A metallic nitrate gives black precipitates with KI solution. On adding excess of KI, an orange-coloured solution is obtained. The ion giving these observations is

Bismuth cation gives black precipitates with KI solution. On adding excess of KI, an orange-coloured solution is obtained.

Due to the fact that potassium iodide is a source of iodide and that bismuth(III) iodide is insoluble in water, an aqueous solution can be used to detect the presence of $B{i}^{3+}$ ions. An effective test is one that produces a black precipitate of bismuth(III) iodide.
When heated in the presence of halide donors, bismuth(III) iodide transforms into iodobismuth(III) anions.

The following reaction is obtained,

2KI+$B{i}^{3+}$​→$K{2}_2\left[Bi{I}_5\right]$(orange solution)

Hence, option C is correct.