University:
Seattle Pacific University
Course:
CHM 1211 | General Chemistry I
Academic year:

2022
Views:

103

Pages:

2
Author:

Maverick Martin

Bonding Basics - Ionic Bonds Answer Key Complete the chart for each element. Element # of Protons # of Electrons # of Valence Electrons Sodium 11 11 -> 10 1 Chlorine 17 17 -> 18 7 Beryllium 4 4 -> 2 2 Fluorine 9 9 -> 10 7 Lithium 3 3 -> 2 1 Oxygen 8 8 -> 10 6 Phosphorus 15 15 -> 18 5 Oxidation Number 1+ 12+ 11+ 23- NOTE: I have the students use a red pen/pencil to change the # of electrons to the amount it would be if the valence electrons were removed or added. They can see the difference between the number of protons (+) and electrons (-), which relates to the charge or oxidation number. If the ion has more protons (+), it would be a positive ion. If it has more electrons (-), it would be a negative ion. Follow your teacher’s directions to complete each ionic bond. Step 2 Step 3 (1) Potassium + Fluorine 1- Write the symbols for each element. 2 - Use Fruity Pebbles (or other cereal/candy with more than one color) to create the Lewis structure for each. K F Step 1 3 - Draw an arrow (or more if needed) to show the transfer of electrons and move the cereal to the new location. K1+F1- 4 - Determine the charge for each ion and write the formula. 5 - Make sure the sum of the oxidation numbers is zero and write the chemical formula. K would have a charge of 1+ since it lost an electron Fluorine would have a charge of 1- since it gained an electron. KF 6 - Have the students use a pencil or crayon to draw the electrons as they remove the pieces of cereal. (2) Magnesium + Iodine I Mg I Mg 2+ 1I2 MgI2 Mg would have a charge of 2+ since it lost two electrons. Each I ion would have a charge of 1- since each gained an electron. A subscript “2” is used to show that two ions were used in the bond. Students will start with one magnesium and one iodine atom. Since the oxidation numbers must equal zero, they will need to add another iodine atom. (3) Sodium + Oxygen Na Na O 1+ 2Na 2 O Na2O Students will start with one sodium and one oxygen atom. Since the oxidation numbers must equal zero, they will need to add another sodium atom. Each Na ion would have a charge of 1+ since each lost an electron. A subscript “2” is used to show that two ions were used in the bond. The O ion would have a charge of 2- since it gained two electrons. (4) Sodium + Chlorine Na Cl Na1+Cl1NaCl Na would have a charge of 1+ since it lost an electron Cl would have a charge of 1- since it gained an electron. (5) Calcium + Chlorine Cl Ca Cl 1Cl 2 2+ Ca CaCl2 Ca would have a charge of 2+ since it lost two electrons. Each Cl ion would have a charge of 1- since each gained an electron. A subscript “2” is used to show that two ions were used in the bond. (6) Aluminum + Chlorine Cl Al Cl Cl 3+ Al 1Cl 3 AlCl3 The Al ion would have a charge of 3+ since it lost three electrons. Each Cl ion would have a charge of 1- since each gained an electron. A subscript “3” is used to show that three ions were used in the bond.