The number of halides that do not exist among the following:${\mathrm{CuF}}_2,{\mathrm{CuCl}}_2,{\mathrm{CuBr}}_2,{\mathrm{CuI}}_2,\mathrm{CuCl},\mathrm{CuBr},{\mathrm{Cu}}_2{\mathrm{I}}_2,\mathrm{CuF}$

${\mathrm{CuBr}}_2,{\mathrm{CuI}}_2,\mathrm{CuF}$ does not exist .Strong reducing agents include iodide ions. Due to this reduction, copper (II) iodide becomes insoluble copper (I) iodide. ${\mathrm{CuI}}_2$ does not exist in solution since it is not stable. Additionally, ${\mathrm{CuBr}}_2$ and $\mathrm{CuF}$ don't exist.