The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are

The reaction between elemental chlorine gas and an aqueous solution of sodium hydroxide at low temperature (cold, dilute NaOH) is an example of disproportionation, that is, the same element is simultaneously oxidised and reduced. First, let us assign oxidation numbers to understand what will happen.

In molecular chlorine $$Cl_2$$, the oxidation number of each chlorine atom is $$0$$. In chloride ion $$Cl^-$$, chlorine is at oxidation number $$-1$$, while in hypochlorite ion $$ClO^-$$ it is at oxidation number $$+1$$. Hence, during the reaction one Cl atom will be reduced from $$0$$ to $$-1$$ and another Cl atom will be oxidised from $$0$$ to $$+1$$.

We now write the two half-reactions separately.

Reduction half-reaction (gain of electrons):
$$Cl_2 + 2e^- \rightarrow 2Cl^-$$

Oxidation half-reaction (loss of electrons):
$$Cl_2 + 2H_2O \rightarrow 2ClO^- + 4H^+ + 2e^-$$

Both half-reactions involve the transfer of $$2e^-$$, so we can directly add them. Adding gives

$$\left(Cl_2 + 2e^- \rightarrow 2Cl^- \right) + \left(Cl_2 + 2H_2O \rightarrow 2ClO^- + 4H^+ + 2e^- \right)$$

Canceling the electrons on both sides we obtain

$$2Cl_2 + 2H_2O \rightarrow 2Cl^- + 2ClO^- + 4H^+$$

Because the actual medium is basic (aqueous NaOH provides $$OH^-$$ ions), we must remove the $$H^+$$ ions by adding the same number of $$OH^-$$ ions to both sides:

$$2Cl_2 + 2H_2O + 4OH^- \rightarrow 2Cl^- + 2ClO^- + 4H^+ + 4OH^-$$

Using the neutralisation relation $$H^+ + OH^- \rightarrow H_2O$$, the right-hand side contains $$4H^+ + 4OH^- = 4H_2O$$. Therefore we rewrite:

$$2Cl_2 + 2H_2O + 4OH^- \rightarrow 2Cl^- + 2ClO^- + 4H_2O$$

Cancel $$2H_2O$$ from both sides (because there are $$4H_2O$$ on the right and $$2H_2O$$ on the left), yielding

$$2Cl_2 + 4OH^- \rightarrow 2Cl^- + 2ClO^- + 2H_2O$$

Dividing the entire equation by $$2$$ gives the simplest stoichiometric form:

$$Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O$$

Re-inserting the spectator sodium cations that accompany the anions in solution, the overall molecular equation can be written as

$$Cl_2 + 2NaOH \rightarrow NaCl + NaClO + H_2O$$

Thus, the two ionic products formed are $$Cl^-$$ (chloride ion) and $$ClO^-$$ (hypochlorite ion). These correspond exactly to Option B.

Hence, the correct answer is Option 2.