The quantity of charge required to obtain one mole of aluminum from $Al_2O_3$ is:

3 F

The correct answer is Option (4) → 3 F.

To determine the quantity of charge required to obtain one mole of aluminum from \( \text{Al}_2\text{O}_3 \), we first need to look at the electrochemical process involved in the reduction of aluminum oxide to aluminum.

The reduction of aluminum oxide (\( \text{Al}_2\text{O}_3 \)) can be represented by the following reaction in an electrolytic cell:

\(\text{Al}_2\text{O}_3 + 3\text{C} \rightarrow 4\text{Al} + 3\text{CO}_2\)

From this reaction, we see that 1 mole of \( \text{Al}_2\text{O}_3 \) produces 2 moles of aluminum (\( \text{Al} \)).

The reduction of aluminum ions (\( \text{Al}^{3+} \)) to aluminum metal involves the following half-reaction:

\(\text{Al}^{3+} + 3\text{e}^- \rightarrow \text{Al}\)

This indicates that 3 moles of electrons (3 Faradays, \( 3F \)) are required to reduce 1 mole of aluminum ions to form 1 mole of aluminum.

Since from 1 mole of \( \text{Al}_2\text{O}_3 \), we can produce 2 moles of aluminum, we need to calculate the charge for 2 moles of aluminum:

\(\text{Total charge} = \text{moles of aluminum} \times \text{moles of electrons per mole of aluminum}\)

For 1 mole of aluminum:

\(\text{Charge for 1 mole of Al} = 3F\)

Conclusion

Therefore, the quantity of charge required to obtain one mole of aluminum from \( \text{Al}_2\text{O}_3 \) is: 3 F