 35.5, Ca = 40, Fe = 56, Cu = 63.5. Avogadro’s number: NA = 6.022x1023 Q1. What are the values of (x,y) that make the following reaction balanced ? 2 CxHy + 25 O2 → 16 CO2 + 18 H2O a) (8,9) b) (8,18) c) (4,18) d) (8,25) Q2. The number of moles in 275 g of calcium carbonate (CaCO3) is a) 17.5 moles b) 27.5 moles c) 2.75 moles d) 100 moles Q3. The molar mass of (NH4)2SO4 is: a) 114 g/mol b) 100 g/mol c) 132 g/mol d) 13.20 g/mol Q4. The mass percent of sulfur (mass% S) in copper sulfate (CuSO4) is a) 20 % b) 75 % c) 55% d) 40 % Q5. One mole of H2O contains a) 1 mole of O b) NA = 6.022x1023 molecules of H2O c) 2 × 6.022x1023 atoms of H d) All of these Q6. From the balanced equation for the combustion of methane (CH4) CH4 + 2 O2 → CO2 + 2 H2O The mass of oxygen (O2) needed for the combustion of 16 g of methane is a) 16 g b) 36 g c) 32 g d) 64 g Q7. The simplest formula of a compound is CH2O and its molar mass is 180 g/mol. The molecular formula of the compound is a) C2H4O b) C2H6O c) C6H12O6 d) C2H4O2 Q8. How many molecules are there in 9.8 g of sulfuric acid (H2SO4)? a) 602 x 1022 b) 6.022 x 1022 c) 6.022 x 1023 d) 6.022 x 1024 Q11. The mass of one atom of oxygen (O) is a) 16 amu b) 16 g/mol c) 16/NA g d) both of a) and c) are correct Q12. 1 amu equals a) 1 g b) 1 mol c) 1/NA g d) 12/NA g Q13. Which of the following is a balanced chemical equation? a) 2C2H6 +7O2 → 4CO2 + 6H2O b) 5H2 + N2 → 5NH3 54 c) 2HgO → 2Hg + 1/2O2 d) Ca2+ + 2HCO3- → CaCO3 + 2CO2 + H2O Homework - CALCULATIONS Atomic masses: H = 1, C = 12, N = 14, O = 16, Na = 23, Mg = 24, S = 32, Ca = 40, Fe = 56, Zn = 65. Avogadro’s number: NA = 6.022x1023 Question 1 1. Calculate the molar mass of CaSO4·2H2O. 2. Calculate the number of moles of H2SO4 in 49 g of this compound. 3. Calculate the mass of 0.04 mol of CaCO3. 4. Determine the mass percent of nitrogen (N) in zinc nitrate Zn(NO3)2. 5. Determine the number of molecules of H2O2 in 17 g of this compound. 6. Determine the mass of one molecule of C2H6 in grams. 7. Determine the mass of one atom of magnesium (Mg) in grams. 8. The empirical formula of a compound is CH2O and its molar mass is 60 g/mol. What is the molecular formula of this compound? Question 1 2 3 4 5 6 7 8 Answer 172 0.5 4g 14.8% 3.011× 1023 30/NA g 24/NA g C2H4O2 g/mol mol molecules Question 2 Octane (C8H18), one of many components of gasoline, mixes with oxygen from the air and burns to form carbon dioxide and water vapor. 1. Write the balanced chemical equation for this reaction. 2. Calculate the mass of oxygen (O2) needed for the combustion of 220 g of octane. ANSWER 1) 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O 2) 771.93 g 55 Chapter 4 At the end of this chapter, the student will be able to ❑ Recognize the wave-particle duality of light and matter Electron ❑ Describe the Bohr’s model of the Configuration and hydrogen atom ❑ Calculate the wavelength of the line Properties of Atoms corresponding to the transition of an electron from one energy level to another. ❑ define the quantum numbers ❑ define the different rules for the electron configuration ❑ write the electron configuration of atoms ❑ determine the valence shell and valence electrons and the number of unpaired electrons in an atom ❑ describe the periodic trends in atoms’ properties using the periodic table of elements. 