A-Level Chemistry Revision Notes - Atomic Structure and Stoichiometry (PDF)

CHAPTER 1: Atomic Structure 1.1 Inside the Atom 1.2 Isotopes Learning outcomes: (a) identify and describe protons, neutrons and electrons in terms of their relative charges and relative masses. (b) deduce the behaviour of beams of protons, neutrons and electrons in electric fields. (c) describe the distribution of mass and charges within an atom. (d) deduce the numbers of protons, neutrons and electrons present in both atoms and ions given proton and nucleon numbers (and charge). (e) (i) describe the contribution of protons and neutrons to atomic nuclei in terms of proton number and nucleon number. (ii) distinguish between isotopes on the basis of different numbers of neutrons present (iii) recognise and use the symbolism where x is the nucleon number and y is the proton number. 1.1 Inside the Atom Sub-atomic particles 1) Electrons revolve around in region of space called orbitals. 2) Electrons do not move in fixed orbits. 3) The nucleus is made up of protons and neutrons which contains almost all the mass of the atom. This is because the mass of electrons is very small compared to others. 4) The nucleus is positively-charged because of the protons. Electrons, being negatively-charged, surround the nucleus. Particles Relative mass Relative Charge Charge / C -19 Protons, p 1 +1 +1.6 x 10 Neutron, n 1 0 0 1 -19 Electron, e -1 -1.6 x 10 1836 Behaviour of sub-atomic particles in electric field 1) Proton will be deflected towards the negative plate because it is positively-charged. 2) Electron will be deflected towards the positive plate because it is negatively-charged. 3) Neutron will not be deflected and continue in their direction of motion because it is neutral(not charged). 4) Angle of deflection of electron > Angle of deflection of proton because the mass of electron is smaller than proton. (angle of deflection is inversely proportional to charge/mass ratio) 5) Conclusion: i. Protons are positively-charged ii. Electrons are negatively-charged iii. Neutrons are neutral iv. Protons are much heavier than electron Nucleon number and proton number 1) Proton number is the total number of protons in an atom. 2) Nucleon number is the total number of protons and neutrons in an atom. 3) Proton number is also known as atomic number while nucleon number is also known as mass number. 4) In a neutral atom, the total number of protons equals to the total number of electrons. 5) When an atom gains or loses electrons, a cation or anion will be formed. 6) Cation is a positively-charged ion. It is formed when an atom loses electron(s). In cation, the number of protons is more than the number of electrons. 7) Anion is a negatively-charged ion. It is formed when an atom gains electron(s). In anion, the number of electrons is more than the number of protons. 8) An atom or ion is said to be i. isoelectronic if they have the same number of electrons. ii. isotonic if they have the same number of neutrons. iii. isotopic if they have the same number of protons. To deduce the number of protons, neutrons and electrons in an atom/ion 1.2 Isotopes Isotopes 1) Isotopes are atoms of the same element with the same number of proton but different number of neutron. Example: 2) Isotopes have the same: i. number of protons and electrons ii. electronic configuration iii. chemical properties(because they have the same number of electrons) 3) Isotopes have different: i. number of neutrons and nucleon number ii. mass iii. density iv. molecular speed 4) Isotopes can be stable or unstable. Unstable isotopes are called radioactive isotopes(radioisotopes). CHAPTER 2: Atoms, Molecules and Stoichiometry 2.1 Mass of Atoms and Molecules 2.2 Mass Spectrometer 2.3 Amount of Substance 2.4 Empirical Formula and Molecular Formula 2.5 Stoichiometry and Equations Learning outcomes: (a) define and use the terms relative atomic, isotopic, molecular and formula masses, based on the C-12 scale. (b) define and use the term mole in terms of the Avogadro constant. (c) analyse mass spectra in terms of isotopic abundances [knowledge of the working of the mass spectrometer is not required]. (d) calculate the relative atomic mass of an element given the relative abundances of its isotopes, or its mass spectrum. (e) define and use the terms empirical and molecular formulae. (f) calculate empirical and molecular formulae, using combustion data or composition by mass. (g) write and/or construct balanced equations. (h) perform calculations, including use of the mole concept, involving: (i) reacting masses (from formulae and equations). (ii) volumes of gases (e.g. in the burning of hydrocarbons). (iii) volumes and concentrations of solutions. When performing calculations, candidates’ answers should reflect the number of significant figures given or asked for in the question. When rounding up or down, candidates should ensure that significant figures are neither lost unnecessarily nor used beyond what is justified. (i) deduce stoichiometric relationships from calculations such as those in (h). 2.1 Mass of Atoms and Molecules Concept of relative mass 1) Relative mass is an indication of how heavy is an atom compared to another atom which is used as a standard model. 2) Relative mass is expressed in atomic mass unit(a.m.u). 3) C-12 was chosen to be the standard model because: i. it is the most abundant isotope of carbon. ii. it is a solid, easy to handle and easily available. 4) C-12 was assigned a mass of exactly 12 a.m.u.. This is known as C-12 scale. 5) For example, an atom which is 3.5 times heavier than a C-12 atom would have a relative mass of (3.5 x 12) = 42 a.m.u.. That means, this atom is 42 times heavier than the mass of (1/12 x the mass of C-12 atom). Relative isotopic mass 1) Relative isotopic mass is the mass of an isotope measured on a scale in which a carbon-12 atom has a mass of exactly 12 units. Relative atomic mass, Ar 1) Relative atomic mass, Ar is the weighted average relative masses of all its isotopes measured on a scale in which a carbon-12 atom has a mass of exactly 12 units. Average mass of one atom of the element Relative atomic mass, Ar = X 12 Mass of one atom of carbon-12 Example: Ratio of Cl-35 to Cl-37 is 3:1. If you have 4 typical atoms of chlorine, total mass is (35 x 3) + (37 x 1) = 142. So, the average mass of the isotopes is 142/4 = 35.5. This implies that 35.5 is the relative atomic mass of chlorine while 35 is the relative mass of Cl-35 and 37 is the relative mass of Cl-37. Relative molecular mass, Mr 1) Relative molecular mass, Mr is the weighted average of the masses of the molecules measured on a scale in which a carbon-12 atom has a mass of exactly 12 units. 2) It should only be applied to substances which exist as molecules. 3) It is found by adding up all the relative atomic masses of all the atoms present in the molecule. 4) Examples: i. Mr (H2O) = 2(1) + 16 = 18 ii. Mr (CHCl3) = 12 + 1 + 3(35.5) = 119.5 Relative formula mass, Mr 1) Relative formula mass, Mr is the weighted average of the masses of the formula units measured on a scale in which a carbon-12 atom has a mass of exactly 12 units. 2) It works for both ionic and covalent compounds. 3) Examples: i. Mr (NaCl) = 23 + 35.5 = 58.5 ii. Mr (CuSO4 H2O) = 64 + 32+ 4(16) +5[2(1) + 16] = 249.5 2.2 Mass Spectrometer What is mass spectrometer? 1) A mass spectrometer is used to determine: a. relative isotopic mass b. relative abundance of isotopes c. relative atomic mass d. relative molecular mass e. structural formula of compounds Determination of relative atomic mass using mass spectrometer 1) Five steps: i. Vaporisation - atoms are vaporised to form gaseous atoms. ii. Ionisation - gaseous atoms are bombarded with high energy electrons to form positive ions. iii. Acceleration - the ions are accelerated so that they have the same kinetic energy. iv. Deflection - ions are deflected by a magnetic field. The amount of deflection depends on: 1) the mass of the ion 2) the amount of positive charge on it - the larger the mass, the smaller the deflection. - the higher the charge, the larger the deflection. - the two factors combine into mass/charge ratio (m/e or m/z). - the smaller the value of m/e, the larger the deflection. v. Detection - the beam of ions are detected electrically. - the data are fed into the computer and the mass spectrum is produced. Side note 1) Ionisation chamber is vacuum so that the ions produced can run freely without knocking air molecules. Mass spectrum (How to calculate relative atomic mass, Ar from it?) Relative abundance c b a m/e m₁ m₂ m₃ (m₁ x a) + (m₂ x b) + (m₃ x c) Ar = a+b+c Example: The mass spectrum of boron, B is as shown, given the relative abundances: B-10 : 23 B-11 : 100 (23 x 10) + (100 x11) Ar = 23 + 100 = 10.8 2.3 Amount of Substance The mole and the Avogadro constant 1) A mole of a substance is the amount of substance that contains the same amount of stated elementary units as there are atoms is 12 g of C-12. 2) The number of atoms is 12 g of C-12 is 6.02 x 10²³. This number is also known as the Avogadro's constant, L. 3) Examples: i. 1 mol of He contains 6.02 x 10²³ He atoms. ii. 1 mol of CO2 contains 6.02 x 10²³ CO2 molecules but 3 x (6.02 x10²³) atoms. iii. 1 mol of NaCl contains 6.02 x 10²³ NaCl units, Na⁺ and Cl⁻ ions. Moles and mass Mass / g No. of mole / mol = -1 Molar mass / g mol Moles and volumes 1) Volume occupied by a gas depends on the amount of gas, temperature and pressure. In other words the volume of a gas is not fixed. 2) Avogadro's Law states that for equal volumes of all gases, under the same conditions, contain the same number of moles. 3) Hence, equal number of moles of any gas, under the same conditions, would occupy the same volume. It does not depend on the nature of gas. 4) At room temperature of 20 ℃ and a pressure of 1 atm, one mole of any gas occupies 24 dm³. 5) At standard temperature and pressure (s.t.p), which is 0 ℃ and 1 atm, one mole of any gas occupies 22.4 dm³. Volume of a gas / dm³ No.of mole / mol = Molar volume / dm³ mol -1 6) i. Complete combustion of hydrocarbon produces water and carbon dioxide. The general equation is as follow: ii. In incomplete combustion, the possible products are carbon dioxide, carbon monoxide, carbon soot and water. Moles and concentration of solutions 1) A solution is a homogeneous mixture of two or more substance. 2) The substance presents in small quantity is called the solute while the substance present is larger quantity is called the solvent. 3) Concentration is the amount of solute present in a fixed quantity of solution. 4) Concentration is expressed in terms of g dm⁻³. Concentration in mol dm⁻³ is called molar concentration or molarity. Mass of solute / g Concentration / g dm⁻³ Volume of solution / dm³ Concentration / g dm⁻³ Molarity / mol dm⁻³ Concentration / g dm⁻³ Molar mass of solute / g mol -1 Molarity / mol dm⁻³ Molar mass of solute / g mol -1 Volume / cm³ X Molarity / mol dm⁻³ No. of moles / mol = 1000 Volume / cm³ X Molarity / mol dm⁻³ No. of moles / mol = 1000 2.4 Empirical Formula and Molecular Formula Percentage composition by mass Percentage composition Ar x No.of mole of that element by mass / % = X 100% Molar mass of compound Empirical formula 1) Empirical formula is a chemical formula that shows the simplest ratio of the atoms that combine to form a molecule. 