Question

part 3a 12 marks choose 4 of 5 (lesson 3a) draw the following ionic bonds and state the formula a) lithium and phosphorus b) calcium and chlorine c) magnesium and nitrogen d) aluminum and chlorine e) potassium and bromine

Explanation:

Step1: Determine charges of ions

Lithium (Li) is in Group 1, so it forms a +1 ion ($Li^+$). Phosphorus (P) is in Group 15, so it forms a -3 ion ($P^{3 -}$).

Step2: Determine the formula

To balance the charges, we need 3 lithium ions for every 1 phosphorus ion. The formula is $Li_3P$.

Step3: Draw the ionic bond (simplified description)

Lithium loses its outer - shell electron to become a positively charged ion. Phosphorus gains 3 electrons to become a negatively charged ion. The electrostatic attraction between the $Li^+$ and $P^{3 -}$ ions forms the ionic bond.

Step1: Determine charges of ions

Calcium (Ca) is in Group 2, so it forms a +2 ion ($Ca^{2+}$). Chlorine (Cl) is in Group 17, so it forms a -1 ion ($Cl^-$).

Step2: Determine the formula

To balance the charges, we need 2 chlorine ions for every 1 calcium ion. The formula is $CaCl_2$.

Step3: Draw the ionic bond (simplified description)

Calcium loses 2 outer - shell electrons to become a positively charged ion. Each chlorine atom gains 1 electron to become a negatively charged ion. The electrostatic attraction between the $Ca^{2+}$ and two $Cl^-$ ions forms the ionic bond.

Step1: Determine charges of ions

Magnesium (Mg) is in Group 2, so it forms a +2 ion ($Mg^{2+}$). Nitrogen (N) is in Group 15, so it forms a -3 ion ($N^{3 -}$).

Step2: Determine the formula

To balance the charges, we need 3 magnesium ions and 2 nitrogen ions. The formula is $Mg_3N_2$.

Step3: Draw the ionic bond (simplified description)

Magnesium atoms lose 2 outer - shell electrons each to become positively charged ions. Nitrogen atoms gain 3 electrons each to become negatively charged ions. The electrostatic attraction between the $Mg^{2+}$ and $N^{3 -}$ ions forms the ionic bond.

Step1: Determine charges of ions

Aluminum (Al) is in Group 13, so it forms a +3 ion ($Al^{3+}$). Chlorine (Cl) is in Group 17, so it forms a -1 ion ($Cl^-$).

Step2: Determine the formula

To balance the charges, we need 3 chlorine ions for every 1 aluminum ion. The formula is $AlCl_3$.

Step3: Draw the ionic bond (simplified description)

Aluminum loses 3 outer - shell electrons to become a positively charged ion. Each chlorine atom gains 1 electron to become a negatively charged ion. The electrostatic attraction between the $Al^{3+}$ and three $Cl^-$ ions forms the ionic bond.

Answer:

a) Formula: $Li_3P$