Setup of Born-Haber cycle, calculation lattice energy of `MgO_((s))`. The given that-enthalpy of formation of `MgO_((s))= -602`, sublimation of `Mg_((s))=148 , 1^(st) & 2^(nd)` ionization energy of `Mg= 738 & 1450` respectively. For Oxygen bond dissocitation energy `=498, 1^(st)&2^(nd)` electron gain enthalpy `= -141 & 844` respectively (all unit in `kJ "mole"^(-1)`).

Answer

To calculate the lattice energy of `MgO` using the Born-Haber cycle, we can follow these steps:

### Step 1: Write the Reaction for Formation of `MgO`
The formation of magnesium oxide from its elements can be represented as:
\[ \text{Mg (s)} + \frac{1}{2} \text{O}_2 (g) \rightarrow \text{MgO (s)} \]

### Step 2: Identify the Given Data
We have the following data:
- Enthalpy of formation of `MgO (s)`: \( \Delta H_f = -602 \, \text{kJ/mol} \)
- Sublimation energy of `Mg (s)`: \( \Delta H_{\text{subl}} = 148 \, \text{kJ/mol} \)
- First ionization energy of `Mg`: \( IE_1 = 738 \, \text{kJ/mol} \)
- Second ionization energy of `Mg`: \( IE_2 = 1450 \, \text{kJ/mol} \)
- Bond dissociation energy of `O2`: \( D = 498 \, \text{kJ/mol} \)
- First electron gain enthalpy of `O`: \( E_{a1} = -141 \, \text{kJ/mol} \)
- Second electron gain enthalpy of `O`: \( E_{a2} = 844 \, \text{kJ/mol} \)

### Step 3: Set Up the Born-Haber Cycle
1. **Sublimation of Magnesium**:
\[ \text{Mg (s)} \rightarrow \text{Mg (g)} \quad \Delta H = +148 \, \text{kJ/mol} \]

2. **Ionization of Magnesium**:
\[ \text{Mg (g)} \rightarrow \text{Mg}^{2+} (g) + 2e^- \quad \Delta H = IE_1 + IE_2 = 738 + 1450 = 2188 \, \text{kJ/mol} \]

3. **Dissociation of Oxygen**:
\[ \frac{1}{2} \text{O}_2 (g) \rightarrow \text{O} (g) \quad \Delta H = \frac{1}{2} D = \frac{498}{2} = 249 \, \text{kJ/mol} \]

4. **Electron Gain by Oxygen**:
\[ \text{O} (g) + 2e^- \rightarrow \text{O}^{2-} (g) \quad \Delta H = E_{a1} + E_{a2} = -141 + 844 = 703 \, \text{kJ/mol} \]

5. **Formation of `MgO` from Ions**:
\[ \text{Mg}^{2+} (g) + \text{O}^{2-} (g) \rightarrow \text{MgO (s)} \quad \Delta H = -U \, \text{(Lattice Energy)} \]

### Step 4: Write the Energy Balance Equation
Using Hess's law, we can write the equation for the Born-Haber cycle:
\[ \Delta H_f = \Delta H_{\text{subl}} + (IE_1 + IE_2) + \frac{1}{2}D + (E_{a1} + E_{a2}) - U \]
Substituting the values:
\[ -602 = 148 + 2188 + 249 + 703 - U \]

### Step 5: Solve for Lattice Energy (U)
Rearranging the equation to solve for \( U \):
\[ -602 = 148 + 2188 + 249 + 703 - U \]
\[ -602 = 3288 - U \]
\[ U = 3288 + 602 = 3890 \, \text{kJ/mol} \]

### Final Result
The lattice energy of `MgO` is:
\[ U = 3890 \, \text{kJ/mol} \]