2) Steps to find empirical formula: i. Find the mass of each element. ii. Find the number of mole of each element (divide by its Ar). iii. Find the simplest ratio (divide by the smallest number). iv. Construct the empirical formula using the simplest ratio. [ If a decimal or fraction exists, round up or eliminate the fraction ] [ Never assume a formula ] 3) Some facts: i. The formula for an ionic compound is always its empirical formula. ii. The empirical formula and molecular formula for simple inorganic molecules are often the same. iii. Organic molecules have different empirical and molecular formula. Molecular formula 1) Molecular formula is a chemical formula that shows the actual number of atoms that combine to form the compound. 2) In order to deduce the molecular formula of a compound, we need to know: i. the relative formula mass of the compound. ii. the empirical formula of the compound. Principle of conservation of mass 1) Mass is neither created nor destroyed during a chemical reaction. Therefore the total mass of the reactants is equal to the total mass of the products in a closed system. 2) For example, the total mass of iodine in the reactants is equal to the total mass of iodine in the products. 3) This can be used to solve problems in calculating the empirical formula. 2.5 Stoichiometry and Equations Stoichiometry 1) Stoichiometry is the proportion of things either reacting or combining. 2) In compounds, it refers to the ratio in which the atoms are combined together. For example, water, H2O has a stoichiometry of 2 hydrogen to 1 oxygen. 3) It also refers to the reacting proportions in a chemical equation. For example: 2H2 + O2 → 2H2O The stoichiometry shows that 2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water. Ionic equations 1) Steps to construct net ionic equations: i. Write the balanced molecular equation. ii. Write the complete ionic equation by splitting it into ions(if possible). iii. Cancel out the spectator ions. (Spectator ions are ions that present in the mixture but do not participate in the reaction.) iv. Write down the 'leftovers', that is the net ionic equation. Chemical equation: Complete ionic equation: Net ionic equation: FAQ 1: When to split compounds into FAQ 2: How to identify spectator ions? ions? 1) The ions present on both sides of 1) Only split aqueous ionic compounds. For the equation are spectator ions. example, NaCl(aq) and HCl(aq) 2) Do not split solid ionic compounds and covalent compounds, as well as metals. For example, NaCl(s), H2O(l), Mg(s) and HCl(g) CHAPTER 3: Electrons in Atoms 3.1 Sub-shells and Atomic Orbitals 3.2 Electronic Configuration 3.3 Ionisation Energy Learning outcomes: (a) describe the number and relative energies of the s, p and d orbitals for the principal quantum numbers 1, 2 and 3 and also the 4s and 4p orbitals. (b) describe the shapes of s and p orbitals. (c) state the electronic configuration of atoms and ions given the proton number (and charge), using the convention 1s²2s²2p⁶ etc. (d) (i) explain and use the term ionisation energy. (ii) explain the factors influencing the ionisation energies of elements. (iii) explain the trends in ionisation energies across a Period and down a Group of the Periodic Table. (e) deduce the electronic configurations of elements from successive ionisation energy data. (f) interpret successive ionisation energy data of an element in terms of the position of that element within the Periodic Table. 3.1 Sub-shells and Atomic Orbitals Principle quantum shell 1) Electrons are arranged outside the nucleus in energy levels or principle quantum shell, n. 2) The principal quantum shells are numbered according how far are they from the nucleus. 3) The lowest energy level, n = 1 is closest to the nucleus, the energy level n = 2 is further out, and so on. 4) The electrons in energy level further away from the nucleus have more energy and held less tightly by the nucleus. The circles represent 5) Electrons do not move in fixed circular paths, they energy levels occupy a space called the atomic orbitals. 6) The total number of electrons that can occupy any principal shell is 2n², where n is the principal quantum number. Quantum sub-shells 1) The principal quantum shells, apart from the first, are split into sub-shells. Each principle quantum shell contains a different number of sub-shells. 2) The first energy level contains one sub-shell, the second energy level contains two and so on. 3) The sub-shells are distinguished by letters s, p, d, f and so on. 4) The energy of electrons in the sub-shells increases in the order s < p < d < f. Principal Maximum number of Number of Name of quantum shell electrons sub-shells sub-shells K, n = 1 2 1 1s L, n = 2 8 2 2s, 2p M, n = 3 18 3 3s, 3p, 3d 32 4 4s, 4p, 4d, 4f N, n = 4 The impossibility of drawing orbits for electrons 1) Heisenberg Uncertainty Principle says, you cannot know with certainty where an electron is and where it is going next. 2) This makes it impossible to draw out an orbit or pathway in which the electrons move. Atomic orbitals 1) An atomic orbital is a region of space around the nucleus where the probability of finding a particular electron is maximum (>95%). 2) The sub-shells are split further into orbitals where the electrons are placed. 3) The number of orbital in each sub-shell depends on the sub-shells. s - one orbital {s} p - three orbitals {px, py, pz} d - five orbitals {dxy, dyz, dxz, dx²-y², dz²} 4) Orbitals having the same energy are called degenerate orbitals. For example, px, py and pz are degenerate orbitals. 5) The concept of orbitals arises from the fact that an electron has dual nature. It behaves as a particle as well as a wave. 6) In the nth principal quantum shell, there are n sub-shells, n² orbitals and a maximum of 2n² electrons. The s orbital 2s 1) All the s orbitals are spherical and non-directional. 2) The shaded region represents the region in which the chance of finding the s electron is more than 95%. 3) The size of the s orbital increases in the order 1s < 2s < 3s < 4s. The p orbital 1) All the p orbitals are dumb-bell shaped and directional. 2) p orbitals are only available from the second principal quantum shell and onwards. 3) There are 3 types of p orbitals, px, py and pz. All 3 different types of p orbitals are perpendicular to each other along the x, y and z axes. 4) Going to a higher energy level, the 'lobes' of the p orbital become longer. 3.2 Electronic Configuration Ways to represent electronic configuration Electronic configuration describes how the electrons in an atom/ion are arranged in their shells, sub-shells and orbitals. 1) Using 'electrons-in-boxes': 2) Using energy levels: 3) Using s, p, d and f notation: 4) Using the noble gas 'core': Filling in the orbitals 1) Three general rules of filling the orbitals are: i. Aufbau Principle states that in the ground state of an atom, the electrons must occupy the orbitals in the order of increasing energy. [ Note: 4s has a slightly lower energy than 3d, therefore electrons are filled in 4s orbital first before the 3d orbitals. ] ii. Pauli's Exclusion Principle states that an orbital can only accommodate a maximum of two electrons only. The two electrons must have opposite spins. iii. Hund's Rule states that in a set of degenerate orbitals, electrons must occupy the orbital singly first before pairing. The reason is because, two electrons occupying the same orbital will experience repulsion. rather than 2) Electronic configuration of the elements (up to Z = 38) 3) The odd ones (K, Sc, Cr and Cu): i. For potassium, 4s is filled before 3d because 4s has a lower energy level than 3d. ii. For scandium (to zinc, the d-block elements), the energy level of 3d and 4s are reversed. 4s is at a higher energy level now. This is because once the 3d orbital(s) is/are filled, the 3d electrons repel the 4s electrons to a higher energy level. iii. For chromium, the electronic configuration is [Ar]3d⁵4s¹ instead of [Ar]3d⁴4s². This is because orbitals that are fully filled or half filled have extra stability due to their symmetrical charge distribution. iv. For copper, the electronic configuration is [Ar]3d¹⁰4s¹ instead of [Ar]3d⁹4s². The reason is the same as stated in (iii). Electronic configuration of ions 1) In the formation of cation, the electrons are removed in the order of decreasing energy (the reverse of filling in). 2) For the d-block elements, the electrons in 4s is removed first before 3d. This is because once the 3d orbital is filled, the 4s electrons are repelled to a higher energy level than 3d. 3) In the formation of anion, the electrons are added in the same manner as filling the electrons. Orbitals and the Periodic Table 1) The elements in the Periodic Table can be divided into four blocks according to their valence shell electronic configuration. 2) The s-block elements have their valence electron(s) in the s orbital. 3) The p-block elements have their valence electrons in the p orbital. 4) The d-block elements have d orbitals filling. 5) The f-block elements have f orbitals filling. Some useful facts 1) The valence electrons always appear at the end of the electronic configuration. (but not necessary the last one). 2) The valence electrons are large responsible for the chemical properties of an element. 3) The number of valence electron will indicate the group number of that element in the Periodic Table. 4) The outermost quantum shell number will indicate the period of that element in the Periodic Table. 3.3 Ionisation Energy What is ionisation energy? 1) The 1st ionisation energy, ΔHi1 is the energy needed to remove one electron from each atom in one mole of the atoms of the element in the gaseous state to form one mole of gaseous 1+ ions. 2) The general unit for ionisation energy is kJ mol⁻¹. Ca(g) → Ca⁺(g) + e⁻ ; ΔHi1 = +590 kJ mol⁻¹ 3) If a second electron is removed from the gaseous 1+ ions, it is the 2nd ionisation energy, ΔHi2. Ca⁺(g) → Ca²⁺(g) + e⁻ ; ΔHi2 = +1150 kJ mol⁻¹ 4) The 2nd ionisation energy, ΔHi2 is the energy needed to remove one electron from each gaseous 1+ ion in one mole of the ions to form one mole of gaseous 2+ ion. 5) The continuos removal of electrons until the nucleus is left only will result in successive ionisation energies. Factors affecting the ionisation energy 1) Charge on the nucleus (Number of proton) - The greater the number of proton in the nucleus, the greater the amount of positive charge. - The greater the positive charge, the greater the attractive force between the nucleus and outer electrons. - More energy is needed to overcome the attractive force. So, the ionisation energy is higher. - The greater the nuclear charge, the higher the ionisation energy. 2) Distance between nucleus and outer electrons (Size of atom/ion) - The larger the size of the atom, the greater the distance between the nucleus and the outer electrons. - The greater the distance between the nucleus and the outer electrons, the weaker the attractive force between nucleus and outer electrons. - Furthermore, the outer electrons experience greater shielding effect by the inner electrons. - Less energy is required to overcome the attractive force. So, the ionisation energy is lower. - The greater the distance between nucleus and outer electrons, the lower the ionisation energy. 