Questions: Use Coulomb's law to calculate the energy of a magnesium ion and an oxide ion at their equilibrium ion-pair separation distance. Ion Radius (pm) Mg2+ 72 O2- 140

Solution Steps

Coulomb's law describes the electrostatic interaction between two charged particles. The energy \( E \) of interaction between two ions can be calculated using the formula:

\[ E = \frac{k \cdot q_1 \cdot q_2}{r} \]

where:

• \( k \) is Coulomb's constant, approximately \( 8.9875 \times 10^9 \, \text{N m}^2/\text{C}^2 \),
• \( q_1 \) and \( q_2 \) are the charges of the ions,
• \( r \) is the distance between the centers of the two ions.

For a magnesium ion (\(\text{Mg}^{2+}\)) and an oxide ion (\(\text{O}^{2-}\)):

• The charge \( q_1 \) for \(\text{Mg}^{2+}\) is \( +2e \),
• The charge \( q_2 \) for \(\text{O}^{2-}\) is \( -2e \),
• The elementary charge \( e \) is approximately \( 1.6022 \times 10^{-19} \, \text{C} \).

The radii of the ions are given as:

• \(\text{Mg}^{2+}\) radius = 72 pm,
• \(\text{O}^{2-}\) radius = 140 pm.

The equilibrium ion-pair separation distance \( r \) is the sum of the radii of the two ions:

\[ r = 72 \, \text{pm} + 140 \, \text{pm} = 212 \, \text{pm} = 212 \times 10^{-12} \, \text{m} \]

Substitute the values into Coulomb's law:

\[ E = \frac{(8.9875 \times 10^9 \, \text{N m}^2/\text{C}^2) \cdot (2 \times 1.6022 \times 10^{-19} \, \text{C}) \cdot (-2 \times 1.6022 \times 10^{-19} \, \text{C})}{212 \times 10^{-12} \, \text{m}} \]

Simplifying the expression:

\[ E = \frac{(8.9875 \times 10^9) \cdot (4 \times (1.6022 \times 10^{-19})^2)}{212 \times 10^{-12}} \]

\[ E = \frac{(8.9875 \times 10^9) \cdot (4 \times 2.5669 \times 10^{-38})}{212 \times 10^{-12}} \]

\[ E = \frac{(8.9875 \times 10^9) \cdot 1.0268 \times 10^{-37}}{212 \times 10^{-12}} \]

\[ E = \frac{9.2265 \times 10^{-28}}{212 \times 10^{-12}} \]

\[ E = -4.352 \times 10^{-19} \, \text{J} \]

Final Answer