G10 Advanced Chemistry CHM51 Revision PDF

G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 1 of 50  Unit 1: Structure & Properties of Matter 1.1 – Electrons in Atoms Inspire Chemistry – Module 4 – Lesson 2: Quantum theory and the atom Describe Bohr's model of the hydrogen atom The hydrogen atom has...

G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 1 of 50  Unit 1: Structure & Properties of Matter 1.1 – Electrons in Atoms Inspire Chemistry – Module 4 – Lesson 2: Quantum theory and the atom Describe Bohr's model of the hydrogen atom The hydrogen atom has a certain allowable energy states. CHM.5.1.01.001.05. Explain what is meant by an energy levels and energy sublevels Each possible electron orbit in Bohr’s model has a fixed energy. The fixed energies an electron can have are called energy levels. The electrons in an atom cannot exist between energy levels. Energy sublevel is an energy level contained with a principal energy level. CHM.5.1.01.001.06. Describe the relationship between energy levels and energy sublevels of hydrogen The number of energy sublevels in a principle energy level increases as n increases. CHM.5.1.01.001.07. Differentiate between the ground and excited states of an atom Ground state: It is the lowest allowable energy state of an atom. Excited state: When an atom gains energy, it is said to be inn an excited state. CHM.5.1.01.001.08. Describe the atomic orbitals and their shapes Atomic orbital It is a three-dimensional region around the nucleus. Each orbital can contain, at most, two electrons Different atomic orbitals are denoted by letters (s, p, d and f) The s orbitals are spherical. The p orbitals are dumbbell-shaped. Not all the d or f orbital have the same shape G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 2 of 50  CHM.5.1.01.002.02. Calculate the number of orbitals and electrons in the energy levels Number of orbitals: use the formula n2 Number of electrons: use the formula 2 n2 Where n is the number of energy level Energy level, n Number of orbitals, n2 Number of electron, 2n2 1 1 2 2 4 8 3 9 18 4 16 32 5 25 50 6 36 72 G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 3 of 50  CHM.5.1.01.002.03. Relate the principal quantum number to the total number of orbitals of the principal energy levels and the number of orbitals in sub-energy levels Summary of Principal Energy Levels and Sublevels Principal Sublevels Total number of orbital Number of orbital Quantum (Types of related to Principal related to sublevel Number (n) Orbital Present) Energy Level (n2) n=1 s 1 orbital (1s) 1 s 1 orbital (2s) n=2 4 p 3 orbitals (2p) s 1 orbital (3s) n=3 p 3 orbitals (3p) 9 d 5 orbitals (3d) s 1 orbital (4s) p 3 orbitals (4p) n=4 16 d 5 orbitals (4d) f 7 orbitals (4f) Note: - The number of orbital related to each sublevel is always an odd number. - The maximum number of orbital related to each principal energy level is 2n2. Inspire Chemistry – Module 4 – Lesson 3: Electron configuration CHM.5.1.01.003.01. Define electron configuration Electron configuration is the arrangement of electrons in an atom. CHM.5.1.01.003.02. Explain the Aufbau principle, the Pauli exclusion principle, and Hund’s rule Aufbau principle It states that electrons occupy the lowest energy orbital available. Feature Example All orbitals related to an All three 2p orbitals are of energy sublevel are of equal equal energy. energy. In a multi-electron atom, the The three 2p orbitals are of energy sublevels within a higher energy than the 2s principal energy level have orbital. different energies. In order of increasing If n = 4, then the sequence energy, the sequence of of energy sublevel is 4s, 4p, energy sublevels within a 4d and 4f. principal energy level is s, p, d and f. Orbitals related to energy The orbital related to the sublevels within one atom’s 4s sublevel has a principal energy level can lower energy than the five overlap orbitals related to orbitals related to the 3d energy sublevels within sublevel. another principal level. G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 4 of 50  Pauli exclusion principle It states that only two electrons may share an atomic orbital, and only when they have opposite spins. Spin is a quantum mechanical property of electrons and may be thought of as clockwise or counterclockwise. A vertical arrow indicates an electron and its direction of spin ( or ). An orbital containing paired electrons is written as Hund’s rule It states that single electrons with the same spin must occupy each equal-energy orbital before additional electrons with opposite spins can enter the same orbitals. CHM.5.1.01.003.03. Describe orbital diagrams Electrons in orbitals can be represented by arrows in boxes. Each box is labelled with the principal quantum number and sublevel associated with the orbital. CHM.5.1.01.003.04. Describe the noble-gas notation Noble-gas notation uses bracketed symbols to represent the electron configurations of noble gases. The electron configuration for an element can be represented using the noble-gas notation for the noble gas in the previous period and the electron configuration for the additional orbitals being filled. Examples G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 5 of 50  CHM.5.1.01.003.05. Write the electron configuration, orbital diagram, and noble gas notation of different elements Example: Sodium, Na  Electron configuration 1s2 2s2 2p6 3s1  Orbital notation  Noble gas notation [Ne] 3s1 CHM.5.1.01.003.06. Explain the electron configuration in transition metals Chromium, Cr Copper, Cu Expected noble-gas notation (Incorrect): Expected noble-gas notation (Incorrect): [Ar]4s2 3d4 [Ar]4s2 3d9 The correct noble-gas notation: The correct noble-gas notation: [Ar]4s2 3d5 [Ar]4s1 3d10 The electron configurations for the Cr and Cu illustrate the increased stability of half-filled and filled sets of s and d orbitals. CHM.5.1.02.001.01. Define valence electrons while determining their importance Valence electrons are electrons in the atom’s outermost orbital. They determine the chemical properties of an element. CHM.5.1.02.001.02. Determine the number of valence electrons in atoms Example: Sulfur, 6 (16 electrons) Electron configuration: [Ne] 3s2 3p4 Only six