Fluorination of an aromatic ring is easily accomplished by treating a diazonium salt with $$HBF_4$$. Which of the following conditions is correct about this reaction?

We start from an aromatic amine, say $$C_6H_5NH_2$$. By the usual diazotisation procedure (treating with $$NaNO_2$$ and $$HCl$$ at $$0^\circ{\rm C}$$) we obtain the diazonium chloride:

$$C_6H_5NH_2 + NaNO_2 + 2HCl \rightarrow C_6H_5N_2^+Cl^- + NaCl + 2H_2O$$

For introducing fluorine, we replace the chloride ion by the tetrafluoroborate ion simply by adding aqueous $$HBF_4$$. A straightforward metathesis gives the stable diazonium tetrafluoroborate salt:

$$C_6H_5N_2^+Cl^- + HBF_4 \rightarrow C_6H_5N_2^+BF_4^- + HCl$$

The key transformation, called the Balz-Schiemann reaction, is the thermolysis of this diazonium tetrafluoroborate. On gentle heating the diazonium group is expelled as nitrogen gas while the aromatic ring captures a fluoride ion. The complete stoichiometric equation is

$$C_6H_5N_2^+BF_4^- \xrightarrow{\; \text{heat} \;} C_6H_5F + BF_3 + N_2 \uparrow$$

There is no need for any metal catalyst such as copper, nor any external reagents like $$NaNO_2$$ or $$NaF$$ at this stage. The reaction proceeds cleanly by simply warming the dry diazonium tetrafluoroborate; the generation of the gaseous products $$N_2$$ and $$BF_3$$ drives the equilibrium to completion.

So among the given choices, the process requires nothing more than heating the isolated diazonium tetrafluoroborate. Hence, the statement that fits correctly is “Only Heat”.

Hence, the correct answer is Option C.