The correct statement of the following is :

I^- ion is a stronger reducing agent therefore, it reduces Fe³⁺ to Fe²⁺ at room temperature. Therefore, FeI₃ is not stable.

$2{\mathrm{FeI}}_3\to 2{\mathrm{FeI}}_2+{\mathrm{I}}_2$

Lead acetate forms yellow ppt with K₂Cr₂O₇ without presence of any acid or in the absence of the acidic medium

$\mathrm{Pb}{\left({\mathrm{CH}}_3\mathrm{COO}\right)}_2+{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\to \underset{\mathrm{Yellow} \mathrm{ppt}}{{\mathrm{PbCrO}}_4\downarrow }+2{\mathrm{CH}}_3\mathrm{COOK}$

Iodide ion has a large size and is a strong reducing agent. So, when copper cation comes in contact with iodide anion, iodide get oxidized to iodine molecule and formation of complex is not takes place

In aqueous solution, Cu and Cu²⁺ are present but not Cu⁺.