electrons occupy the outermost 3s and 3p orbitals Sulfur has 6 valence electrons CHM.5.1.02.001.03. Define Lewis (electron-dot) structure The Lewis structure is the structure consisting of the element’s symbol, which represents the atomic nucleus and inner-level of electrons, surrounded by dots representing the atom’s valence electrons. G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 6 of 50  CHM.5.1.02.001.04. Represent the electron-dot structure of atoms G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 7 of 50  Inspire Chemistry – Module 5 – Lesson 2: Classification of the elements CHM.5.1.01.004.10. Use the electron configuration notation, orbital notation, and noble gas notation of an element (Atomic number, Z 1 – 36) to identify the name and symbol of an element Example: Consider the information given below about the three elements X, Y and Z. X Y Z 1s2 2s2 2p6 3s2 3p6 4s2 3d3 Identify elements X, Y and Z. X Vanadium or V Y Chlorine or Cl Z Silicon or Si CHM.5.1.01.004.11. Use the electron configuration notation, orbital notation, and noble gas notation of an element (Atomic number, Z 1 – 36) to determine number of valence electrons of an element Example: Consider the information given below about the three elements X, Y and Z. Y Z Identify the number of valence electrons for elements X, Y and Z. Y 7 Z 4 G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 8 of 50  CHM.5.1.01.008.01. Identify the characteristics of the elements in each of the block (s, p, d, and f) of the periodic table s-b lock elements - It consists of elements of groups 1 and 2 and the element helium. - Group 1 elements have partially filled s orbitals containing one valence electron and electron configurations ending in s1. - Group 2 elements have completely filled s orbitals containing two valence electrons and electron configurations ending in s2. - s-block spans two groups of the periodic table (since s orbitals hold two electrons at most). p-b lock elements - It consists of elements of groups 13 through 18. - It contains elements with filled or partially filled p orbitals. - There are no p-block elements in period 1 because the p sublevel does not exist for the first principal energy level (n = 1). - The first p-block element is boron, B, which is in the second period. - p-block spans six groups of the periodic table (since the three p orbitals can hold a maximum of six electrons). Note: The s – and p – blocks compromise the representative elements. d-b lock elements - It contains the transition metals and it is the largest of the blocks. - With some exceptions, d-block elements are characterized by a filled outermost s orbital of energy level n, and filled or partially filled d orbitals of energy level n – 1. As moving across a period, electrons fill the d orbitals. - According to the Aufbau principle, the 4s orbital has lower energy than the 3d orbital, thus the 4s orbital is filled before the 3d orbital. - d-block spans ten groups of the periodic table (since the five d orbitals can hold a maximum of ten electrons) f-b lock elements - It contains the inner transition metals. - The f-block elements are characterized by a filled, or partially outermost s orbital, and filled or partially filled 4f and 5f orbitals. G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 9 of 50  CHM.5.1.01.008.02. Use the electron configuration notation, orbital notation, and noble gas notation of an element (Atomic number, Z 1 – 36) to identify the location of an element in the periodic table (period, group and block) CHM.5.1.01.008.03. Use the electron configuration notation, orbital notation, and noble gas notation of an element (Atomic number, Z 1 – 36) to identify the type of element (metal, nonmetal, transition metal or metalloid) Example: 1s2 2s2 2p6 3s2 3p3 Name of the element: Phosphorus Symbol: P Number of valence electrons: 5 Location: Period: 3 Group: 15 Block: p-block Type of element: Non-metal CHM.5.1.01.004.12. Conclude the electron configuration of elements based on their position in the periodic table (period and group) Example: Sodium, Na  Electron configuration 1s2 2s2 2p6 3s1  Orbital notation  Noble gas notation [Ne] 3s1 CHM.5.1.01.004.13. Determine the position of the elements in the periodic table based on their electronic configuration Example: 1s2 2s2 2p6 3s2 Location: Period: 3 Group: 2 Block: s-block Inspire Chemistry – Module 5 – Lesson 3: Periodic Trends CHM.5.1.01.009.01. List the five periodic trends The five periodic trends are: 1) Atomic radius 2) Ionic radius 3) Ionization energy 4) Electronegativity 5) Electron affinity G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 10 of 50  CHM.5.1.01.009.02. Define electron shielding Electron shielding is where electrons in an atom can shield each other from the pull of the nucleus. It describes the decrease in attraction between an electron and the nucleus in any atom with more than one electron shell. CHM.5.1.01.009.03. Define effective nuclear charge (ENC) Effective nuclear charge is the net positive charge experienced by valence electrons. It can be approximated by the equation: Z eff = Z – S Where Z is the atomic number and S is the number of shielding electrons or core electrons. CHM.5.1.01.009.04. Explain the effective nuclear charge periodic trend across a period and down a group Across a period: ENC increases across a period from left to right due to increasing nuclear charge with no accompanying increase in the shielding effect. Down a group: ENC decreases down a group; although nuclear charge increases down a group, shielding effect more than counters its effect. CHM.5.1.01.009.05. Define atomic radius For Non-metals For Metals (That commonly occur as molecules) The atomic radius is one-half the distance The atomic radius is half the distance between adjacent nuclei in a crystal of the between nuclei of identical atoms that are element. chemically bonded together. G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 11 of 50  CHM.5.1.01.009.06. Explain the periodic trend of atomic radii across a period and down a group Across a period from left to right, the atomic radii decreases. Explanation: For the same shielding effect, the effective nuclear charge increases, thus the