Electrolysis of molten \(NaCl\) gives:

\(Na\) at cathode and \(Cl_2\) at anode

The correct answer is option 4. \(Na\) at cathode and \(Cl_2\) at anode.

The electrolysis of molten NaCl can be represented through the following chemical equations using LaTeX coding:

At the Cathode (Reduction):

The reduction half-reaction at the cathode involves the conversion of sodium ions (Na⁺) into sodium metal (Na) by gaining electrons:

\(\text{Cathode: } \quad 2\text{Na}^+ + 2\text{e}^- \rightarrow 2\text{Na}\)

At the Anode (Oxidation):
The oxidation half-reaction at the anode involves the conversion of chloride ions (Cl⁻) into chlorine gas (Cl₂) by losing electrons:

\(\text{Anode: } \quad 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-\)

Overall Reaction:

The overall reaction for the electrolysis of molten NaCl can be represented as the combination of these two half-reactions:

\(2\text{Na}^+ + 2\text{Cl}^- \rightarrow 2\text{Na} + \text{Cl}_2\)

However, it's important to note that in practice, the presence of water (H₂O) can lead to the production of sodium hydroxide (NaOH) and hydrogen gas (H₂) instead of the overall reaction you've mentioned. This is because water can be electrolyzed before it reacts with NaCl, leading to a different set of reactions:

\(2\text{NaCl} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2 + \text{Cl}_2\)

So, in industrial electrolysis of molten NaCl, steps are taken to minimize the presence of water to ensure the desired production of sodium metal and chlorine gas.