Chemistry 101 In-Class Assignment 1 (Due by Sept. 24) Ions & Ionic Compounds and an Introduction to Lewis Dot Structures 1. Draw lewis dot structures for an atom of each of following elements: (a) K (c) Ar (b) Si (d) As 2. Identify the ion likely to be formed by the following elements: (a) Ca (e) Cl (b) P (f) Al (c) Rb (g) Cs (d) Ba (h) Se 3. Predict the formulas of the ionic compounds that would be formed by the combination of each pair of elements. (a) Ca and Br (c) Al and O (b) Li and F (d) Mg and N 4. Name the following compounds: (a) RbCl (b) Na2 O (c) AlCl3 (d) BaO 5. What are the chemical formulas for: (a) potassium ﬂuoride (b) magnesium oxide (c) calcium chloride (d) aluminum bromide Naming binary ionic compounds. Binary ionic compounds (binary = composed of 2 elements) are named in the following matter: (a) The name of the positive metal ion comes ﬁrst. This is followed by the name of the negative ion. (b) For positive ions, we typically use the name of the element. (c) For negative ions, we use the element name with the ending changed to -ide. examples: • NaCl = sodium chloride • Mg3 N2 = magnesium nitride Chemistry 101, Fall 2003 In-Class Assignment 4, page 2 Metals that form more than one type of ion (text section: 4.4) Many transition metals commonly form more than one type of ion. Iron, for instance, forms a +2 ion as well as a +3 ion. Fe2+ and Fe3+ Just a few main group metals do this as well - such as tin and lead. To identify compounds containing such elements we must specify the charge on the metal ion. This is done in two diﬀerent ways. The ﬁrst is through common names that have been used historically for the diﬀerent ions. The other is through a formal naming system which uses a roman numeral after the element name, indicating the charge: Element Ions common name (historical) systematic name Iron Fe2+ ferrous iron (II) Fe3+ ferric iron (III) Copper Cu+ cuprous copper (I) Cu2+ cupric copper (II) Gold Au+ aurous gold (I) Au3+ auric gold (III) Tin Sn2+ stannous tin (II) Sn4+ stannic tin (IV) Lead Pb2+ plumbous lead (II) Pb4+ plumbic lead (IV) The cation name is used in combination with the anion name to name the compound. By balancing the charges of the ions, you can also determine the formula of the compound. For example: • – ferrous chloride = iron (II) chloride = FeCl2 ferrous = Fe2+ , need 2 Cl− ’s to balance charge – ferric chloride = iron (III) chloride = FeCl3 ferric = Fe3+ , need 3 Cl− ’s to balance charge 6. Try this by completing the following table: . common name . . formal name . . formula . ferrous chloride iron (II) chloride FeCl2 cupric chloride AuBr3 tin (IV) sulﬁde plumbic oxide SnCl2 tin (IV) oxide Chemistry 101, Fall 2003 In-Class Assignment 4, page 3 Covalent Bonds (Text section: 4.5) An atom can achieve a noble gas conﬁguration by sharing electrons with other atoms. We have seen that metals and nonmetals combine to form ionic compounds. However, when two nonmetals combine, they typically share electrons in covalent bonds and form what are know as covalent compounds. For example, consider two hydrogen atoms. Each one has one valence electron. However, both hyrogens would like to have two electrons, to obtain the same conﬁguration as [He]. To do this, each H atom shares .H an electron with the other, forming a covalent bond. In H2 , each H atom has 2 electrons. H . + .H H . H H H2 a hydrogen molecule H’s share e−’s (diatomic = 2 atoms) forming a covalent each H has 1 e− bond shared pairs can and wants 1 more be represented as lines between atoms Here’s another example, this time for chlorine. A Cl atom has 7 valence electrons, so it needs 1 more to achieve an octet. . . Cl . . . . . . . . . . . . . . . . . . . . . . . .Cl + . . . . Cl Cl . Cl Cl . Cl 2 . . . . . . . . a chlorine molecule each Cl shares 1 e−’s forming a covalent each Cl has 7 e− bond consisting of 2 electrons the shared electron pair and wants 1 more is represented as a line Each chlorine now has 8 electrons, because you can count the shared ones: this Cl now has an this Cl atom also has an octet . . . . octet of 8 electrons . . . . . . Cl Cl . . . . Sometimes you need to share more than one pair of electrons to achieve an octet: . .O . . . . . . . O + . . .. . . .. . O O . . O O . . O2 . . . . . . . . . . an oxygen molecule each O shares 2e−’s forming a double each O has 6 e− covalent bond the four shared electrons and wants 2 more are represented as two lines between the atoms Chemistry 101, Fall 2003 In-Class Assignment 4, page 4 We can check to see that each oxygen has the appropriate number of electrons: each oxygen atom total # valence . .. . . .. . O O now has an octet electrons = 12 (note the 12 dots) . . . . For problems 7-12, calculate the total number of valence electrons in each of the following molecules. Then, draw lewis dot structures for each: 7. Br2 # valence e− 8. HCl # valence e− 9. HBr # valence e− 10. N2 # valence e− 11. NO # valence e− This one is tricky, there is no way to achieve an octet. So, get each atom as close to an octet as possible without going over. 12. SiH4 # valence e− Another tricky one. Take Si to be the central atom and arrange the H atoms around it. Try to get an octet around silicon and two electrons around each of the hydrogens.