Aryl fluoride may be prepared from arene diazonium chloride using:
(A) $HB{F_4}/\Delta $
(B) $HB{F_4}/NaN{O_2},Cu$
(C) $CuF/HF$
(D) $Cu/HF$

Hint: Under the presence of nitrous acid, aromatic amines are diazotized. aromatic diazonium salts are considered as one of the intermediates in the preparation of a variety of aromatic compounds including dyes. The Diazonium group is a very good leaving group, so it can be substituted by other groups.

Complete step-by-step answer:We know that aryl fluoride can form by using arene diazonium chloride by using a well known chemistry reaction- Balz-Schiemann reaction. When aryl diazonium chloride is reacted with fluoroboric acid, arene diazonium fluoroborate is precipitated which on further heating decomposes to yield aryl fluoride.
So, by writing the whole reaction we get the following data, which is as follows,


As in the above reaction we can see that when diazonium chloride is reacted as in the presence of $HB{F_4}/\Delta $ it forms the Aryl Fluoride as we see all the steps in the above reaction as with this we get the answer as $HB{F_4}/\Delta $ .
Therefore, the correct answer from this is $HB{F_4}/\Delta $ .

Option ‘A’ is correct

Note: A Point should be noted that Aryl Fluoride forms from Arene diazonium chloride in the presence of $HB{F_4}/\Delta $ . In this reaction the Diazonium group is substituted by the F-group. This is an exothermic reaction as heat is used in this process.