56  Electromagnetic Radiation (light) - Wave like Wavelength,  is the distance from one crest to the next in the wave. Measured in units of distance. Frequency,  is the number of complete cycles per sec., expressed in s-1 or Hz Speed of Light, C, same for all electromagnetic radiation. C= 3 x 108 m/s in vacuum = c/ Q2. The yellow light given by the sodium lamp has  = 589 nm. What is the frequency of this radiation? C = 3 × 108 m/s Answer:  = 589 nm = 589 × 10-9 m  = C/ = 3 × 108 / (589 x 10-9) = 5.09 x 1014 s-1 57 The continuous spectrum of light Continuous spectrum of light in the visible region The line spectrum of hydrogen The spectrum of hydrogen element is a discontinuous spectrum (line spectrum) Why? 58  The wave-particle duality of light ❑ Planck: Atoms can gain or lost only certain quantities of energy, E = nh (where n is a positive integer, 1, 2, 3, etc) by absorbing or emitting radiation. This means energy is quantized. ❑ The wave-particle duality of light (proposed by Einstein): Light consisted of packets of energy called photons (particle-like) that had wave-like properties as well. Ephoton = h =hc/=Eatom Planck's constant h = 6.63 x 10-34 Js 59  Bohr's Model of the Hydrogen Atom ❑ Hydrogen atom has only certain allowable energy levels for the electron orbits. (quantized energy) Ground state: the lowest possible energy level an electron can be at (n=1). Excited state: an energy level higher than the ground state (n>1). The energy level En = - RE / n2 = - 2.18 x 10-18 / n2 where n = 1, 2, 3,.... The energy constant RE = 2.18 x 10-18 J E Ground state (lowest energy) ❑The electron traveled in circular orbits. Each orbit has certain radius rn rn = 0.5 x n2 Å Example: For n=3, rn = 0.5 × 32 = 4.5 Å 60 ❑When the electron moves from one orbit to another, it absorbs or emits a photon whose energy equals the difference in energy between the two orbits. Electron moves from a -2.18 x 10-18 / (ninitial)2 (In Joule) higher energy level orbit to a lower energy level orbit -2.18 x 10-18 / (nfinal)2 Emits a photon Electron moves from a lower energy level orbit -2.18 x 10-18 / (nfinal)2 to a higher energy level orbit -2.18 x 10-18 J / (ninitial)2 absorbs a photon 61 Practice: Calculate the wavelength () of the line corresponding to the transition of an electron from n = 4 to n = 2 in the hydrogen atom. RE = 2.18 × 10-18 J, C= 3 x 108 m/s and h = 6.63 x 10-34 Js ni = 4 E4 = -2.18 x 10-18 / (42) nf = 2 E2 = -2.18 x 10-18 / (22) E= Efinal – Einitial = E2 – E4 = - 2.18 x 10-18 [ (1/22) – (1/42 ) ] = - 2.18 x 10-18 [ (1/4) – (1/16 ) ] = - 4.1 x 10-19 J. The sign (-) indicates a lose of energy since light is emitted from an atom when the electron moves from a higher level to a lower level (nfinal< ninitial) E = h C/  =h C/ E, The wavelength  of the light emitted is = (6.63 x 10-34 x 3 x 108) / 4.1 x 10-19 = 4.86 x 10-7 m = 486 nm (green light) 62  Explanation of the line spectrum of hydrogen The lines of the hydrogen spectrum correspond to the transition of electrons from a higher energy level to a lower energy level: for the visible series, nf = 2 and ni = 3, 4, 5,... And  =h C/ E 63 de Broglie hypothesis:The Wave-Particle Duality of Matter ❑de Broglie hypothesizes particles could have a wave behavior. ❑The electrons have wave-like properties ❑The de Broglie relation:  = h/mv = h/p h: Planck’s constant: h= 6.62x10-34 Js m: mass of the particle v: velocity of the particle p: momentum of the particle (p = mv) ❑ Every particle has a wave nature as well but it is only truly evident when a particle