3) Shielding effect by the inner electrons - All electrons are negatively-charged, so they repel each other. Electrons in full inner shells will repel the outer electrons and so prevent the full nuclear charge being felt by the outer electrons. This is called shielding or screening. - The greater the shielding effect, the weaker the attractive force between the nucleus and outer electrons. - Less energy is required to overcome the attractive force. So, the ionisation energy os lower. - The greater the shielding effect, the lower the ionisation energy. Pattern of ionisation down a Group 1) The first ionisation energy decreases down a Group. 2) This is because the atomic size increases and hence the distance between the nucleus and outer electrons increases. The outer electrons also experience a greater shielding effect. 3) These two factors outweigh the increasing nuclear charge. 4) The above factors causes the attractive force between nucleus and outer electrons to decrease, less energy is required to overcome the weaker attractive force. Hence, the ionisation energy is lower. Pattern of ionisation energy across a Period 1) The general trend of ionisation energy across a Period is increasing. 2) This is because, across a Period, the number of proton in the nucleus increases by one therefore the nuclear charge increases. 3) However, the distance between the nucleus and outer electrons decreases across a Period and the outer electrons experience the same amount of shielding. 4) The above factors causes the attractive force between nucleus and outer electrons to increase, more energy is required to overcome the stronger attractive force. Hence, the ionisation energy is higher. The drop between (Be-B) and (Mg-Al) 1) There is a slight decrease in first ionisation energy between beryllium-born and magnesium-aluminium. 2) This is because the fifth electron in boron is located in the 2p sub-shell, which is slightly further away from the nucleus. The outer electron in boron is shielded by the 1s² as well as 2s² electrons. Be : 1s²2s² B : 1s²2s²2p¹ 3) The decrease in first ionisation energy between magnesium and aluminium has the same reason, except that everything is happening at the third energy level. Mg : 1s²2s²2p⁶3s² Al : 1s²2s²2p⁶3s²3p¹ The drop between (N-O) and (P-S) 1) There is a slight decrease in first ionisation energy between nitrogen-oxygen and phosphorus-sulphur. 2) This is because the electron being removed in oxygen is from the orbital which contains a pair of electrons. The extra repulsion between the pair of electrons results in less energy needed to remove the electron. This is called spin-pair repulsion. N : 1s²2s²2px¹2py¹2pz¹ O : 1s²2s²2px²2py¹2pz¹ 3) The decrease in first ionisation energy between phosphorus and sulphur has the same reason, except that everything is happening at the third energy level. P : 1s²2s²2p⁶3s²3px¹3py¹3pz¹ S : 1s²2s²2p⁶3s²3px²3py¹3pz¹ Successive ionisation energy 1) The following data can be obtained from successive ionisation energy: i. Total number of electrons in an atom. - is equal to the number of separate ionisation energies possessed by the atom. ii. Number of principal quantum shells occupied and the number of electrons in each. -by looking the big difference between two successive ionisation energies. iii. Number of sub-shells occupied and the number of electrons in each. 2) Successive ionisation energies get larger because removing an electron from a positive ion with increasing positive charge is going to be more difficult due to the increasing attractive force. 3) There is a relatively big increase in ionisation energy between the first and second electron being removed. This suggests that the second electron being removed is from a principal quantum shell closer to the nucleus. 4) The big jump occurs three times, so there are four principal quantum shells occupied by this atom. 5) After the big jumps, there is a steady increase in ionisation energy, this suggests that the electrons being removed come from the same principal quantum shell. 6) The electronic configuration for this atom can be written as 2,8,8,1. 7) Two more examples: This element comes from Group 2 This element comes from Group 14 of the Periodic Table of the Periodic Table Looking at the pattern in more detail 1) The electronic configuration of chlorine is 1s²2s²2p⁶3s²3px²3py²3pz¹. 2) Between the second and third ionisation energy, there is a slight increase in difference in ionisation energy. This is because the first two electrons being removed come from the orbitals which contain a paired electrons. The extra repulsion between the electrons result in the ionisation energy being lower. 3) There is also a slight increase in difference in ionisation energy between the fifth and sixth electron being removed. This is because the sixth electron being removed comes from the 3s sub-shell, which is slightly closer to the nucleus. 4) The drastic increase in ionisation energy between the seventh and eighth electrons suggests that the eighth electron comes from a principal quantum shell closer to the nucleus. CHAPTER 4: Chemical Bonding 4.1 Ionic Bonding 4.2 Covalent Bonding 4.3 Shapes of Molecules 4.4 Electronegativity, Bond Polarity, Bond Length and Bond Energy 4.5 Intermolecular Forces 4.6 Metallic Bonding 4.7 Bonding and Physical Properties of Substances Learning outcomes: (a) describe ionic (electrovalent) bonding, as in sodium chloride and magnesium oxide, including the use of ‘dot-and-cross’ diagrams. (b) describe, including the use of ‘dot-and-cross’ diagrams, (i) covalent bonding, as in hydrogen, oxygen, chlorine, hydrogen chloride, carbon dioxide, methane, ethene. (ii) co-ordinate (dative covalent) bonding, as in the formation of the ammonium ion and in the Al2Cl6 molecule. (c) explain the shapes of, and bond angles in, molecules by using the qualitative model of electron-pair repulsion (including lone pairs), using as simple examples: BF3 (trigonal), CO2 (linear), CH4 (tetrahedral), NH3 (pyramidal), H2O (non-linear), SF6 (octahedral), PF5 (trigonal bipyramid). (d) describe covalent bonding in terms of orbital overlap, giving σ and π bonds, including the concept of hybridisation to form sp, sp² and sp³ orbitals. (e) explain the shape of, and bond angles in, ethane and ethene in terms of σ and π bonds. (f) predict the shapes of, and bond angles in, molecules analogous to those specified in (c) and (e). (g) describe hydrogen bonding, using ammonia and water as simple examples of molecules containing N-H and O-H groups. (h) understand, in simple terms, the concept of electronegativity and apply it to explain the properties of molecules such as bond polarity and the dipole moments of molecules. (i) explain the terms bond energy, bond length and bond polarity and use them to compare the reactivities of covalent bonds. (j) describe intermolecular forces (van der Waals’ forces), based on permanent and induced dipoles, as in CHCl3(l); Br2(l) and the liquid noble gases. (k) describe metallic bonding in terms of a lattice of positive ions surrounded by mobile electrons. (l) describe, interpret and/or predict the effect of different types of bonding (ionic bonding, covalent bonding, hydrogen bonding, other intermolecular interactions, metallic bonding) on the physical properties of substances. (m) deduce the type of bonding present from given information. (n) show understanding of chemical reactions in terms of energy transfers associated with the breaking and making of chemical bonds. 4.1 Ionic Bonding Formation of ionic bond 1) Ionic bond is the electrostatic force of attraction between oppositely-charged ions formed by the complete transfer of electrons from an atom to another atom. 2) Ionic bond is also called electrovalent bond. 3) i. An atom(usually a metal) that loses electron(s) will form a positive ion(cation). ii. The electron is then transferred to another atom. iii. The atom that gains the electron(usually a non-metal) will form a negative ion(anion). iv. The cations and anions are then attracted by strong electrostatic force of attraction. The force of attraction constitutes the ionic bond. 4) The force of attraction between cation and anion is very strong, therefore ionic bond is a very strong bond. 5) Ionic bonds are non-directional, each cation will attract any neighbouring anion and vice versa to form a huge ionic lattice. 6) The compound formed as a result of ionic bond is called ionic compound. An example is sodium chloride, NaCl. Dot-and-cross diagram 1) A dot-and-cross diagram shows: i. the outer electron shell only. ii. that the charge of the ion is spread evenly using a square bracket. iii. the charge of each ion. 2) It is also called the Lewis diagram. 3) Using dot-and-cross diagram to represent the formation of ionic bond: Strength of ionic bonds 1) The strength of ionic bond is a measure of the electrostatic force of attraction between the ions. (Q⁺) (Q⁻) 2) The force of attraction between the oppositely-charged E∝ ions is proportional to the charge on the ions and d² inversely proportional to the square of distance between the ions. 3) The strength of ionic bond is manifested in the melting point of the ionic compound. 4) i. For instance, the melting point of NaCl is higher than NaBr. ii. This is because the size of Br⁻ ion is larger than Cl⁻ ion. Therefore the distance between Br⁻ and Na⁺ is larger than that of between Cl⁻ and Na⁺. iii. As a result, the electrostatic force of attraction between Na⁺ and Cl⁻ is stronger than that of between Na⁺ and Br⁻ ion. 5) i. The melting point of NaCl is lower than MgCl2. ii. This is because Mg²⁺ ion has a higher charge than Na⁺ ion. Besides that, the size of Mg²⁺ ion is smaller than Na⁺ ion. iii. The above two factors causes the electrostatic force of attraction between Mg²⁺ and Cl⁻ to be stronger than that of between Na⁺ and Cl⁻. 4.2 Covalent Bonding Formation of covalent bond 1) Covalent bond is the electrostatic force of attraction that two neighbouring nuclei have for a localised pair of electrons shared between them. 2) Covalent bond is formed without transferring electrons, instead, the atoms share their valence electron(s) to achieve duplet/octet electronic configuration. 3) The shared pair of electrons constitutes the covalent bond. 4) Using dot-and-cross diagram to represent the formation of covalent bond: Formation of hydrogen Formation of oxygen gas, gas, H₂ O₂ Single bond 1) Single bond is formed when one pair of electrons is shared between two atoms. 2) Examples of compounds with single bonds: Double bond 1) Double bond is formed when two pairs of electrons are shared between two atoms. 2) Examples of compounds with double bonds: Triple bond 1) Triple bond is formed when three pairs of electrons are shared between two atoms. 