attraction of the nucleus to the valence electrons increases and atomic radii decreases. Down a group, the atomic radii increases. Explanation: Down a group, the principal quantum number, n, increases, and size of atom increases. Thus, as we move down, the attractive force of the nucleus dissipated as the electrons spend more time further from the nucleus, thus the atomic radii increases. CHM.5.1.01.009.07. Define ionization energy (first ionization energy) Ionization energy It is the amount of energy needed to remove an electron from a gaseous atom. First ionization energy It is the amount of energy needed to remove an electron from a gaseous atom to form a single positively charged gaseous ion. X(g) + IE → X1+(g) + e– Atoms with large ionization energy values are less likely to form positive ions. Low ionization energy indicates an atom loses an outer electron easily. G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 12 of 50  CHM.5.1.01.009.08. Explain why second ionization energy is greater than first ionization energy Energy required for each successive ionization energy always increases. The second ionization energy involves the removal of an electron from a positively charged ion where the attraction of the electron to the nucleus is higher, thus more energy is needed. CHM.5.1.01.009.09. Describe the relation between the ionization energy values and number of valence electrons and the group the element belongs to The ionization energy at which the large increase in energy occurs is related to the atom’s number of valence electrons. The increase in ionization energy shows that atoms hold onto their inner core electrons much more strongly than they hold onto their valence (outermost) electrons. Example: First large jump in IE Number of Element belongs to Element is between valence electrons which group Li IE1 and IE2 1 1 Be IE2 and IE3 2 2 B IE3 and IE4 3 3 N IE5 and IE6 5 15 O IE6 and IE7 6 16 F IE7 and IE8 7 17 Ne IE8 and IE9 9 18 G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 13 of 50  CHM.5.1.01.009.10. Explain the trend of first ionization energy across a period (Exceptions between Groups 2 & 3 and 5 & 6 are included), and down a group of the periodic table Across a period from left to right, the ionization energy increases. For the same shielding effect, the effective nuclear charge increases, and the attraction of the valence electron to the nucleus increases, thus more energy is needed to remove an electron and ionization energy increases. Exceptions: i. Groups 2 and 3 IE1 of group 2 elements involves the removal of an ns electron while that of group 3 involves the removal of a np electron. p-electron is more energetic than s-electron and easier to remove, thus IE1 of group 2 elements is greater than IE1 of group 3 elements ii. Groups 5 and 6 IE1 of group 5 involves the removal of np3 electron while that of group 6 involves the removal of np4. np4 has higher electron-electron repulsion thus less energy is needed to remove the electron, thus IE1 of group 5 elements is greater than IE1 of group 6 elements Down a group, ionization energy decreases As you go down the group, the atomic size increases, the valence electron occupies a higher principle quantum number, n, and is further from the nucleus thus less energy will be needed to remove the valence electron. CHM.5.1.01.009.11. Explain why noble gases have the highest ionization energy Since noble gases are stable, and have filled outer shell configurations, it is difficult to remove an electron from any of the noble gases. CHM.5.1.01.009.12. Explain why each successive ionization of an electron requires a greater amount of energy Each successive electron removed from an ion feels an increasingly stronger effective nuclear charge, thus more energy is needed to remove an electron. G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 14 of 50  CHM.5.1.01.009.13. Explain the difference in size between an atom and its ion (anion or cation) Cations: The radii of cations are smaller than those of their atoms, because positive ions are formed by removing one or more electrons from the outermost shell. Since these electrons are furthest from the nucleus; their absence will make the cation significantly smaller. The removal of an electron causes a decrease in the amount of electron-electron repulsion which also causes the cation to be smaller than the neutral atom. Anions: The radii of anions are larger than those of the corresponding neutral atom When an electron is added to an atom to form an anion, there is an increase in the electron-electron repulsion. This causes electrons to spread out as much as possible, and so anions have a larger radius than the corresponding atoms. CHM.5.1.01.009.14. Describe the trend of ionic radii across a period and down a group of the periodic table Across a period from left to right In general, across a period from left to right, the size of the positive ions gradually decreases. Then beginning in group 15 or 16, the size of the much larger negative ions also gradually decreases Down a group In general, the ionic radii of both positive and negative ions increases down a group, because the outer electrons are in orbitals corresponding to higher principle energy levels. G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 15 of 50  CHM.5.1.01.009.15. Define isoelectronic species while listing some examples of it according to the relative size of ion or atom Isoelectronic species are species that contain the same number of electrons. The species with the largest atomic number has the smallest radii. Example: Al3+ < Mg2+ < Na1+ < Ne < F1− < O2− CHM.5.1.01.011.01. Define electronegativity Electronegativity is the relative ability of atoms to attract electrons in a chemical bond. CHM.5.1.01.011.02. Explain the periodic trend of electronegativity across a period and down a group of the periodic table Across a period from left to right electronegativity increases For the same shielding effect, the effective nuclear charge increases, and the attraction of the nucleus to the incoming electron increases Down a group electronegativity decreases or remains about the same Metals at the far left of the periodic table have low values. By contrast, nonmetals