In the metallurgy of \(Al\), purified\(Al_2O_3\) is mixed with \(Na_3AlF_6\) or \(CaF_2\). The role of \(Na_3AlF_6\) or \(CaF_2\) is to:

Lower the melting point of \(Al_2O_3\)

The correct answer is option 2. Lower the melting point of \(Al_2O_3\).

In the metallurgy of aluminum (\(Al\)), the Hall-Héroult process is commonly used. In this process, purified \(Al_2O_3\) (alumina) is mixed with cryolite (\(Na_3AlF_6\)) or fluorspar (\(CaF_2\)). The primary role of \(Na_3AlF_6\) or \(CaF_2\) in this process is to Lower the melting point of \(Al_2O_3\)

Alumina (\(Al_2O_3\)) has a very high melting point (over 2000°C), making it impractical to reduce it directly using conventional methods.
By adding cryolite (\(Na_3AlF_6\)) or fluorspar (\(CaF_2\)), the mixture forms a eutectic mixture with a lower melting point. The eutectic mixture has a lower overall melting point than pure alumina, making the process more energetically efficient.

This lower melting point allows the reduction of alumina to aluminum to occur at a more reasonable temperature, saving energy in the overall process.

So, the correct option is 2. Lower the melting point of \(Al_2O_3\).