2) Examples of compounds with triple bonds: Lone pair and bond pair of electrons 1) The pair of electrons used in covalent bonding is called the bond pair while the pair of electrons not used in covalent bonding is called the lone pair. Octet-deficient and expanded octet species 1) In general, atoms tend to share their electrons to a achieve a duplet/octet electronic configuration - the octet rule. 2) i. In octet-deficient species, the central atom has less than eight electrons. ii. Some examples are boron trifluoride, BF3 and nitrogen monoxide, NO. 4) i. In expanded octet species, the central atom has more than eight electrons. ii. An example is phosphorus(V) chloride, PCl5. iii. This is possible only for Period 3 elements and beyond, this is because starting from Period 3, the atoms have empty d orbitals in the third energy level to accommodate more than eight electrons. Octet deficient Expanded octet Co-ordinate bond (dative covalent bond) 1) A co-ordinate bond is formed when one atom provides both the electrons needed for a covalent bond. 2) Conditions of forming a co-ordinate bond: i. one atom has a lone pair of electrons. ii. another atom has an unfilled orbital to accept the lone pair, in other words, an electron-deficient species. 3) Once the bond is formed, it is identical to the other covalent bonds. It does not matter where the electrons come from. 4) In a displayed formula, a co-ordinate bond is represented by an arrow, the head of the arrow points away from the lone pair which forms the bond. 5) An example is the reaction between ammonia and hydrogen chloride. In this reaction, ammonium ion is formed by the transfer of hydrogen ion(an octet deficient species) from hydrogen chloride to the lone pair of electrons in the ammonia molecule. 6) Another example is aluminium chloride. At high temperature, it exists as AlCl3. At low temperature(around 180-190°C), it exists as Al2Cl6, a dimer(two molecules joined together). This is possible because lone pairs of electrons from the chlorine atom form co-ordinate bonds with the aluminium atom. Tips to draw dot-and-cross diagram for covalent molecules 1) Identify the central atom and terminal atom(s). For example, in ammonia, the nitrogen is the central atom while the hydrogens are the terminal atoms. 2) During the sharing of electrons, the terminal atoms must attain octet configuration(or duplet for hydrogen) but not necessarily for the central atom. 3) i. If the central atom is from Period 2 of the Periodic Table, the total number of electrons surrounding it cannot exceed eight(but can less than eight). ii. If the central atom is from Period 3 and beyond, the total number of electrons surrounding it can exceed eight. 4) i. For polyatomic anions, the negative charge will be distributed among the most electronegative atom(s). This is to decrease the charge density on a particular atom and to stabilise the ion. ii. For polyatomic cation, the positive charge will be distributed among the less electronegative atom(s). The reason is same as above. 5) If the terminal atom already has octet configuration(for example, Cl⁻), it will contribute two electrons to the central atom to form a co-ordinate bond. 4.3 Shapes of Molecules Valence shell electron pair repulsion(VSEPR) theory 1) All electrons are negatively-charged, so they will repel each other when they are close together. 2) So, a pair of electrons in the bonds surrounding the central atom in a molecule will repel the other electron pairs. This repulsion forces the pairs of electrons apart until the repulsive forces are minimised. 3) The amount of repulsion is as follow: 4) General steps to determine the shape of a molecule: i. determine the number of valence electrons in the central atom. ii. find the total number of electrons surrounding the central atom by adding the number of shared electrons to it. (Dot-and-cross diagram might be necessary) iii. find the number of electron pairs by dividing the total number of electrons by two. iv. determine how many pairs is/are bond pairs and lone pairs. (A double bond or triple bond is counted as one bond pair) v. refer to the table to obtain the shape of the molecule Effect of lone pair on bond angle 1) For methane, ammonia and water, the electron pair geometries are tetrahedral. However, the molecular geometries are different. 2) In methane, all the bonds are identical, repulsion between the bonds is the same. Thus, methane has a perfect tetrahedral structure with bond angle 109.5°. 3) In ammonia, the repulsion between the lone pair and the bond pairs is stronger than in methane. This forces the bond angle to decrease slightly to 107°. 4) In water, there are two lone pairs and thus the repulsion is the greatest, the two bond pairs are pushed closer to one another and the bond angle is reduced to 104.5°. Effect of electronegativity on bond angle 1) Water and hydrogen sulfide have the same general shape with the same number of bond pairs and lone pairs. However, their bond angles are different. 2) This is because oxygen has a higher electronegativity than sulphur. The bond pairs of electrons are closer to the oxygen atom compared to the sulfur atom. 3) This results in greater repulsion in the O-H bonds than in the S-H bonds. Therefore, the bond angle increases from 92.5° to 104.5°. Sigma(σ) bond and pi(π) bond 1) A sigma bond is formed by orbitals from two atoms overlapping end-to-end. 2) In a sigma bond, the electron density is concentrated between the two nuclei. 3) A pi bond is formed by the p orbitals from two atoms overlapping sideways. 4) In a pi bond, there are two regions of high electron density alongside the nuclei. 5) A pi bond is weaker than a sigma bond because the overlapping of charge clouds is less than in a sigma bond. 6) In covalent molecules, single bonds are sigma bonds(σ), a double bond consists of one sigma bond and one pi bond(1σ, 1π), and a triple bond consists of one sigma bond and two pi bonds(1σ, 2π). Hybridisation 1) Hybridisation is the mixing of atomic orbitals to produce a new set of hybrid orbitals of equivalent energies. This is a theory used to account for the discrepancies in explaining the shapes of covalent molecules. 2) There is a problem with simple view on methane, CH4. Methane has two unpaired electrons only in its outer shell to share with the hydrogen atoms, but the formula of methane is not CH2. 3) The concept of hybridisation is used to account for this discrepancy. 4) General steps in hybridisation: i. promotion of electron. ii. mixing of orbitals to produce a new set of hybrid orbitals of equivalent energies(sp, sp² or sp³ hybrid orbitals) iii. forming of a new molecular orbital. sp³ hybridisation 1) An example of compound which undergoes sp³ hybridisation is methane, CH4. C: 2) The carbon atom uses some energy to promote one of its electron from 2s to empty 2p orbital so that there are four unpaired electrons for covalent bonding. C* : 3) The carbon now is said to be in an excited state(C*). 4) The orbitals then 'mix' or hybridise to produce four hybrid orbitals of equivalent energies. The new orbitals are called sp³ hybrid orbitals because they are made from one s orbital and three p orbitals. 5) Each hybrid orbital has one big lobe and one small lobe. They rearrange themselves so that they are as far as possible to form a tetrahedral geometry. The hybrid orbitals are 109.5° apart. 6) The s orbitals from the hydrogen atoms then overlap with the four hybrid orbitals to form four sigma bonds because the overlapping is end-to-end. All the bonds are identical. 7) Another example is ethane, C2H6. The two carbon atoms undergo sp³ hybridisation to form four hybrid orbitals. The two carbon atoms are bonded by overlapping one of their hybrid orbitals. The remaining ones then overlap with the s orbitals of the hydrogen atoms. 8) The bond angle is approximately 109.5°. This is an approximation because all the bonds are not identical. sp² hybridisation 1) An example of compound which undergoes sp² hybridisation is ethene, C2H4. 2) The same thing happens as in sp³ hybridisation, except that this time the carbon atoms 'mix' or hybridise three of the four orbitals only because the carbon atom is bonding with three other atoms only. 3) This produces three sp² hybrid orbitals because they are made from one s orbital and two p orbitals. Another p orbital remains unchanged and it is perpendicular to the plane containing the hybrid orbitals. 4) The hybrid orbitals rearrange themselves so that they are as far as possible, that is, a trigonal planar arrangement, the hybrid orbitals are 120° apart. 5) The hybrid orbitals then overlap with s orbitals from the hydrogen atoms and another hybrid orbital from the other carbon atom to form five sigma bonds. The remaining p orbitals overlap sideways to form a pi bond. A double bond is formed between the two carbon atoms. 6) Another example is boron trichloride, BCl3. The boron atom undergoes sp² hybridisation to produce three sp² hybrid orbitals. The hybrid orbitals rearrange themselves to form a trigonal planar geometry. The p orbitals from chlorine atoms then overlap with the hybrid orbitals to form three sigma bonds. sp hybridisation 1) An example of compound which undergoes sp hybridisation is ethyne, C2H2. 2) The same thing happens as in sp³ and sp² hybridisation, except that this time the carbon atoms 'mix' or hybridise two of the four orbitals only because the carbon atom is bonding with two other atoms only. 3) This produces two sp hybrid orbitals because they are made from one s orbital and one p orbital. The other two p orbitals remain unchanged and they are perpendicular to each other and to the two hybrid orbitals. 4) The hybrid orbitals rearrange themselves so that they are as far as possible, that is, a linear arrangement, the hybrid orbitals are 180° apart. 5) The hybrid orbitals overlap with the s orbitals from the hydrogen atoms and to the hybrid orbital from the other carbon atom to form three sigma bonds. The remaining p orbitals overlap sideways to form two pi bonds. A triple bond is formed between the two carbon atoms. Example of covalent molecule with multiple hybridisations 1) In carbon dioxide, CO2, the carbon atom undergoes sp hybridisation while the oxygen atoms undergo sp² hybridisation. The overlapping of the hybrid and p orbitals are shown in the diagram. 4.4 Electronegativity, Bond Polarity, Bond Length and Bond Energy Electronegativity 1) Electronegativity is the ability of an atom which is covalently bonded to the other atom to attract the bond pair of electrons towards itself. 2) The more electronegative an atom is, the higher the tendency of that atom to attract the bond pair of electrons towards itself. 3) The Pauling scale is commonly used to quantify the value of electronegativity of a particular element. 4) Fluorine is the most electronegative element because of its small size, followed by oxygen and nitrogen. Trends of electronegativity values in the Periodic Table 1) i. The value of electronegativity increases across a Period(from left to right). ii. This is because the number of proton increases across a Period. Therefore the amount of positive charge in the nucleus also increases. iii. The shielding effect by inner electrons remains constant. iv. Therefore the attraction towards the bond pair of electrons increases, making it more electronegative. 