at the far right (excluding noble gases) have high values. G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 16 of 50  Values among transition metals are not as regular. The least electronegative element is francium, with an electronegativity of 0.70. It has the least tendency to attract electrons. When it reacts, it tends to lose electrons and form cations. The most electronegative element is fluorine, with a value of 3.98. Because fluorine has such a strong tendency to attract electrons, when it is bonded to any other element it either attracts the shared electrons or forms an anion. In a chemical bond, the atom with the greater electronegativity more strongly attracts the bond’s electrons. CHM.5.1.01.011.03. Describe what is unusual about the electronegativity of noble gases The noble gases are unusual in that some of them do not form compounds and thus cannot be assigned an electronegativity value CHM.5.1.01.011.04. Define electron affinity Electron affinity is the amount of energy released when an electron is added to the atom in its gaseous state. CHM.5.1.01.011.05. Explain the periodic trend of electron affinity across a period and down a group of the periodic table, while explaining why it becomes more negative across the period Across a period from left to right, the electron affinity becomes more negative (increases) because the higher effective nuclear charge increases the nuclear attraction for the incoming electron Down a group, the electron affinity does not change very much because the lower-electron nucleus attraction down the group is evenly counterbalanced by a simultaneous lowering in the electron- electron repulsion CHM.5.1.01.011.06. Explain exceptions to the electron affinity trend Exceptions to the electron affinity trend down a group are the noble gases (He, Ne, Ar, Kr and Xe). They have positive electron affinities (meaning very low), because if they were to accept another electron, that electron would have to go into a new, higher-energy subshell, and this is energetically unfavorable Unit 2: Chemical bonding and Reaction 2.1 – Ionic Compounds and Metals Inspire Chemistry – Module 6 – Lesson 1: Ion formation CHM.5.1.02.022.01. Define chemical bond while explaining why elements bond Chemical bond is the force that holds two atoms together. Elements bond to become more stable and attain the noble gas configuration of the nearest noble gas. G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 17 of 50  CHM.5.1.02.022.02. Explain the octet rule The octet rule states that atoms lose, gain or share electrons in order to be stable like the nearest noble gas (i.e. has eight valence electrons) CHM.5.1.02.022.03. Describe how ions (cations and anions) form to fulfill the octet rule Cation (Positive Ion) Anion (Negative Ion) A cation is a positively charged ion. An anion is a negatively charged ion. A cation is formed when an atom loses one or An anion is formed when an atom gains one or more electron to be stable like the nearest more electron to be stable like the nearest noble noble gas gas The number of protons do not change during The number of protons do not change during the formation of a cation. the formation of an anion. CHM.5.1.02.022.04. Relate the charge of an ion to the group number of an element in the periodic table Noble gas notation of the Group Charge of the Ion formed element Group 1 [Noble gas] ns1 1+ ns1 electrons are lost Group 2 [Noble gas] ns2 2+ ns2 electrons are lost Group 13 [Noble gas] ns2 np1 3+ ns2 np1 electrons are lost Group 15 [Noble gas] ns2 np3 3− When 3 electrons are gained Group 16 [Noble gas] ns2 np4 2− When 2 electrons are gained G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 18 of 50  Group 17 [Noble gas] ns2 np5 1− When 1 electron is gained CHM.5.1.02.022.05. Write the electron configuration notation, noble gas notation and Lewis structure of different anions and cations Example: Chloride ion, Cl– Example: Sodium ion, Na+ Electron configuration: 1s2 2s2 2p6 3s2 3p6 Electron configuration: 1s2 2s2 2p6 Noble gas notation: [Ar] Noble gas notation: [Ne] Lewis structure: Lewis structure: Inspire Chemistry – Module 6 – Lesson 2: Ionic bonds and ionic compounds CHM.5.1.02.022.06. Define ionic bond while identifying the type of elements involved and movement of electrons Ionic bond is the electrostatic force of attraction between oppositely charged ions. Elements involved in an ionic bond are metals and non-metals. An ionic compound is a compounds that contains ionic bonds. A metal atom tends to lose electrons to become stable like the nearest noble gas while a nonmetal atom tends to gain the electrons lost by the metal atom to be stable like the nearest noble gas. G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 19 of 50  CHM.5.1.02.022.07. Use the Lewis diagram (electron-dot structure) to explain how elements from the periodic groups combine to form an ionic compound Examples: o Calcium Chloride, CaCl2 o Magnesium sulfide, MgS o Sodium Chloride, NaCl o Potassium Oxide, K2O o Aluminum oxide, Al2O3 G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 20 of 50  CHM.5.1.02.022.08. Explain the physical properties of ionic compounds as melting point and boiling point, conductivity when solid, molten or aqueous, and its solubility in water Melting point and boiling point High melting point and boiling point – Large amount of energy is needed to overcome the strong electrostatic force of attraction between oppositely charged particles Conductivity when solid, molten or aqueous Poor conductivity when solid because ions are not free to move and carry electric current. High conductivity when molten or aqueous because ions are free to move and carry electric current Solubility in water Soluble in water, because ionic compound is polar and water is polar (Like dissolves like) An ionic compound whose aqueous solution conducts an electric current is called an electrolyte. G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 21 of 50  i.e. The lower-energy product is more stable than the original reactants. CHM.5.1.02.022.12. Define the crystal lattice (using diagram) The crystal lattice is a three-dimensional geometric arrangement of particles. In a crystal lattice, each