is very light such as an electron (m = 9.11 ×10-28 g). ❑ However, we cannot observe both the wave nature and the particle nature of the electron at the same time. ❑ According to the Heisenberg Uncertainty Principle, we cannot know both the position and the speed of the electron at a given time. 64  Quantum Numbers and Atomic Orbitals Atomic Orbital: A region in space in which there is high probability of finding an electron. Quantum numbers specify the properties of atomic orbitals and their electrons, i.e., an atomic orbital is specified by three quantum numbers (n, l, ml). The spin of the electron is specified by the quantum number ms (or s) Principal Quantum Number, n Indicates main energy levels: n = 1, 2, 3, 4… Each main energy level has sub-levels A main energy level 65  Orbital Quantum Number, ℓ (Angular Momentum Quantum Number/ Azimuthal quantum number) ℓ indicates the shape of orbital sublevels ℓ = 0, 1, …., (n-1) Example: When n=4, ℓ = 0, 1, 2 and 3. ℓ=0 ℓ=1 ℓ=2 ℓ=3 s p d f 4 sublevels for N=4 A sublevel 66  Magnetic Quantum Number, mℓ Indicates the orientation of the orbital in space. Equals to integer values, including zero ranging from - ℓ to +ℓ An orbital The sublevel s contains one orbital, p contains 3 orbitals, d contains 5 orbitals and f contains 7 orbitals. In a main energy level n, there are n2 orbitals, i.e., for n = 4 there are 42 = 16 orbitals. For each main energy level n, the number of mℓ values represents the number of orbitals. 67  Electron Spin Quantum Number, (ms or s) Indicates the direction in which an electron spins in an orbital (clockwise or counterclockwise). Values of ms = -1/2, +1/2 PRACTICE Q1. When the principle quantum number n=4, the orbital quantum number ℓ takes the values a) 0,1,2,3,4 b) -4,-3,-2,-1, 0, 1,2,3,4 c) 0,1,2,3 d) -1/2, 1/2 Q2. The name of the sub-level with n= 3 and l = 2 is a) 2d b) 3d c) 3p d) 2s Q3. When the orbital quantum number ℓ = 1, the magnetic quantum number mℓ takes the values a) 0,1 b) -1, 0, 1 c) 0 d) -1/2, 1/2 Q4. The quantum number that takes only the values -1/2 and 1/2 is the magnetic quantum number (mℓ). a) True b) False 68  Electronic configurations Electronic configuration is how the electrons are distributed among the various atomic orbitals in an atom. There are 3 rules to build up electronic configurations: 1. Aufbau (building up) Principle Electrons enter the orbitals in order of ascending energy. ✓Increased n+ ℓ, increased energy ✓Same value of n+ ℓ, lowest energy for lowest n orbital 1s 2s 2p 3s 3p 4s 3d n +ℓ 1+0 2+0 2+1 3+0 3+1 4+0 3+2 Energy 2p and 3s have the same value of n+ ℓ, 2p has a lower energy than 3p General pattern for filling the sublevels 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s … 69 2. Hund’s Rule : When in orbitals of equal energy, electrons will try to remain unpaired. The electrons occupy degenerate orbitals singly to the maximum extent possible and with their spins parallel. Placing two electrons in one orbital means that, as they are both negatively charged, there will be some electrostatic repulsion between them. Placing each electron in a separate orbital reduces the repulsion and the system is more stable. YES No 2 electrons 2p 3 electrons 2p 70 3. Pauli’s exclusion principle : No two electrons in the same atom have the same set of the four quantum numbers (n , ℓ, mℓ , ms). This means that: Orbitals of the same energy must be occupied singly and with the same