Answer
$ClF_3$
Work Step by Step
The given reaction can be written in the form of chemical equation as
$Cl_2(g)+3F_2(g)\rightarrow 2Cl_aF_b(g)$
To balance the above equation, we have to write $1$ as subscript of $Cl$ and $3$ for $F$ in the product so that both the substances can have equal number of atoms on both sides, that is two of $Cl$ and six of $F$.
Now, the above equation will be written as
$Cl_2(g)+3F_2(g)\rightarrow 2ClF_3(g)$ and the required formula is $ClF_3$