2) i. The value of electronegativity decreases down a Group. ii. This is because the size of the atoms increases down a Group. Therefore the distance between the nucleus and the bond pair of electrons also increases. iii. The shielding effect by inner electrons is also greater. iv. Therefore the attraction towards the bond pair of electrons decreases, making it less electronegative. Bond Polarity 1) i. When two covalently-bonded atoms have the same electronegativity, the electron cloud is evenly distributed between the two atoms. ii. The bond is described as a 'pure' covalent bond or non-polar bond. iii. Some examples are H2, Cl2 and Br2. 2) i. However, when an atom is more electronegative than the other, the more electronegative atom will attract the bond pair of electrons more towards itself. The electron cloud is not evenly distributed or distorted. ii. The more electronegative end acquires a partial negative charge while the less electronegative end acquires a partial positive charge. iii. The bond is said to be polarised, or, a polar bond. iv. Some examples of compound which contain polar bond(s) are HCl and CH4. v. The molecule is described as 'covalent with some ionic character'. 3) i. When the electronegativity difference between the two atoms is very great, the less electronegative atom will lose its electrons and the more electronegative atom will gain the electrons. ii. An ionic bond will be formed. Polar and non-polar molecules 1) A molecule is polar, and thus, has a dipole moment(μ ≠ 0) if: i. the bonds are polarised ii. the dipole of the bonds do not cancel out each other(in other words, it is asymmetrical) 2) The dipole moment, μ is the product of charges and the distance between the centre of the charges. 3) Examples: 4) A liquid containing polar molecules can be deflected by a charged rod brought near to it. This is because there is a positive end and a negative end in polar molecules. So, irregardless of the charge on the rod, one end of the molecule will always be attracted towards the charged rod. 5) When polar molecules are placed in an electric field, the positive end of the molecule will face towards the negative terminal while the negative end of the molecule will face towards the positive terminal. 6) Generally, polar molecules are more reactive than non-polar molecules because many chemical reactions are started by a reagent attacking an electrically- charged end of the polar molecule. An example is CO is more reactive than N2 although both of them have triple bonds because CO is polar while N2 is not. Bond length and bond energy 1) Bond length is the distance between two nuclei of two atoms joined by a covalent bond. 2) Bond energy is the energy needed to break one mole of covalent bonds between two atoms in the gaseous state. A-B(g) → A(g) + B(g) ∆H° = +x J 4) i. The shorter the bond length, the stronger is the bond. ii. The greater the bond energy, the stronger is the bond. iii. Hence, the shorter and bond length, the greater the bond energy. 4.5 Intermolecular Forces Intermolecular and intramolecular forces 1) i. Intramolecular force is the force of attraction that hold individual atoms together in a molecule. ii. Commonly referred to as covalent bonds. 2) Intermolecular force is the force of attraction between one molecule and the neighbouring molecule. 3) There are several types of intermolecular forces: Intermolecular forces van der Waals' forces Hydrogen bond - Permanent dipole-dipole forces (Keesom forces) - Temporary dipole-induced dipole forces (Dispersion forces or London forces) 4) Intermolecular forces are much weaker than intramolecular forces. 5) Intermolecular forces are responsible for the melting and boiling points of substances, as well as their physical states. Permanent dipole-dipole forces(Keesom forces) Permanent dipole-dipole forces 1) Polar molecules have a negative end and a positive end. The oppositely-charged ends will attract one another. The forces of attraction is called permanent dipole- dipole forces. 2) Only polar molecules will experience permanent dipole-dipole forces. Temporary dipole-induced dipole forces(Dispersion forces or London forces) 1) Electrons are mobile, and constantly revolving around the nucleus. Even in non- polar molecules, there is a high possibility that at any given instant, the electron density is higher on one side than the other. 2) This will result in the formation of temporary dipole(or instantaneous dipole) because it only lasts for just a tiny fraction of time. 3) In the next instance, the distribution of electron density will change and the molecule has a new temporary dipole. 4) The temporary dipole set up can distort the electron charge clouds of the neighbouring molecules, giving rise to induced dipoles. 5) The forces of attraction between temporary dipoles and induced dipoles give rise to dispersion forces. 6) All molecules(polar and non-polar) will experience dispersion forces. For polar molecules, permanent dipole-dipole forces exist in addition with dispersion forces. Factors affecting strength of dispersion forces 1) Number of electrons in the molecule(Size of molecule) - As the number of electrons increases, the size of the molecule also increase. - This causes the attraction between the nucleus and the electrons to get weaker, the electrons become progressively easier to be distorted. - This causes more temporary dipoles to be set up and the dispersion forces get stronger. - The boiling point of halogens increases going down the Group. The size of the molecules increases and thus the van der Waals' forces become stronger. 2) Number of contact points between the molecules(Surface area) - For two molecules of the same number of electrons, the one which has a larger surface area has more contact points between the molecules. - The more contact points between molecules, more temporary dipoles can be set up and the dispersion forces become stronger. - Butane is a long chain molecule while 2-methypropane is more spherical (more compact). This causes the number of contact point between butane molecules to be more than in 2-methypropane. - Therefore the van der Waals' forces of attraction in butane is stronger than in 2-methypropane. This causes the boiling point of butane to be higher. Hydrogen bonding 1) Hydrogen bond is the electrostatic force of attraction between a hydrogen atom (which is covalently bonded to a small and highly electronegative atom) and the lone pair of electrons of another small and highly electronegative atom. 2) The 'small and highly electronegative atoms' are fluorine, oxygen and nitrogen. 3) The conditions necessary for forming hydrogen bonds: i. a hydrogen atom bonded directly to a small and electronegative atom, causing it to acquire a significant amount of partial positive charge. ii. a lone pair of electrons from the small and electronegative atom. 4) The attraction between the hydrogen atom and the lone pair of electrons constitutes the hydrogen bond. 5) The evidence of hydrogen bonding: - H2O, HF and NH3 have exceptionally high boiling points compared to other compounds with greater number of electrons. - This suggests that there must be an additional intermolecular force other than van der Waals' forces exists between them. - This additional intermolecular force is called the hydrogen bond. 6) H2O has a higher boiling point compared to NH3 and HF. This is because a H2O molecule can form, on average four hydrogen bonds. In NH3, the number of hydrogen bonds is restricted by the one lone pair of electrons in the NH3 molecule. In HF, it is restricted by the number of hydrogen atoms. The peculiar properties of water 1) Solid ice is less dense than liquid water - Most solids are denser than their liquids, this is because in the solid state, the molecules are packed closer and occupies smaller volume. - However, solid ice is less dense than liquid water. - This is because in the solid state, the water molecules are bonded to each other through hydrogen bonding to form a giant 3-dimensional tetrahedral structure. This structure has a lot of empty spaces in between the water molecules. - When heat is supplied, some of the hydrogen bond breaks down and the rigid structure collapses, filling up the spaces in between them, thus decreasing the volume occupied and increasing its density. 2) Solubility of substances in water - Most covalent compounds are not soluble water while ionic compounds are. - However, some covalent compounds are soluble in water because they can form hydrogen bonds with water, some examples are ammonia and alcohols 3) High boiling point of water - The boiling point of water is exceptionally high compared to other Group 16 hydrides(see graph at page 21). - This is because water molecules are capable of forming hydrogen bonds between themselves. This additional type of intermolecular forces increases its boiling point significantly. 4) High surface tension and viscosity - Water has a high viscosity. This is because hydrogen bonding reduces the ability of water to slide over each other, making it more viscous. - Water has a high surface tension. This is because hydrogen bonds also exert a significant downward force at the surface of liquid. 4.6 Metallic Bonding Metallic bonds 1) Metallic bond is the electrostatic force of attraction between the delocalised electrons and the positive metal ions. 2) Take sodium as an example, in the solid state, sodium atoms are packed so closely together that the 3s orbitals(containing an electron) overlap with one another to form a giant molecular orbital. 3) The 3s electrons is then free to move throughout the lattice structure, it is no longer bound to any sodium atom. The electrons are said to be delocalised. The delocalised electrons are sometimes referred to as 'sea of delocalised electrons'. 4) The attraction between the delocalised electrons and the positive metal ions constitutes the metallic bond. Factors affecting the strength of metallic bonds 1) Number of electrons in the sea of delocalised electrons - The more the number of electrons delocalised, the stronger the metallic bond. - For example, the melting point of magnesium is higher than sodium. - This is because one magnesium atom can donate two electrons from its outer shell into the sea of delocalised electrons while sodium can only donate one. - There is twice amount of delocalised electrons in magnesium than in sodium. So, the metallic bond in magnesium is stronger due to the stronger attraction. 2) Size of the metal - The smaller the size of the metal, the stronger the metallic bond. - For example, the melting point of magnesium is lower than beryllium although both of them have two electrons delocalised per atom. - This is because the size of magnesium is bigger, causing the attraction between the delocalised electrons and the positive nucleus to be weaker. The shielding effect in magnesium is also more than in beryllium. - So, the metallic bond in magnesium is weaker than in beryllium. 4.7 Bonding and Physical Properties of Substances Physical state at room temperature 1) Ionic compounds - Ionic compounds are solids at room temperature, this is because the ionic bond holding the oppositely-charged ions is very strong, a lot of energy is required to overcome the strong forces of attraction. 