spin before pairing up of electrons occurs. Electrons occupying the same orbital must have opposite spins. An orbital is occupied au maximum by 2 electrons. An orbital containing paired electrons is presented as : YES No 4 electrons 2p 5 electrons 2p 6 electrons 2p 71  Practice 1: Determine the electronic configurations of 4Be, 17Cl, 26Fe, and 35Br. 4Be: 1s22s2 17Cl: 1s22s22p63s23p5 26Fe: 1s22s22p63s23p64s23d6 35Br: 1s22s22p63s23p64s23d104p5 Practice 2: How many unpaired electrons are there in 10Ne , 8O, and 15P? 10Ne: 1s2 2s2 2p6 zero unpaired electrons 8O: 1s2 2s2 2p4 2 unpaired electrons 15P: 1s2 2s2 2p6 3s2 3p3 3 unpaired electrons 72  Exceptions in some transition metals: Cr and Cu 1 electron ❑ Chromium 24Cr: Expected electronic configuration:1s2 2s2 2p6 3s2 3p6 4s2 3d4 ✓ Real electronic configuration: 1s2 2s2 2p6 3s2 3p6 4s1 3d5 1 electron ❑ Copper 29Cu: Expected electronic configuration:1s2 2s2 2p6 3s2 3p6 4s2 3d9 ✓ Real electronic configuration: 1s2 2s2 2p6 3s2 3p6 4s1 3d10 The reason for these anomalies is the greater stablity of d subshells that are either half-filled (d5) or completely filled (d10) 73  Valence-Shell and Valence Electrons ❑The most outer sublevels (for the highest n) are called valence-shell ❑Electrons in the most outer sublevels are called valence electrons ❑Transition metal: metal whose atom has an incomplete d subshell or which can give rise to cations with an incomplete d subshell. For transition metals, the valence shell is ns (n-1)d Examples: Elements Valence shell Valence electrons 9F: 1s2 2s2 2p5 2s 2p 7 11Na: 1s2 2s2 2p6 3s1 3s 1 18Ar: 1s2 2s2 2p6 3s2 3p6 3s 3p 8 35Br: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 4s 4p 7 Transition metals Valence shell Valence électrons 26Fe: 1s2 2s2 2p6 3s2 3p6 4s2 3d6 4s 3d 8 28Ni : 1s2 2s2 2p6 3s2 3p6 4s2 3d8 4s 3d 10 74  ELECTRONIC CONFIGURATION OF IONS Electrons are removed first from the highest occupied orbitals (except for transition metals). SODIUM 11Na 1s2 2s2 2p6 3s1 1 electron removed from the 3s orbital 11Na 1s2 2s2 2p6 + CHLORINE 17Cl 1s2 2s2 2p6 3s2 3p5 1 electron added to the 3p orbital 17Cl 1s2 2s2 2p6 3s2 3p6 - Isoelectronic configuration Atoms or ions having the same electronic configuration Example: 8O2-, 11Na+, 9F- 2-: 1s2 2s2 2p6 8O + 2 2 6 11Na : 1s 2s 2p - 2 2 6 9F : 1s 2s 2p 75  Electronic configuration of transition metal ions Electrons in the ns orbital are removed before any electrons in the (n-1)d orbitals. Examples: TITANIUM 22 Ti: 1s2 2s2 2p6 3s2 3p6 4s2 3d2 Ti+: 1s2 2s2 2p6 3s2 3p6 4s1 3d2 (1 electron is removed from 4s) Ti2+: 1s2 2s2 2p6 3s2 3p6 3d2 (2 electrons are removed from 4s) Ti3+: 1s2 2s2 2p6 3s2 3p6 3d1 (1 electron is removed from 3d) Ti4+: 1s2 2s2 2p6 3s2 3p6 (2 electrons are removed from 3d) Practice. Determine the electronic configurations of Fe, Fe2+ and Fe3+ Answer: Iron 26Fe: 1s2 2s2 2p6 3s2 3p6 4s2 3d6 Fe2+:1s2 2s2 2p6 3s2 3p6 3d6 (2 electron are removed from 4s) Fe3+: 1s2 2s2 2p6 3s2 3p6 3d5 (1 electron is removed from 3d) 76  Short cut electronic configuration Noble gases: [2He], [10Ne], [18Ar], [36Kr], [54Xe] [86Rn], [118Og] Short cut electronic configuration 10Ne: 1s2 2s2 2p6 Cl: 1s 2 2s2 2p6 3s2 3p5 17Cl: [Ne] 3s2 3p5 17 18Ar: 1s2 2s2 2p6 3s2 3p6 28 Ni: [Ar] 3d8 4s2 2 2 6 2 6 2 8 28Ni : 1s 2s 2p 3s 3p 4s 3d Practice: Which of the following is the short cut electronic configuration of 40Zr? Atomic numbers: 18Ar, 36Kr and 54Xe a) [Ar] 4d25s2 