This is a very good question. These compounds have been studied for long enough that most people working with them "know" the answers to these questions, but the answers are buried in physical chemistry articles from 20 or more years ago. The imidazolium tetrafluoroborate salts are popular RTILs because 1) they have the "right" properties, and 2) they are easy to make.
Concerning these compounds having the "right" properties:
Most RTILs require two characteristics in the their constituent ions to minimize the intermolecular order which would cause them to solidify. First, they need a cation-anion pair that cannot associate strongly, and second they need to have lower symmetry. Immidazolium tetrafluoroborate (BF4-) salts have both of these. The aromaticity of imidazole is important. The positive charge of the imidazolium cation is distributed through the resonance of the aromatic pi system. The negative charge in BF4- is distributed inductively over the four fluorine atoms. Additionally, imidazolium is planar and BF4- is spherical. Both the diffuse charge distribution and poor shape match prevent these two ions from having a strong association. The two different alkyl groups (ethyl and methyl) on the imidazolium ring are enough to break the symmetry of the system and prevent more efficient packing, which would lead to the substance being solid.
Concerning their ease of synthesis:
The imidazolium salts are easy to prepare from imidazole and various alkyl halides, all commercially available and inexpensive. The predominance of the BF4- anion is also a result of the synthesis. The synthesis of an imidazolium salt RTIL begins with imidazole and and an alkyl halide, say chloromethane. The product of this step is 1-methylimidazolium chloride. This salt is reacted with another alkyl halide, say chloroethane, in the presence of a base, say triethylamine. The products of the this step are 1-ethyl-3-methylimidazolium chloride and triethylammonium chloride. 1-ethyl-3-methylimidazolium chloride is a solid, because the chloride ion is small, with concentrated charge. The tetrafluoroborate salt is prepared by salt metathesis. 1-ethyl-3-methylimidazolium chloride is dissolved in water and an aqueous solution of silver tetrafluoroborate (AgBF4) is added. Silver chloride, AgCl, is virtually insoluble in water (Ksp = 1.77 x 10-10) and precipitates. The solution is filtered, and the filtrate is concentrated to produce 1-ethyl-3-methylimidazolium tetrafluoroborate (EMIM BF4-). AgBF4 and AgPF6 are commercially available, so you will see those ions more frequently. The other ions, while they have good properties for ionic liquids, are not usually available as silver salts, so those reagents have to be prepared. You are already doing three reactions to get to EMIM BF4-. Who wants to do more synthesis just to make their solvents?
