One mole of magnesium in the vapor state absorbed `1200 kJ mol^(-1)` of enegry. If the first and second ionization energies of `Mg` are `750` and `1450 kJ mol^(-1)`, respectively, the final composition of the mixture is

Options

A. `86%Mg^(+)+14%Mg^(2+)`
B. `69%Mg^(+) +31%Mg^(2+)`
C. `14%Mg^(+)+86%Mg^(2+)`
D. `31%Mg^(+)+69%Mg^(2+)`

Answer

To solve the problem, we need to analyze the energy absorbed by one mole of magnesium in the vapor state and how it relates to the ionization energies of magnesium. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Ionization Energies
The first ionization energy of magnesium (Mg) is the energy required to remove one electron from a neutral magnesium atom to form a magnesium ion (Mg⁺):
\[ \text{Mg} \rightarrow \text{Mg}^+ + e^- \quad (IE_1 = 750 \text{ kJ/mol}) \]

The second ionization energy is the energy required to remove a second electron from the Mg⁺ ion to form Mg²⁺:
\[ \text{Mg}^+ \rightarrow \text{Mg}^{2+} + e^- \quad (IE_2 = 1450 \text{ kJ/mol}) \]

### Step 2: Calculate Total Ionization Energy Required
To fully ionize one mole of magnesium to Mg²⁺, we need to sum the first and second ionization energies:
\[ \text{Total Ionization Energy} = IE_1 + IE_2 = 750 \text{ kJ/mol} + 1450 \text{ kJ/mol} = 2200 \text{ kJ/mol} \]

### Step 3: Determine the Energy Absorbed
We know that one mole of magnesium in the vapor state absorbed 1200 kJ of energy. This energy will be used to ionize magnesium.

### Step 4: Calculate Energy Consumed for Ionization
The energy absorbed (1200 kJ) is less than the total energy required to fully ionize magnesium (2200 kJ). Therefore, we can find out how much energy is used for the first ionization:
\[ \text{Energy Consumed for First Ionization} = 1200 \text{ kJ} - 750 \text{ kJ} = 450 \text{ kJ} \]

### Step 5: Calculate the Percentage of Mg²⁺ Formed
Now, we need to determine how much of the absorbed energy (450 kJ) contributes to the second ionization:
\[ \text{Percentage of Mg}^{2+} = \left( \frac{\text{Energy Consumed for Second Ionization}}{\text{Second Ionization Energy}} \right) \times 100 \]
Since the second ionization energy is 1450 kJ/mol, we can find the fraction of Mg⁺ that gets converted to Mg²⁺:
\[ \text{Fraction of Mg}^{2+} = \frac{450 \text{ kJ}}{1450 \text{ kJ}} \approx 0.3103 \]
Thus, the percentage of Mg²⁺ is:
\[ \text{Percentage of Mg}^{2+} \approx 31\% \]

### Step 6: Calculate the Percentage of Mg⁺
The remaining percentage will be for Mg⁺:
\[ \text{Percentage of Mg}^+ = 100\% - 31\% = 69\% \]

### Final Composition
Thus, the final composition of the mixture is:
- Mg⁺: 69%
- Mg²⁺: 31%

### Conclusion
The correct answer is that the final composition of the mixture consists of 69% Mg⁺ and 31% Mg²⁺.

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