positive ion is surrounded by negative ions, and each negative ion is surrounded by positive ions. Crystal ions vary in shape due to the sizes and relative numbers of the ions bonded. CHM.5.1.02.022.13. Define lattice energy Lattice energy is the energy required to separate one mole of the ions of an ionic compound. CHM.5.1.02.022.14. Relate lattice energy to ionic bond strength The strength of the electrical forces holding ions in place is reflected by the lattice energy. The greater the lattice energy, the stronger the force of attraction. CHM.5.1.02.022.15. Describe the relationship between lattice energy and the charge of ions The ionic bond formed from the attraction of ions with larger positive or negative charges generally has a greater lattice energy. Example: The lattice energy of MgO is almost four times greater than that of NaF because the charge of the ions in MgO (2+ and 2−) is greater than the charge of the ions in NaF (1+ and 1−). G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 22 of 50  CHM.5.1.02.022.16. Describe the relationship between lattice energy and the size of ions - Lattice energy is directly related to the size of ions bonded. - Smaller ions form compounds with more closely spaced ionic charges. - Because the electrostatic force f attraction between opposite charges increases as the distance between the charges decreases, smaller ions produce stronger attractions and greater lattice energies. - Example: The lattice energy of a lithium compound is greater than that of a potassium compound with the same anion because a lithium ion is smaller than a potassium ion. CHM.5.1.02.022.17. Perform an experiment to study the properties of ionic compounds Inspire Chemistry – Module 6 – Lesson 3: Names and formulas for ionic compounds CHM.5.1.02.022.18. Describe how formula units relate to the composition of an ionic compound - Formula unit is the simplest ratio of the ions represented in an ionic compound. - Example: The formula unit of magnesium chloride is MgCl2 because the magnesium and chloride ions exits in a 1:2 ratio. CHM.5.1.01.013.01. Write the chemical name of an ionic compound containing monoatomic and polyatomic ions (including oxyanions) - A monoatomic ion is a one-atom ion. - A polyatomic ion is an ion made up of more than one atom. - An oxyanion is a polyatomic ion composed of an element, usually a nonmetal, bonded to one or more oxygen atoms. Rules for naming ionic compounds: 1) Name the cation followed by the anion. 2) For monoatomic cations, use the element name 3) For monoatomic anions, use the root of the element name plus the suffix – ide 4) For transition metals and metals on the right side of the periodic table (does not apply to Group 1 and 2 cations), the name of the chemical formula must indicate the oxidation number of the cation. The oxidation number is written as a Roman numeral in parentheses after the name of the cation 5) When the compound contains a polyatomic ion, simply use the name of the polyatomic ion in place of the anion or cation To indicate more than one polyatomic ion in a chemical formula, place parentheses around the polyatomic ions and use a subscript G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 23 of 50  Common Monoatomic Ions Group Atoms that commonly form ions Charge of ion 1 H, Li, Na, K, Rb, Cs 1+ 2 Be, Mg, Ca, Sr, Ba 2+ 15 N, P, As 3− 16 O, S, Se, Te 2− 17 F, Cl, Br, I 1− Monoatomic Metal Ions Group Common Ions Group Common Ions 3 Sc3+ Scandium (III) 9 Co2+ Cobalt (II) Y3+ Yttrium (III) Co3+ Cobalt (III) La3+ Lanthanum (III) 4 Ti2+ Titanium (II) 10 Ni2+ Nickel (II) Ti3+ Titanium (III) Pd2+ Palladium (II) Pt2+ Platinum (II) Pt4+ Platinum (IV) 5 V2+ Vanadium (II) 11 Cu+ Copper (II) V3+ Vanadium (III) Cu2+ Copper (II) Ag+ Silver Au+ Gold Au3+ Gold (III) 6 Cr2+ Chromium (II) 12 Zn2+ Zinc Cr3+ Chromium (III) Cd2+ Cadmium Hg22+ Mercury (I) Hg2+ Mercury (II) 7 Mn2+ Manganese (II) 13 Al3+ Aluminum Mn3+ Manganese (III) Ga2+ Gallium (II) Ga3+ Gallium (III) In+ Indium (I) In2+ Indium (II) In3+ Indium (III) Tl+ Thallium (I) Tl3+ Thallium (III) 8 Fe2+ Iron (II) 14 Sn2+ Tin (II) Fe3+ Iron (III) Sn4+ Tin (IV) Pb2+ Lead (II) Pb4+ Lead (IV) G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 24 of 50  Common Polyatomic Ions Ion Name Ion Name NH4+ ammonium IO4― periodate NO2― nitrite C2H3O2― acetate NO3― nitrate H2PO4― dihydrogen phosphate OH― hydroxide CO32― carbonate CN― cyanide SO32― sulfite MnO4― permanganate SO42― sulfate HCO3― hydrogen carbonate S2O32― thiosulfate ClO― hypochlorite O22― peroxide ClO2― chlorite CrO42― chromate ClO3― chlorate Cr2O72― dichromate ClO4― perchlorate HPO42― hydrogen phosphate BrO3― bromate PO43― phosphate IO3― iodate AsO43― arsenate Oxyanions Ion Name Ion Name NO3― nitrate BrO4― perbromate NO2― nitrite BrO3― bromate SO42― sulfate BrO2― bromite SO32― sulfite BrO― perbromite ClO4― perchlorate IO4― periodate ClO3― chlorate IO3― iodate ClO2― chlorite IO2― iodite ClO― hypochlorite IO― periodite G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 25 of 50  CHM.5.1.01.013.02. Write the chemical formula (using the stock method) of an ionic compound containing monoatomic and polyatomic ions Example: Write the chemical formula for each of the following compounds. Chemical name Chemical formula Barium hydroxide Ba(OH)2 Ammonium phosphate (NH4)3PO4 Iron (II) Phosphate Fe3(PO4)2 Magnesium nitrate Mg(NO3)2 Potassium dichromate K2Cr2O7 Manganese (IV) chlorate Mn(ClO3)4 Cesium nitride Cs3N Example: Write the chemical name for each of the following species. Chemical formula Chemical name Cu(NO3)2 Copper (II) nitrate SO32― Sulfite ion (Sulphite ion) Ca3P2 Calcium phosphide PbBr4 Lead (IV) bromide Na2O Sodium oxide (NH4)2SO4 Ammonium sulfate AlP Aluminum phosphide KMnO4 Potassium permanganate Example: Write the chemical formula of the ionic compound formed between each of the following pairs of elements: Elements Chemical formula Na and F NaF Ca and Cl CaCl2 Mg and N Mg3N2 Al and O Al2O3 G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 26 of 50 G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 27 of 50 G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 28 of 50  2.2 – Covalent Bonding Inspire Chemistry – Module 7 – Lesson 1: The covalent