b) [Xe] 4d25s2 c) [Kr] 4d15s2 d) [Kr] 4d25s2 77  Periodic table of elements and properties of atoms The vertical columns of the table (Numbered from 1 to 18 ) are called groups or families. Element in the same group have similar but not identical characteristics. Elements in the same group have the same outer shell electron configuration (same number of outer shell electrons), and hence similar chemical properties. The horizontal rows of the table (Numbered from 1 to 7) are called periods. Each period contains elements with electrons in the same outer shell. 78  Effective nuclear charge (Zeff) The effective nuclear charge (Zeff) is the net positive charge experienced (felt) by an electron in a multi-electron atom Zeff: The effective nuclear charge, Zeff = Z - S Z: Atomic number of the atom, S: Number of inner electrons (in the core) Practice: Calculate Zeff for 11Na and 11Na+ Answer: 11Na: 1s22s22p63s1 Z=11 and S= 10 so Zeff = 11-10 = 1 11Na : 1s 2s 2p Z=11 and S= 2 so Zeff = 11-2 = 9 + 2 2 6 ELEMENTS OF THE SAME PERIOD ELEMENTS OF THE SAME GROUP Elements in the same period Elements in the same group Element S Zeff=Z-S Element S Zeff=Z-S 22s22p63s1 10 22s1 2 11Na:1s 11-10=1 3Li:1s 3-2=1 22s22p63s23p1 10 22s22p63s1 10 13Al:1s 13-10=3 11Na:1s 11-10=1 22s22p63s23p3 10 22s22p63s23p64s1 18 15P:1s 15-10=5 19K:1s 19-18=1 37Rb:1s 2s 2p 3s 3p 4s 3d 4p 5s 22s22p63s23p5 10 2 2 6 2 6 2 10 6 1 36 17Cl:1s 17-10=7 37-36=1 ❑ When we move from the left to the right along a period, in the periodic table, Zeff increases ❑ When we move along a group in the periodic table, Zeff remains constant 79  Atomic size or Radius The atomic size (or radius, r) is the half distance of closest approach between two identical atoms.  Cation is smaller than atom from which it is formed. Excess of protons in the ion draws the outer electrons closest to nucleus.  Anion is larger than atom from which it is formed. More repulsion between electrons and the ionic radius increases. Å Å 80  Ionization energy (I) The ionization energy is the amount of energy it takes to detach one electron from a gaseous neutral atom If it is easy to detach an electron, it has low ionization energy. If it is hard to detach an electron, it has a high ionization energy The larger the atom the easier it is to detach an electron. The smaller the atom the harder to detach an electron Electronegativity (EN) Electronegativity is the ability of an atom to attract a pair of electrons towards itself in a chemical bond. Same electronegativity A:A B is more electronegative than A A :B 81  Periodic Trends Zeff increases 82 When we move from the left to the right along a period, in the periodic table, ✓ Zeff increases ✓ The atomic size decreases ✓ The electronegativity increases ✓ The ionization energy increases When we move from the top to the bottom along a group, in the periodic table, ✓ Zeff remains constant ✓ The atomic size increases ✓ The electronegativity decreases ✓ The ionization energy decreases Practice: Arrange the following elements 12Mg, 16S, 13Al and 17Cl from the highest to the lowest 1. atomic radius 2. ionization energy 3. Electronegativity 4. Effective nuclear charge (Zeff) Answer: 1. decreasing atomic radius: Mg > Al > S >Cl 2. decreasing ionization energy: Cl > S > Al > Mg 3. decreasing electronegativity: Cl > S > Al > Mg 4. decreasing Zeff: Cl > S > Al > Mg 83  Practice on chapter 4 Q1. The wave-particle duality of light means