2) Covalent compounds - Most simple covalent molecules(like water and ammonia) are liquids or gases at room temperature. This is because the intermolecular force between the molecules is weak, little energy is required to overcome it. - Some simple covalent molecules(like iodine) are solids because the intermolecular force between them is strong enough. - For giant covalent structures(like diamond and silicon dioxide), they are solids at room temperature. This is because the covalent bonds holding the atoms are very strong, a lot of energy is required to overcome it. 3) Metals - Metals(except mercury) are solids at room temperature. This is because the metallic bond holding the metal ions is very strong, a lot of energy is required to overcome it. Electrical conductivity 1) Ionic compounds - Ionic compounds do not conduct electricity in the solid state, this is because the ions are not free to move. - In the molten or aqueous state, it conducts electricity because the ions are free to move(mobile ions are present). 2) Covalent compounds - Simple covalent molecules do not conduct electricity because the ions are not free to move. - Some giant covalent structures(like graphite) are able to conduct electricity because the electrons are free to move(delocalised). (See also Chapter 5) 3) Metals - Metals conduct electricity because the delocalised electrons are free to move. Solubility 1) Ionic compounds - Most ionic compounds are soluble in water(polar solvents) and insoluble in non-polar solvents. - This is because in non-polar solvents, the molecules are held together by weak intermolecular forces. The ionic bonds in ionic compounds are much stronger and the energy needed to break them is high. If ion-solvent bonds are formed, the energy released is not enough to compensate the energy absorbed, making the entire structure unstable. [ Note: To form ion-solvent bonds, the ion-ion(ionic bond) and solvent-solvent (van der Waal's forces) bonds must be broken first. ] - While in polar molecules, there is an attraction between polar molecules and the ions. So, the energy released when ion-solvent bonds are formed is enough to compensate the energy needed to break the strong ionic bonds. 2) Covalent compounds - This is because when a covalent molecule is dissolved in a non-polar solvent, the molecule-solvent attraction is strong enough to compensate the energy needed to break the weak van der Waal's forces between covalent molecules. - While when dissolved in polar solvents, the energy needed to break the attractions in polar solvents is too high. The energy released when molecule- solvent attraction set up is not enough to compensate it. This makes the structure to gain energy overall, making it less stable. - Some covalent compounds react with water rather than dissolving in it. For example, hydrogen chloride reacts with water to form hydrogen ions and chloride ions, and the ions are soluble. 3) Metals - Metals do not dissolve in polar and non-polar solvents. - However, some metals like sodium and calcium can react with water. CHAPTER 5: States of Matter 5.1 The Gaseous State 5.2 The Liquid State 5.3 The Solid State 5.4 Ceramics 5.5 Conserving Materials Learning outcomes: (a) state the basic assumptions of the kinetic theory as applied to an ideal gas. (b) explain qualitatively in terms of intermolecular forces and molecular size: (i) the conditions necessary for a gas to approach ideal behaviour. (ii) the limitations of ideality at very high pressures and very low temperatures. (c) state and use the general gas equation pV = nRT in calculations, including the determination of Mr. (d) describe, using a kinetic-molecular model: the liquid state, melting, vaporisation, vapour pressure. (e) describe, in simple terms, the lattice structure of a crystalline solid which is: (i) ionic, as in sodium chloride, magnesium oxide. (ii) simple molecular, as in iodine. (iii) giant molecular, as in silicon(IV) oxide and the graphite and diamond allotropes of carbon. (iv) hydrogen-bonded, as in ice. (v) metallic, as in copper. [the concept of the ‘unit cell’ is not required] (f) explain the strength, high melting point and electrical insulating properties of ceramics in terms of their giant molecular structure. (g) relate the uses of ceramics, based on magnesium oxide, aluminium oxide and silicon(IV) oxide, to their properties (suitable examples include furnace linings, electrical insulators, glass, crockery). (h) discuss the finite nature of materials as a resource and the importance of recycling processes. (i) outline the importance of hydrogen bonding to the physical properties of substances, including ice and water (for example, boiling and melting points, viscosity and surface tension). (j) suggest from quoted physical data the type of structure and bonding present in a substance. 5.1 The Gaseous State Kinetic theory of gases 1) Assumptions made in the kinetic theory of gases as applied to ideal gases: - The gas particles have zero intermolecular forces between them. - The gas particles behave as point particles which have negligible volume. - The gas particles are in constant random motion, colliding with each other and the wall of the container frequently. - All collisions between the gas particles are perfectly elastic. - Pressure is due to the collision of gas particles with the wall of container. 2) In the gaseous state, the particles can move freely and are far apart. 3) A gas has no fixed shape and volume, it takes the shape of container and always fills it. The volume of a gas depends on its pressure, temperature and number of moles. Ideal gas and real gas 1) A real gas is most like an ideal gas at: i. low pressures - At low pressures, the distance between gas particles is large and the volume of the gas particles is negligible(very small compared to the volume of the container). - Intermolecular forces are also negligible at low pressures. ii. high temperatures(well above its boiling point) - At high temperatures, the gas particles have negligible intermolecular forces between them because they have sufficient energy to overcome it. 2) However, an ideal gas does not exist, because: i. there are intermolecular forces between the gas particles. ii. the volume occupied by the gas particles is not zero. 3) A real gas shows biggest deviation from an ideal gas at: i. high pressures - At high pressures, the gas particles are packed close together, thus the volume occupied by the gas particles is not negligible. - The intermolecular forces between them is also not negligible. ii. low temperature - At low temperatures, the gas particles are packed close together, thus the volume occupied by the gas particles is not negligible. - The intermolecular forces between them is also not negligible because they do not have sufficient energy to overcome it. The general gas equation Side note 1) 1 atm = 1.01 x 10⁵ Pa 2) (x) °C = (x + 273) K 3) 1 m³ = 1000 dm³ = 1000000 cm³ 4) At s.t.p: - p = 1.01 x 10⁵ Pa - T = 273 K ← Boyle's Law ← Charles' Law 5.2 The Liquid State Kinetic theory of liquids 1) In liquids, the particles are packed quite closely together but in a fairly random arrangement(gaps are present between them). So, the particles have limited movements. 2) A liquid has fixed volume but do not have fixed shape. 3) Intermolecular forces are present between the particles, its strength is stronger than in the gaseous state but weaker than in the solid state. Melting and freezing 1) When a solid is heated, the energy is absorbed by the particles and they vibrate about their fixed positions more vigorously. 2) Then, a point is reached where the particles have energy high enough to overcome the attractive forces that hold them in fixed positions. They break away from their fixed positions and move freely. 3) The solid then becomes a liquid, this process is called melting. The temperature at which this process happens is called the melting point. 4) In freezing, the reverse happens. The liquid particles lose energy until they do not have enough energy to move freely. They are held together in fixed positions again. The liquid solidifies. Boiling(vapourisation) and condensation 1) When a liquid is heated, the vapour pressure of the liquid increases until eventually it is equal to the atmospheric pressure. Bubbles of vapour will form in the body of liquid. 2) The bubbles then rise to the surface of the liquid, burst open and escape into the atmosphere as a gas. The liquid boils. 3) This happens because when a liquid is heated, the particles absorb energy until it is sufficient to overcome the forces of attraction between them. The particles break away from the fairly close arrangement of the liquid and boils. 4) This process is called boiling. The temperature at which this process happens is called the boiling point. i. Boiling point depends on external pressure. If a liquid boils under a pressure lower than the atmospheric pressure(1 atm), it will boil faster. ii. Conversely, if it boils at a pressure higher than the atmospheric pressure, it will boil slower. 5) In condensation, the reverse happens, the particles lose energy and experience increasing attractive force. They move slower and become closer together when temperature is sufficiently low. The gas liquefies. Evaporation and vapour pressure 1) The energy distribution of particles in the liquid state follows a shape similar to the normal distribution. 2) The average energy is governed by the temperature. The higher the temperature, the higher the average energy. 3) Some particles have energy higher than the average while some have lower. The more energetic particles at the surface of the liquid can be moving fast enough and eventually overcome the attractive forces and escape into the atmosphere. They evaporate to form vapour(Vapour is the gas form of a particle below its boiling point). 4) Unlike boiling, evaporation only takes place on the surface of the liquid. 5) In an open container, the liquid will evaporate until none is left. 6) However, a different thing happens when the liquid is evaporated in a closed container. 7) At first, liquid particles with higher energy escape from the surface of the liquid to become vapour. The vapour particles will collide with the wall of container. The collisions exert a pressure called vapour pressure. 8) As more and more particles escape, the vapour particles become close together. Eventually the particles with lower energy will not be able to overcome the attractive forces between them. The vapour begins to condense and return to liquid. 9) Eventually the vapour particles return to liquid at the same rate as the liquid particles evaporate to form vapour. An equilibrium is reached. At this equilibrium, the concentration of liquid particles and vapour particles remains constant. 10) In this situation, the vapour pressure is maximum and is called the saturated vapour pressure. 11) Vapour pressure will increase when temperature is increased. At higher temperature, more liquid can undergo evaporation. The vapour particles are more energetic and collide with the wall of container harder and more frequently. This causes the pressure exerted to be higher. 5.3 The Solid State Kinetic theory of solids 1) In solids, very strong forces of attraction hold the particles in fixed positions and close to each other. Hence, particles in the solid state can only rotate and cannot translate. 