bond CHM.5.1.01.011.07. Define, using energy diagram, covalent bond Covalent bond is a chemical bond that results from the sharing of valence electrons. When two atoms approach each other, electron-electron repulsion and proton-proton repulsion both act to try to keep the atoms apart. However, proton-electron attraction can counterbalance this, pulling the two atoms together so that a bond is formed. CHM.5.1.01.011.08. Identify type of elements involved in the covalent bond with the movement of electrons It occurs when two or more non-metal atoms bond together by the equal sharing of one or more pair of electrons. Covalent bond generally can occur between elements that are near each other on the periodic table. CHM.5.1.02.001.05. Describe how the octet rule applies to covalent bonds Atoms share valence electrons; the shared electrons complete the octet of each atom. G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 29 of 50  CHM.5.1.02.001.06. Describe, using Lewis structure and ball and stick diagram, the difference among single, double and triple bonds Single Covalent Bond Double Covalent Bond Triple Covalent Bond One pair of electrons are Two pairs of electrons are Three pairs of electrons are shared between two non- shared between two non- shared between two non- metal atoms. metal atoms. metal atoms. CHM.5.1.02.001.07. Differentiate between sigma and pi bonds Sigma bond (𝜎) is a single covalent bond between two atoms that share an electron pair in an area centered between the two atoms. pi bond (𝜋) is a covalent bond formed when parallel orbitals overlap to share electrons. CHM.5.1.02.003.01. Identify, in different compounds, the number of sigma and pi bonds Single Covalent Bond Double Covalent Bond Triple Covalent Bond 1 sigma bond 1 sigma bond 1 sigma bond 1 pi bond 2 pi bonds G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 30 of 50  Example: For each of the following Lewis structures, identify the number of sigma and pi bonds. Lewis Structure Number of sigma bonds Number of pi bonds 3 1 2 2 3 0 2 2 8 0 1 2 5 1 G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 31 of 50  CHM.5.1.02.007.01 Explain the properties of covalent compounds in the solid phase in terms of conductivity, hardness and melting point Conductivity Nonconductor, there are no free ions or electrons that can move and carry electric current. Hardness Soft, due to the weak forces of attraction between the molecules. Melting point Low melting point (compared to ionic solids), due to the weak forces of attraction between the molecules. CHM.5.1.02.007.02. Explain, using examples, the difference between a covalent molecular solid and a covalent network solid Covalent molecular solid (as iodine, sucrose, glucose) is soft and has low melting point because of weak intermolecular forces. Covalent network solids (as quartz and diamond) have high melting points and are very hard because of the strength of the network of covalent bonds. CHM.5.1.02.007.03. Define bond dissociation energy Bond dissociation energy is the amount energy needed to break a covalent bond. The triple bond has the largest bond dissociation energy while the single covalent bond has the smallest bond dissociation energy. CHM.5.1.02.007.04. Identify the relationship between the type of a covalent bond (single, double and triple) and its bond length, bond strength and bond dissociation energy Bond length is the distance from the center of one nucleus to the center of the other nucleus in two bonded atoms. The shorter the bond length, the stronger the bond. Bond dissociation energy indicates the strength of a chemical bond because of the inverse relationship between bond energy and bond length. The smaller the bond length is, the greater the bond dissociation energy. Single Covalent Bond Double Covalent Bond Triple Covalent Bond Bond Length Longest Shortest Bond Strength Weakest Strongest Bond dissociation Smallest Largest energy Inspire Chemistry – Module 7 – Lesson 2: Naming molecules G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 32 of 50  CHM.5.1.01.014.01. Define binary molecular compound A binary molecular compound is a molecule composed of only two non-metal atoms CHM.5.1.01.014.02. Name a binary molecular compound based on its molecular formula (up to deca-) Steps: 1) The first element in the formula is always named first, using the entire element name 2) The second element in the formula is named using its root and adding the suffix – ide 3) Prefixes are used to indicate the number of atoms of each element that are present in the compound Number of atoms Prefix Number of atoms Prefix 1 mono- 6 hexa- 2 di- 7 hepta- 3 tri- 8 octa- 4 tetra- 9 nona- 5 penta- 10 deca- Exceptions 1) The first element in the compound name never uses the mono- prefix Example: CO It is called carbon monoxide and not monocarbon monoxide 2) If using a prefix results in two consecutive vowels, one of the vowels is usually dropped to avoid problems in pronunciation Example: CO It is called carbon monoxide and not monoocarbon monoxide Common names of some molecular compounds: Scientific Name Molecular Compound Common Name (not to be used) NH3 Ammonia Nitrogen trihydride N2H4 Hydrazine Dinitrogen tetrahydride NO Nitric oxide Nitrogen monoxide H2O Water Dihydrogen monoxide N2O Nitrous oxide Dinitrogen monoxide G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 33 of 50  CHM.5.1.01.014.03. Determine the chemical formula of a compound from its name Example: Write the chemical formula for each of the following species. Chemical name Chemical formula Dinitrogen trioxide N2O3 Phosphorous pentachloride PCl5 Oxygen difluoride OF2 Sulfur hexafluoride SF6 CHM.5.1.01.014.04. Describe the difference between a binary acid and an oxyacid A binary acid contains hydrogen and one other element (non-metal). An oxyacid contains hydrogen, another element (non-metal) and oxygen. CHM.5.1.01.014.05. Name an acid (binary acid and oxyacid) given its chemical formula and vice versa Examples NOT Limited to: Binary acids: HCl Hydrochloric acid HBr Hydrobromic acid HI Hydroiodic acid HF Hydrofluoric acid H2S Hydrosulfuric acid HCN Hydrocyanic acid Oxyacids: HNO3 Nitric acid HNO2 Nitrous acid H2SO4 Sulfuric acid H2SO3 Sulfurous acid