that a)light consists of particles called photons that have wave-like properties b) electrons behave as a wave b) light shining on certain metal plates caused a flow of electrons. d) all of these Q2. The wavelength () of the line corresponding to the transition of an electron from n = 3 to n = 2 is (RE = 2.18 × 10-18 J, C= 3 x 108 m/s and h = 6.63 x 10-34 Js) a) 486 nm b) 656 nm c) 1250 nm d) 32 nm Q3. Which of the following sets of quantum numbers (n, l, ml, ms) describes an electron in an orbital a)(2, 3, 1, 1/2) b) (3, 2, 3, 1/2) c) (2, 1, 1, -1/2) d) (3, 1, 0, -1/4) Q4. Choose the correct order of the following sublevels from the lowest to the highest energy. a) 3p < 4p < 3d < 4s b) 3p < 3d < 4s < 4p c) 3d < 3p < 4s < 4p d) 3p < 4s < 3d < 4p Q5. Which one of the following about d orbitals is correct? a) They are found in all principal energy levels. b) There are 5 types of d orbitals c) They are spherical in shape d) Each d orbital can hold up to 3 electrons Q6. What is the number of sublevels associated with n = 2 a)1 b) 2 c) 4 d) 9 Q7. What is the number of orbitals associated with n = 2 a)1 b) 2 c) 4 d) 9 Q8. The electronic configuration of copper 29Cu is a)1s22s22p63s23p64s23d9 b) 1s22s22p63s23p64s13d5 c) 1s22s22p63s23p64s13d10 d) s22s22p63s23p64s13d54p3 Q9. The electronic configuration of 17Cl is a) 1s22s22p63s23p6 b) 1s22s22p63s23p5 c) 1s22s22p63p53s2 d) 1s22p62s23p53s2 Q10. Iron (Fe) is a transition metal. The electronic configuration of 26Fe2+ is a) 1s22s22p63s23p64s23d6 b) 1s22s22p63s23p64s23d4 c) 1s22s22p63s23p64s13d5 d) 1s22s22p63s23p63d6 Q11. How many unpaired electrons are there in 20Ca2+? a)0 b) 2 c) 10 d) 20 2- + Q12. The ions 8O , 11Na and 9F - are isotopes b) have isoelectronic configurations c) are compounds d) are called sublevels a)Q13. The effective nuclear charge (Zeff) of oxygen (8O) is b)a) 2 b) 4 c) 6 d) 8 84 Q14. Arrange the elements 3Li, 11Na, 19K and 37Rb, of the same group, from the highest to the lowest electronegativity a) 3Li >11Na > 19K > 37Rb b) 37Rb >19K >11Na > 3Li c) 3Li >19K >11Na>37Rb d) 37Rb >11Na >19K > 3Li Q15. In the periodic table, when we move from the top to the bottom in a group, the atomic size a) increases b) decreases c) remains constant d) both increases and decreases Q16. The energy of light is a) inversely proportional to frequency b) directly proportional to frequency c) constant d) equal to frequency CALCULATIONS Q1. Two states differ in energy by 3.03 x 10-19 J. Calculate the a) wavelength () in meter and nanometer b) frequency () (h = 6.63 x 10-34 J.s and C = 3 x 108 m/s) Answer: E = 3.03 x 10-19 J a) E = h C/  = h C/ E  = (6.63 x 10-34 x 3 x 108) / 3.03 x 10-19   = 6.56 x 10-7 m = 656 nm. b)  = C /   = 3 x 108 / 6.56 x 10-7 = 4.57 x 1014 s-1 Book Reference https://www.routledge.com/Understanding-General- Chemistry/Korchef/p/book/9781032189147 85 Chapter 5 At the end of this chapter, the student will be able to ❑ Differentiate between polar and nonpolar molecules Molecular Forces ❑ Identify and differentiate between the and Properties of different types of intramolecular forces and intermolecular forces Matter ❑ interpret properties of matter based on the intermolecular forces, ❑ define the thermodynamic equilibrium between the different states of matter ❑ recognize the exothermic and endothermic phase changes ❑ analyze a phase diagram.  