Calculate the `EA` of `O` atom to `O^(2-)` ion from the following data:
i. `Delta_(f)H^(ɵ)[MgO(s)] = -600 kJ mol^(-1)`
ii. `Delta_(u)H^(ɵ) [MgO(s)] = -3860 kJ mol^(-1)`
iii. `IE_(1) + IE_(2)` of `Mg(g) = 2170 kJ mol^(-1)`
iv. `Delta_("diss")H^(ɵ)` of `Mg(s) = + 494 kJ mol^(-1)`
v. `Delta_("sub")H^(ɵ)` of `Mg(s) = +150 kJ mol^(-1)`

Options

A. `+693 kJ mol^(-1)`
B. `-693 kJ mol^(-1)`
C. `+69.3 kJ mol^(-1)`
D. `-69.3 kJ mol^(-1)`

Explanation

To calculate the electron affinity (EA) of an oxygen atom to form the O²⁻ ion using the provided data, we can apply Hess's law and construct a Born-Haber cycle for the formation of MgO. Let's break down the steps:

### Step 1: Write the Formation Reaction
The formation of magnesium oxide (MgO) from magnesium (Mg) and oxygen (O) can be represented as:
\[ \text{Mg(s)} + \frac{1}{2} \text{O}_2(g) \rightarrow \text{MgO(s)} \]
The enthalpy change for this reaction is the standard enthalpy of formation (\( \Delta_f H^\circ \)) of MgO, which is given as -600 kJ/mol.

### Step 2: Identify the Energy Changes
We need to consider the following energy changes involved in the formation of MgO:
1. **Sublimation of Mg(s)**:
\[ \Delta_{\text{sub}} H^\circ = +150 \, \text{kJ/mol} \]
2. **Ionization Energies of Mg**:
\[ \text{IE}_1 + \text{IE}_2 = 2170 \, \text{kJ/mol} \]
3. **Dissociation of O₂**:
\[ \Delta_{\text{diss}} H^\circ = +494 \, \text{kJ/mol} \]
Since we need half a mole of O₂, we divide this by 2:
\[ \frac{494}{2} = 247 \, \text{kJ/mol} \]
4. **Electron Affinity of O**: This is the value we need to calculate (EA).
5. **Lattice Energy of MgO**:
\[ \Delta_u H^\circ = -3860 \, \text{kJ/mol} \]

### Step 3: Set Up the Born-Haber Cycle Equation
Using Hess's law, we can set up the following equation based on the energy changes:
\[ \Delta_f H^\circ = \Delta_{\text{sub}} H^\circ + (\text{IE}_1 + \text{IE}_2) + \frac{1}{2} \Delta_{\text{diss}} H^\circ + \text{EA} + \Delta_u H^\circ \]
Substituting the known values:
\[ -600 = 150 + 2170 + 247 + \text{EA} - 3860 \]

### Step 4: Solve for EA
Now, we can rearrange the equation to solve for EA:
\[ \text{EA} = -600 - 150 - 2170 - 247 + 3860 \]
Calculating the right-hand side:
\[ \text{EA} = -600 - 150 - 2170 - 247 + 3860 \]
\[ \text{EA} = -600 - 150 - 2170 - 247 + 3860 = 693 \, \text{kJ/mol} \]

### Final Result
Thus, the electron affinity of the oxygen atom to form the O²⁻ ion is:
\[ \text{EA} = +693 \, \text{kJ/mol} \]