H2CO3 Carbonic acid HClO4 Perchloric acid HClO3 Chloric acid HClO2 Chlorous acid HClO Hypochlorous acid G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 34 of 50  CHM.5.1.01.014.06. Build molecular models for molecular compounds Inspire Chemistry – Module 7 – Lesson 3: Molecular structures CHM.5.1.02.003.02. Compare among structural formula, space-filling molecular model and ball-and- stick molecular model Space-filling molecular Ball-and-stick molecular Structural formula model model It is a molecular model that It is a molecular model where It is a molecular model where uses letter symbols and bonds the molecular structures are the molecular structures are to show relative positions of given as full-sized spheres given as full-sized spheres atoms (atom of each specific (atom of each specific element are represented by element are represented by spheres of a representative spheres of a representative color) without rods. color) with rods. CHM.5.1.02.002.01. Draw Lewis structures for a number of covalent compounds with single bonds and multiple bonds Steps for drawing Lewis structure for covalent compounds with single and multiple bonds: 1) Predict the location of certain atoms The atom that has the least attraction for shared electrons will be the central atom in the molecule. This element is usually the one close to the left side of the periodic table. The central atom is located in the center of the molecule; all other atoms become terminal atoms. Hydrogen is always a terminal, or end, atom. Because it can share only one pair of electrons, hydrogen can be connected to only one another atom. 2) Determine the number of electrons available for bonding This number is equal to the total number of valence electrons in the atoms that make up the molecule. 3) Determine the number of bonding pairs To do this, divide the number of electrons available for bonding by two. 4) Place the bonding pairs Place one bonding pair (single bond) between the central atom and each of the terminal atoms. G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 35 of 50  5) Determine the number of electron pairs remaining Subtract the number of pairs used in Step 4 from the total number of bonding pairs in Step 3 (i.e. Step 3 – Step 4). The remaining pairs include lone pairs as well as pairs used in double and triple bonds. Place lone pairs around each terminal atom (except H atoms) bonded to the central atom to satisfy the octet rule. Any remaining pairs will be assigned to the central atom. Remember that not all elements form double bonds; only C, N, O, P and S 6) Determine whether the central atom satisfies the octet rule Is the central atom surrounded by four electron pairs? If not, it does not satisfy the octet rule. To satisfy the octet rule, convert one or two of the lone pairs on the terminal atoms into a double bond or a triple bond between the terminal atom and the central atom. These pairs are still associated with the terminal atom as well as the central atom. Carbon, nitrogen, oxygen and sulfur often form double and triple bonds. CHM.5.1.02.002.02. Draw Lewis structures for a number of polyatomic ions Steps for drawing Lewis structure for polyatomic ions: Follow the same steps for drawing the Lewis structure of a covalent compound except for Step 2 To find the total number of electrons available for bonding in polyatomic ion: Find the number of electrons available in the atoms present in the ion. Add the ion charge if the ion is negative, or subtract the ion charge if the ion is positive. Compared to the number of valence electrons present in the atoms that make up the ion, more electrons are present if the ion is negatively charged and fewer are present if the ion is positive. CHM.5.1.02.002.03. Define Resonance Resonance is a condition that occurs when more than one valid Lewis structure exists for the same molecule. The two or more Lewis structures that represent a single molecule or ion are referred to as resonance structures Resonance structures differ only in the position of the electron pairs, never the atom positions. Each molecule or ion that exhibits resonance behaves as if it has only one structure. Experimentally measured bond lengths show that the bonds are identical to each other. They are shorter than single bonds but longer than double bonds. The actual bond length is an average of the bonds in the resonance structures. G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 36 of 50  CHM.5.1.02.002.04. Identify some combinations of resonance (O3, NO&, NO& SO$& and CO$&) 3 $ 3 3 Species Possible Resonating Structures O3 NO3⁻ NO2⁻ SO32− CO32− G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 37 of 50  CHM.5.1.02.002.05. Determine the exceptions of the octet rule (odd number of valence electrons, suboctets and coordinate covalent bonds, expanded octets) 1) Odd number of valence electrons Some molecules might have an odd number of valence electrons and are unable to form an octet around each atom. Examples: NO2 ClO2 NO It has 17 valence electrons It has 19 valence electrons It has 11 valence electrons 2) Suboctets and coordinate covalent bonds a) Suboctets Some compounds form suboctets (stable configurations with less than eight electrons around the atom). These compounds tend to be reactive and can share a pair of electrons donated by another atom. Example: BH3 Boron, B, a group 13 element (a metalloid), forms three covalent bonds with other non-metallic atoms. The boron atom shares only six electrons (does not obey the octet rule). b) Coordinate covalent bonds It is a covalent bond formed when an atom donates a pair of electrons to be shared with an atom or ion that needs two electrons to become stable. Examples: Formation of hydronium ion, H3O+ G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 38 of 50  3) Expanded Octets Some compounds that do not follow the octet rule have central atoms that contain more than eight valence electrons i.e. Expanded octet. An expanded octet can be explained by considering the d orbitals that occur in the energy levels of elements in period three or higher. PCl5 SF6 XeF4 G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 39 of 50  Inspire Chemistry – Module 7 – Lesson 4: Molecular shapes