Molecular Forces Intramolecular forces: Forces are within a molecule (ex: Ionic bond, polar and nonpolar covalent bond). Intermolecular forces occur between molecules: o Keesom forces (between polar molecules) o Debye forces (between a nonpolar molecule and a polar molecule) o London dispersion forces (between nonpolar molecules) o Hydrogen bond (Hydrogen on one molecule attached to (N, O, or F) atoms and either (N, O, or F) on another molecule. In a substance, intramolecular forces (chemical bonds) are stronger than intermolecular forces. 87  Intramolecular Forces Ionic bond results from the transfer of electrons from a metal to a nonmetal. It involves the electrostatic attraction between oppositely charged ions (cation and anion) Covalent bond results from sharing electrons between nonmetals. Nonpolar covalent bond is when nonmetals equally share electrons between them. Polar covalent bond occurs between two nonmetals in which electrons are shared unequally. https://www.angelo.edu/f aculty/kboudrea/general/ shapes/00_lewis.htm Metallic bond attractive force holding pure metals together.  Polar and nonpolar molecules Reminder: Electronegativity is a measure of the tendency (ability) of an atom to attract a bonding pair of electrons towards itself. Nonpolar molecules occur when there is an equal sharing of electrons between their atoms. Examples of nonpolar molecules: ▪ O2, H2, Cl2 , I2, Br2 Cl : Cl ▪ CO2 There is an equal sharing of electrons. ▪ Hydrocarbons: CH4 , C2H6 , C3H8 ,…. Cl2 is a nonpolar molecule. Polar molecules occur when atoms do not share electrons equally. A dipole forms, with part of the molecule carrying a slight positive charge and the other part carrying a slight negative charge. This happens when there is a difference between the electronegativities of the atoms or because of the molecular geometry. H :Cl Examples of polar molecules: ▪ HBr, HCl, HI, H2O, HF, NH3 ▪ SO2 , O3 Cl is more electronegative than H. The two atoms did not share electrons equally. HCl is a polar molecule. HCl molecule forms a permanent dipole.  Intermolecular forces: Keesom Forces (Dipole-dipole interactions) Act between polar molecules. Oppositely charged ends attract and like ends repel. Moderate strength Interaction between permanent dipoles. Examples: HCl, HBr, HI Keesom forces become stronger when ❑ the difference of electronegativity between the atoms forming the molecule increases. ❑ the temperature decreases. ❑ the distance between molecules decreases.  Debye Forces ❑ Occur between a nonpolar molecule and a polar molecule. ❑ The permanent dipole in the polar molecule induces an electric dipole in the nonpolar molecule when it comes extremely close. Examples: Debye forces between O2 and H2O Debye forces When O2 comes extremely close to H2O  London Dispersion Forces ❑ Polarizability is the ease with which the electron distribution around an atom or molecule can be distorted ('How easy is it to form an induced dipole'). ❑Instantaneous dipole, that occurs accidentally in an atom or a molecule due to electrons movement, induces a similar dipole (induced dipole) in a neighboring atom or molecule. ❑ London forces occur in nonpolar molecules. ❑ Weak forces Examples: Homonuclear diatomic molecules: H2, Cl2, I2, O2, F2, Br2, … Hydrocarbons: CH4, C3H8, C6H14 , …. Carbon dioxide CO2 Noble gases : He, Ne, Ar, …. London dispersion forces become stronger when ❑ the polarizability increases. Polarizability increases when ✓ The size of atoms or molecules increases. ✓ The number of electrons increases. ❑ the distance between molecules decreases. ❑ the chain of carbon atoms in hydrocarbons becomes longer.  