2) A solid has fixed shape and volume. 3) The structure of solids can be crystalline or amorphous. i. In crystalline solids, the particles are arranged in a regular and orderly pattern called a lattice structure. ii. In amorphous solids, the particles do not have a regular and orderly arrangement. Examples are rubber and plastic. Ionic lattices 1) An example of ionic lattice is sodium chloride, NaCl, where strong ionic bonds hold the Na⁺ and Cl⁻ ions rigidly in place in the solid lattice. 2) Sodium chloride has a face-centred cubic structure. In this structure, each ion is surrounded by six other oppositely-charged ions. So, sodium chloride is described as 6:6-co-ordinated. 3) Some other examples with face-centred cubic structure are magnesium chloride and magnesium oxide. 4) Another example is caesium chloride, CsCl, which has a body-centred cubic structure. In this structure, each ion is surrounded by eight other oppositely- charged ions. So, caesium chloride is described as 8:8-co-ordinated. 5) This is just like layers of oppositely-charged ions stacking on each other. 6) Properties of ionic compounds:(For more detailed discussions, see Chapter 4) i. They are hard - This is because the ionic bonds within are very strong. A lot of energy is required to split them apart. ii. They have high melting and boiling points - This is because the ionic bonds within are very strong. A lot of energy is required to break the bonds. iii. They are brittle - This is because when a stress is applied, the layers of ions may be displaced by the force. - This brings ions with the same charge together and the repulsion between them causes the crystal to split. iv. They conduct electricity in the molten or aqueous state - This is because in the molten or aqueous state, free moving ions are present to carry the current. v. Many of them are soluble in water(polar solvents) but insoluble in non-polar solvents - This is because when ion-polar solvent attraction is set up, the energy released is enough to compensate the energy needed to break the ionic bonds. Simple molecular lattices 1) An example of simple molecular lattice is iodine, I2. Solid iodine has a face- centred cubic structure. 2) Weak van der Waals' forces of attraction hold the individual iodine molecules together. 3) Properties of simple molecular solids:(For more detailed discussions, see Chapter 4) i. They do not conduct electricity in any state - This is because there are no free electrons to carry the current. ii. They have low melting and boiling points - This is because the van der Waals' forces of attraction between them is weak, little energy is required to overcome it. iii. They are soluble in non-polar solvents but insoluble in water(polar solvents) - This is because the molecule-non-polar solvent attraction set up is enough to compensate the energy needed to break the weak van der Waals' forces of attraction between the simple molecules. Giant covalent lattices 1) In giant covalent structures, all the atoms are covalently-bonded to each other, linking the whole structure. An example of giant covalent lattice is silicon(IV) oxide or silicon dioxide, SiO2. 2) In silicon dioxide, each silicon atom is covalently-bonded to four oxygen atoms in a tetrahedral arrangement while each oxygen atom is covalently- bonded to two silicon atoms in a V-shaped(bent) arrangement. 3) Properties of silicon dioxide: i. Very high melting and boiling points - The covalent bonds holding the atoms are very strong, a lot of energy is required to break these strong bonds. ii. Very hard - This is because of the need to break the strong covalent bonds iii. Does not conduct electricity - This is because there are no delocalised electrons to move around and carry the current. All the electrons are held tightly to the atoms. 4) Another two examples are diamond and graphite. Both of them are allotropes of carbon. 5) Allotropes are different crystalline or molecular forms of the same element in the same physical state, having different atomic arrangement. 6) In diamond, all the carbon atoms undergo sp³ hybridisation and covalently-bonded to four other carbon atoms in a tetrahedral arrangement. This network of carbon extends throughout the whole structure. 7) The properties of diamond are same as the ones in silicon dioxide. 8) In graphite, the carbon atoms undergo sp² hybridisation. Each carbon atom is covalently-bonded to three other carbon atoms in a trigonal planar arrangement within each layer, forming hexagons. 9) The unhybridised p orbital in each planar layers overlap sideways to form a giant molecular orbital above and below each layer. The electrons are then delocalised, free to move between the layers of carbon. However, the electrons cannot move from one layer to another. 10) The distance between carbon atoms between the layers is greater than the distance between carbon atoms in each layer. 11) Properties of graphite: i. Very high melting and boiling points - This is because the covalent bonds holding the carbon atoms with each layer is very strong, a lot of energy is required to overcome the strong forces of attraction. - In fact, it has a higher melting point than diamond because of the additional attraction between the delocalised electrons and the nucleus. Dispersion forces are also present, this is because the delocalised electrons can set up a temporary dipole easily. ii. Soft and slippery - Layers of carbons can slide over easily, this is because the intermolecular forces between the layers are very weak. - Because of this, graphite is often used as pencil 'leads' and lubricants. iii. Conducts electricity - It can conduct electricity in the direction parallel to each layer but not perpendicular to it. - This is because the delocalised electrons are free to move between the layers but not from one layer to another. Metallic lattices 1) An example of metallic lattice is copper, Cu. In copper, each copper atom is surrounded by twelve other copper atoms. Therefore, copper is a 12-co-ordinated metal. Three other copper Six copper atoms atoms surround it up surround it in a layer and down. 3) The copper atoms are bonded to each other through strong metallic bonding. (For more information about metallic bonding, refer Chapter 4) 4) Another example is sodium. In sodium, each sodium atom is surrounded by eight other sodium atoms only. Therefore, sodium is a 8-co-ordinated metal. In the first diagram, no atoms are touching each other in a layer. 5) Properties of metals: i. High melting and boiling points - This is because the metallic bonds are very strong, a lot of energy is required to break these strong bonds. ii. Conducts heat and electricity - This is because delocalised electrons are present and free to move. iii. Ductile and malleable - Metals are ductile(can be pulled into wires) and malleable(can be hammered into different shapes). This is because the layers of metals can slide over each other, the broken metallic bonds can be immediately replaced by the new ones. 5.4 Ceramics Ceramics 1) A ceramic is an inorganic non-metallic solid which is prepared by heating a substance or mixture of substances to a high temperature. 2) Ceramics often contain silicon dioxide, magnesium oxide and aluminium oxide. This gives ceramics their giant covalent or ionic structures. 3) Properties of ceramics: i. Very high melting and boiling points - This is because most ceramics contain giant covalent or ionic structures. The covalent or ionic bonds holding them together is very strong, a lot of energy is required to overcome it. ii. Does not conduct electricity or heat - This is because there are no delocalised electrons or free moving ions present. Therefore most of them are electrical insulators. iii. Chemically unreactive - This is because all the electrons are held firmly in strong covalent bonds and not available for a reaction. iv. Very hard - This is because the ionic or covalent bonds holding them is very strong. Uses of ceramics 1) Ceramics containing magnesium oxide are used: - as electrical insulators in industrial electrical cables. - as a refractory in furnace linings because it has a high melting point. - in fire-resistant wall boards. 2) Ceramics containing aluminium oxide are used: - as a refractory in furnace linings because it has a high melting point. - as an abrasive for grinding hard materials because they do not conduct heat or melt when heat is given off during grinding. - in transparent aluminium oxide-scandium windows. - in high temperature and high voltage electrical insulators. - in the replacement of artificial hip joints. 3) Ceramics containing silicon dioxide are used: - as a refractory in furnace linings because it has a high melting point. - as a abrasive, for example in sandpaper. - in the manufacture of glass. 5.5 Conserving Materials Why conserve materials? 1) There is only a limited supply of metal ores in the Earth. Therefore metals are finite resources. They do not get replaced once they are used up. Ways to conserve materials 1) One way to conserve materials is through recycling. 2) Advantages of recycling: i. Recycling saves new resources. ii. Recycling reduces the amount of waste materials to be disposed off. iii. Recycling saves energy because less energy is needed to recycle metals than to extract it from their ores. iv. Recycling protects the environment because it reduces pollution associated with product manufacture, disposal and littering. For example, the process of extracting metals produces gases which can lead to acid rain. 3) Two metals that can be recycled easily are copper and aluminium. Glass can also be recycled easily. 4) Another way to conserve materials is using renewable resources. Examples are like water, wood, sunlight, wind and etcetera. CHAPTER 6: Chemical Energetics 6.1 Enthalpy Changes 6.2 Standard Enthalpy Changes 6.3 Hess' Law 6.4 Bond Energy Learning outcomes: (a) explain that some chemical reactions are accompanied by energy changes, principally in the form of heat energy; the energy changes can be exothermic (∆H, negative) or endothermic. (b) explain and use the terms: (i) enthalpy change of reaction and standard conditions, with particular reference to: formation, combustion, hydration, solution, neutralisation, atomisation. (ii) bond energy (∆H positive, i.e. bond breaking). (c) calculate enthalpy changes from appropriate experimental results, including the use of the relationship enthalpy change, ∆H = –mc∆T. (d) apply Hess’ Law to construct simple energy cycles, and carry out calculations involving such cycles and relevant energy terms, with particular reference to: (i) determining enthalpy changes that cannot be found by direct experiment, e.g. an enthalpy change of formation from enthalpy changes of combustion. (ii) average bond energies. (e) construct and interpret a reaction pathway diagram, in terms of the enthalpy change of the reaction and of the activation energy. 