CHM.5.1.02.004.01. State the basis for the VSEPR bonding model Valence Shell Electron Pair Repulsion – VSEPR VSEPR is a model based on an arrangement that minimizes the repulsion of shared and unshared electron pairs around the central atom. The model predicts some molecular shapes based on the idea that pairs of valence electrons surrounding an atom repel each other. CHM.5.1.02.005.01. Determine, using the VSEPR model, the bond angle for different molecules or ions (including exceptions to the octet rule) Bond angle is the angle formed by any two terminal atoms and the central atom. CHM.5.1.02.005.02. Determine, using the VSEPR model, the bonding and non-bonding domains for different molecules or ions (including exceptions to the octet rule) CHM.5.1.02.005.03. Determine, using the VSEPR model, the electron domain geometry and molecular geometry for different molecules or ions (including exceptions to the octet rule) Students must be able to Draw the Lewis Structure of each species given Electron Bonding Nonbonding domain Molecular Geometry Domains domains Geometry Linear AX2 Linear 2 0 180o Trigonal Trigonal AX3 3 0 planar Planar 120o Trigonal Bent AX2E 2 1 Planar < 120o Tetrahedral AX4 Tetrahedral 4 0 109.5o G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 40 of 50  Trigonal AX3E Tetrahedral 3 1 Pyramidal 107o Bent, v-shaped or AX2E2 Tetrahedral 2 2 angular 104.5o Trigonal Trigonal AX5 5 0 bipyramidal bipyramidal Trigonal AX4E 4 1 See-saw bipyramidal Trigonal AX3E2 3 2 T-shaped bipyramidal AX6 Octahedral 6 0 Octahedral Square AX5E Octahedral 5 1 pyramidal Square AX4E2 Octahedral 4 2 planar G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 41 of 50  Example: Consider the molecule of water given below to identify each of the following. - Electron domain geometry: Tetrahedral - Number of bonding domains: 2 or Two - Number of non-bonding domains: 2 or Two - Molecular geometry: Bent - Measurement of bond X: 104.5o CHM.5.1.02.005.04. Describe how the presence of a lone electron pair affects the spacing of shared bonding orbitals A lone pair occupies more space than a shared electron pair, thus the presence of a lone pair pushes the bonding pairs closer. CHM.5.1.02.005.05. Compare the size of an orbital that has a shared electron pair with one that has a lone pair The orbital containing a lone electron pair occupies more space than a shared electron pair. G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 42 of 50  - Inspire Chemistry – Module 7 – Lesson 5: Electronegativity and polarity CHM.5.1.01.011.09. Describe polar covalent bond A polar covalent bond is a covalent bond formed by the unequal sharing of the electron pair due to the difference in electronegativity. The polar nature of the bond may be represented by an arrow pointing to the more electronegative atom. The more electronegative element will have a partial negative charge (𝛿−)and the less electronegative will have a partial positive charge (𝛿+). When polar molecules are placed between oppositely charged plates, they tend to become oriented with respect to the positive and negative plates. Example: Hydrogen Fluoride, HF CHM.5.1.01.011.10 Differentiate between polar covalent and non-polar covalent bonds with comparing the location of the shared electrons In a polar covalent bond, there is an unequal attraction for the electron pair resulting in one of the bonded atoms possessing a partial negative charge and the other atom possessing a partial positive charge In a non-polar covalent bond, the electron pair is shared equally by the bonded atoms Polar molecules have unequal distribution of electrons while non-polar molecules have equal distribution of electrons G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 43 of 50  CHM.5.1.01.011.11. Categorize bond types using electronegativity difference into mostly ionic, polar covalent, mostly covalent and non-polar covalent bonds Electronegativity Difference Bond Character > 1.7 Mostly ionic 0.4 – 1.7 Polar covalent < 0.4 Mostly covalent 0 Non-polar covalent Example: By referring to the electronegativity values, identify the type of bond (or bond character) for each of the following compounds. Compound Type of Bond or Bond Character KF EN = 3.98 – 0.82 = 3.16 Mostly ionic SO2 EN = 3.44 – 2.58 = 0.86 Polar covalent NO2 EN = 3.44 – 3.04 = 0.40 Polar covalent HBr EN = 2.96 – 2.20 = 0.76 Polar covalent H2 EN = 2.20 – 2.20 = 0.00 Non-polar covalent Na3N EN = 3.04 – 0.93 = 2.11 Mostly ionic CH4 EN = 2.55 – 2.20 = 0.35 Mostly covalent MgO EN = 3.44 – 1.31 = 2.13 Mostly ionic HCl EN = 3.16 – 2.20 = 0.96 Polar covalent O2 EN = 3.44 – 3.44 = 0.00 Non-polar covalent G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 44 of 50  CHM.5.1.01.011.12. Explain how the electronegativity difference is related to bond character The greater the electronegativity difference, the greater the ionic nature of the bond. A compound with a 50% ionic character has bonds that are mostly ionic. CHM.5.1.02.008.01. Describe conditions needed for a molecule to be polar For a molecule to be polar: 1) must have dipoles 2) the net dipole moment is not equal to zero CHM.5.1.02.008.01. Identify a polar compound from a list of compounds CHM.5.1.02.008.01. Determine, using the VSEPR model, the polarity of different molecules or ions (including exceptions to the octet rule) Electron domain Molecular Geometry Hybridization Polarity Geometry AX2 Linear Linear sp Non-polar AX3 Trigonal Planar Trigonal planar Non-polar sp2 AX2E Trigonal Planar Bent Polar AX4 Tetrahedral Tetrahedral Non-polar AX3E Tetrahedral Trigonal Pyramidal Polar sp3 Bent, v-shaped or AX2E2 Tetrahedral Polar angular AX5 Trigonal bipyramidal Trigonal bipyramidal Non-polar AX4E 3 Trigonal bipyramidal See-saw dsp AX3E2 Trigonal bipyramidal T-shaped AX6 Octahedral Octahedral Non-polar AX5E 2 3 Octahedral Square pyramidal d sp Polar AX4E2 Octahedral Square planar Non-polar This document was created for educational purposes. The document is not meant for publication or mass printing or production. It is made to support UAE Ministry of Education and Applied Technology High School students for AY 2021 – 2022. G10 Advanced Chemistry– CHM51 – Course Specification – Term 1 (Detailed KPIs) Page 45 of 50