PRACTICE Q1: Which of the followings shows the highest London dispersion forces? a) C3H8 b) C4H10 c) C5H12 d) C6H14 Answer : C6H14 shows the highest London dispersion forces C3H8 < C4H10< C5H12 < C6H14 Increasing London forces Q2: According to the following order from the lowest to the highest atomic size: F < Cl < Br < I explain why F2 and Cl2 are gases, Br2 is a liquid and I2 is a solid at room temperature. Answer : F2, Cl2, Br2 and I2 are nonpolar molecules, and they show London dispersion forces. As the size increased, London forces increased: Increased London dispersion forces  Hydrogen Bonding ❑ Hydrogen on one molecule attached to a highly electronegative atom (N, O, or F) and either (N, O, or F) on another molecule. ❑ Strong dipole-dipole force. ❑ Occurs between Polar molecules. Examples: H2O, HF, NH3, CH3COOH Order from the strongest to the weakest intermolecular forces: Hydrogen bond > Keesom force >Debye force > London force Dr Korchef Atef-2021  Properties of Matter Vapor pressure The pressure exerted by the vapor on the surface of the liquid at equilibrium is called the vapor pressure. In a closed system, when the rate at which the liquid is entering the gas phase equals the rate at which the vapor is returning to the liquid phase, the system is at equilibrium. After this time, the liquid level will remain constant. The pressure exerted by the vapor on the surface of the liquid at equilibrium is called the vapor pressure. The vapor pressure of a liquid increases when ✓ the intermolecluar forces decreases. ✓ the temperature increases.  Boiling Point The temperature at which the vapor pressure equals the external pressure (atmospheric pressure) is called the boiling point. This is the point where bubbles of vapor form within the liquid. ❑ The boiling point of a liquid at 1 atm pressure is the normal boiling point. ❑The boiling point increases with increasing external pressure. ❑The molecules that have strong intermolecular forces have high boiling point.  PRACTICE Which of the compounds hexane (C6H14) and propane (C3H8) has the highest boiling point ? Which of these two compounds has the lowest vapor pressure ? Explain your answer. Answer: Hexane C6H14 has a longer chain than propane C3H8. Therefore, C6H14 is held by higher attractive molecular forces via London forces than C3H8. Hexane C6H14 has lower vapor pressure than propane C3H8. Hexane C6H14 has higher boiling point than propane C3H8. The substance having the highest boiling point has the lowest vapor pressure and conversely, the substance having the lowest boiling point has the highest vapor pressure.  Viscosity of a liquid ❑ Viscosity is the resistance to flow of a liquid. When the viscosity of a liquid increases, the liquid flows more slowly. The viscosity of honey is higher than the viscosity of water. Honey flows more slowly than water. Viscosity of a liquid increases when ✓the intermolecular forces increases ✓the temperature decreases  Phase Changes Phase changes are transformations from one phase to another. The energy required to go from one state to another state is called enthalpy change (H). melting, vaporization and sublimation are endothermic processes (need energy, H>0). freezing, condensation and deposition are exothermic processes (release energy, H