6.1 Enthalpy Changes Exothermic and endothermic reactions 1) Most chemical reactions are accompanied by energy changes. Some absorbs energy while some releases it. 2) An exothermic reaction is a reaction that releases energy to the surroundings. Therefore the product contain less energy with respect to the reactants. The energy is released as heat energy, so the surroundings warm up. 3) An endothermic reaction is a reaction that absorbs energy from the surroundings. Therefore the products contain more energy with respect to the reactants. The energy is absorbed as heat energy, so the surroundings cool down. 4) An energy level diagram shows the relative energies of the products and reactants. The higher its energy, the higher its position. An exothermic reaction An endothermic reaction 5) i. Examples of exothermic reactions are neutralisation, combustion, dissolving anhydrous salts and the reaction between acids and metal carbonates. ii. Examples of endothermic reactions are atomisation, photosynthesis, dissolving crystalline salts and thermal decomposition. Enthalpy changes 1) Enthalpy, H is an indication of the total energy of a substance and it cannot be measured directly. However, enthalpy change, ∆H is measurable. 2) Enthalpy change of a reaction, ∆H is the heat change(heat is absorbed or evolved) when a chemical reaction takes place. 3) i. In exothermic reactions, the enthalpy change is always negative.(∆H < 0) ii. In endothermic reactions, the enthalpy change is always positive.(∆H > 0) Energetic stability of a system 1) Exothermic reactions are more energetically favourable than endothermic ones because a system with lower energy content is more stable. 2) Thus, the more negative the ∆H value, the more stable the system is. 3) For example, oxygen,O2 is energetically more stable than ozone, O3. Their relative positions on the energy level diagram are as follow: 4) But ozone does not convert to oxygen immediately in atmosphere, this is because ozone is kinetically stable although it is energetically unstable with respect to the products. The activation energy is not sufficient for the reaction to take place. Measuring enthalpy changes 1) Enthalpy changes can be measured experimentally by measuring the temperature change as a reaction proceeds. Two common examples are measuring the enthalpy change of neutralisation and combustion. 2) The formulae: where m = mass of substance(or solution)/g Heat energy absorbed c = specific heat capacity of solution/J g⁻¹ °C⁻¹ = mc∆T where m = mass of solution/g (c is normally assumed to be 4.18 J g⁻¹ °C⁻¹) or released c = specific heat capacity ∆T = change in temperature/°C of solution (c is normally taken as 4.13 J g⁻¹ K⁻¹ ) heat energy absorbed or released ∆T = change in temperature ∆H = Unit = kJ mol⁻¹ no. of moles of the limiting reagent 3) Assumptions made in this calculation: - The density of all aqueous solution is 1 g cm⁻³. Thus, numerically, 1 cm³ = 1 g. - The solution has the same specific heat capacity as water(4.18 J g⁻¹ °C⁻¹). 4) An example - to calculate the enthalpy change of the reaction between sodium hydroxide, NaOH(aq) and hydrochloric acid, HCl(aq). NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) i. Pour 25 cm³ of NaOH into 25 cm³ of HCl in a polystyrene cup(polystyrene cup is used because it is a good heat insulator). Both solutions have concentrations of 1.0 mol dm⁻³. ii. Measure the initial and maximum temperature reached. iii. For example, the mean initial temperature of both solutions is 18.1 °C and the maximum temperature reached is 24.8 °C. Then, ΔT = 6.7 °C. iv. Mass of both solutions is (25+25) x 1 = 50 g since density = mass/volume. v. Therefore heat released = 50 x 4.18 x 6.7 = 1400 J = 1.4 kJ vi. Number of moles of water = 0.025 mol vii. Therefore ΔH = -(1.4/0.025) = -56 kJ mol⁻¹ 5) Another example - to calculate the enthalpy change of the reaction of combustion of ethanol, C2H5OH. C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g) i. In this experiment, a spirit burner is used to combust the flammable liquid ethanol. The heat evolved is used to heat up the water, the maximum temperature reached is then measured. ii. Mass of water = 250 g Initial temperature of water = 19.5 °C Maximum temperature of water = 23.7 °C Initial mass of ethanol = 41.36 g Final mass of ethanol = 41.18 g iii. Therefore, ΔT = 4.2 °C and m = 250 g. iv. Heat released = 250 x 4.18 x 4.2 = 4389 J = 4.389 kJ v. Number of moles of ethanol burned = 0.18/46 = 0.0039 mol vi. Therefore ΔH = -(4.389/0.0039) = -1100 kJ mol⁻¹ (2 s.f) 6.2 Standard Enthalpy Changes The standard condition 1) To make comparison of enthalpy changes a fair comparison, same conditions must be used. These are called the standard conditions: - A pressure of 100 kPa(approximately atmospheric pressure). - A temperature of 298 K or 25 °C. - Where solutions are involved, a concentration of 1.0 mol dm⁻³. - Every substance involved must be in its normal physical state at 100 kPa and 298 K. For example, water is in liquid, not ice or steam. - If allotropes are involved, the allotrope which is more energetically stable is used. For example, for carbon, graphite is chosen over diamond because graphite is energetically more stable than diamond. 2) Standard enthalpy changes can be represented by the symbol ΔH°. This refers that the reaction is carried out under standard conditions. For example: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ; ΔH° = -890.3 kJ mol⁻¹ Various enthalpy changes 1) Enthalpy changes are described according to the type of reaction. Examples: - standard enthalpy change of reaction, ΔH°r (in general) - standard enthalpy change of formation, ΔH°f - standard enthalpy change of combustion, ΔH°c - standard enthalpy change of neutralisation, ΔH°n - standard enthalpy change of atomisation, ΔH°at - standard enthalpy change of solution, ΔH°sol - standard enthalpy change of hydration, ΔH°hyd Standard enthalpy change of reaction, ΔH°r 1) Standard enthalpy change of reaction, ΔH°r is the enthalpy change when the amount of reactants shown in the equation react to give products under standard condition. The reactants and products must be in their standard states. 2H2(g) + O2(g) → 2H2O(l) ; ΔH°r = -576 kJ mol⁻¹ 2) The equation shows that 576 kJ of energy is released when two moles of hydrogen react with one mole of oxygen to give two moles of water. 3) This is a theoretical reaction, it does not happen in practice. The enthalpy change can be found by applying Hess' law. Standard enthalpy change of formation, ΔH°f 1) Standard enthalpy change of formation, ΔH°f is the enthalpy change when one mole of a compound is formed from its elements under standard condition. The reactants and products must be in their standard states. 2Fe(s) + 1½O2(g) → Fe2O3(s) ; ΔH°f [ Fe2O3(s) ] = -824.5 kJ mol⁻¹ 2) By definition, the standard enthalpy change of formation of an element is zero. 3) The standard enthalpy change of formation can be exothermic or endothermic. Standard enthalpy change of combustion, ΔH°c 1) Standard enthalpy change of combustion, ΔH°c is the enthalpy change when one mole of a substance is burnt in excess oxygen under standard conditions. The reactants and products must be in their standard states. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ; ΔH°c [ CH4(g) ] = -890.3 kJ mol⁻¹ 2) The standard enthalpy change of combustion is always exothermic. Standard enthalpy change of neutralisation, ΔH°n 1) Standard enthalpy change of neutralisation, ΔH°n is the enthalpy change when one mole of water is formed by the reaction of an acid with an alkali under standard conditions. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ; ΔH°n = -57.1 kJ mol⁻¹ 2) For any acid-alkali reaction, the ionic equation is: H⁺(aq) + OH⁻(aq) → H2O(l) 3) The enthalpy change of neutralisation between strong acids and strong bases is a constant(-57.1 kJ mol⁻¹). This is because all strong acids and strong bases dissociate completely in water to form aqueous ions. 4) So, neutralisation between strong acids and strong bases involves the same reaction, that is, H⁺ reacts with OH⁻ to form H2O. The other ions present are just simply spectator ions. They do not take part in the reaction. Hence, the heat released is the same. 5) However, the enthalpy change of neutralisation between sulfuric acid and sodium hydroxide is more exothermic(-66.8 kJ mol⁻¹) than expected. This is because the enthalpy change of dilution of sufuric acid is significant. When sulfuric acid is added to sodium hydroxide, the acid is diluted in the process and heat is released. 6) The enthalpy change of neutralisation involving weak acids or weak bases is less than 57.1 kJ mol⁻¹. For example: CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l) ; ΔH°n = -56.1 kJ mol⁻¹ NaOH(aq) + HCN(aq) → NaCN(aq) + H2O(l) ; ΔH°n = -11.7 kJ mol⁻¹ 7) This is because weak acids and weak bases only partially dissociated in water. For example: CH3COOH(aq) ⇌ CH3COO⁻(aq) + H⁺(aq) ; ΔH° = positive 8) On addition of strong base such as NaOH, the OH⁻ ions react with H⁺ from the dissociation of CH3COOH. The removal of H⁺ ions causes the position of equilibrium to shift to the right and more CH3COOH molecules dissociate. However the dissociation of CH3COOH is an endothermic process, hence, some energy is absorbed and the enthalpy change overall is less negative. 9) The standard enthalpy change of neutralisation is always exothermic. Standard enthalpy change of atomisation, ΔH°at 1) Standard enthalpy change of atomisation, ΔH°at is the enthalpy change when one mole of gases atoms is formed from its element under standard conditions. ½H2(g) → H(g) ; ΔH°at [ ½H2 ] = +218 kJ mol⁻¹ Na(s) → Na(g) ; ΔH°at [ Na ] = +107 kJ mol⁻¹ 2) By definition, the standard enthalpy change of atomisation of the noble gases is zero because all of them exist as monoatomic gases at standard conditions. 3) The standard enthalpy change of atomisation of diatomic gases(example: O2, Cl2, N2 and F2) is equal to half the value of their bond energies. 4) The standard enthalpy change of atomisation is always endothermic. Standard enthalpy change of solution, ΔH°sol 1) Standard enthalpy change of solution, ΔH°sol is the enthalpy change when one mole of solute is dissolved in a solvent to form an infinitely dilute solution under standard conditions. NaCl(s) + aq → NaCl(aq) ; ΔH°sol = +6.0 kJ mol⁻¹ NaOH(s) + aq → NaOH(aq) ; ΔH°sol = -44.5 kJ mol⁻¹ 2) An infinitely dilute solution is one which does not produce any further enthalpy change when more solvent is added. 3) The standard enthalpy change of solution can be exothermic or endothermic. Standard enthalpy change of hydration, ΔH°hyd 1) Standard enthalpy change of hydration, ΔH°hyd is the enthalpy change when one mole of gases ions dissolves in water to form hydrated ions of infinite dilution under standard conditions. Ca²⁺(g) + aq → Ca²⁺(aq) ; ΔH°hyd = -1650 kJ mol⁻¹ 2) Water is a polar molecule, this means that it has a negative end and a positive end. The negative end of the water molecule will be attracted to the cations while the positive end of the water molecule will be attracted to the anions. 3) The attraction set up is called the ion-dipole forces. 4) The standard enthalpy change of hydration is always exothermic. 6.3 Hess' Law Hess' law and Hess' cycle 1) Hess' law states that the total enthalpy change in a chemical reaction is independent of the route whic

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