Formulae for Nessler's reagent is :

Solution

Chemistry ma’am · 4 steps

Build it the way the lab does: HgCl₂, then excess KI, and count the iodides around mercury.

1. First the precipitate. Add KI to mercury(II) chloride and a red precipitate of HgI₂ forms: HgCl₂ + 2KI → HgI₂ v + 2KCl.
2. Excess KI dissolves it. Keep adding KI and the red solid dissolves. Two more iodide ions bind to Hg²⁺ and form the soluble complex ion [HgI₄]²⁻.

   HgI₂ + 2KI → K₂[HgI₄]

3. Check the charge. Hg²⁺ with four I⁻ gives a net charge of -2, so two K⁺ ions balance it. The reagent is this salt, potassium tetraiodomercurate(II), made alkaline with KOH. It gives a brown precipitate with NH₄⁺ ions.
4. So. Nessler's reagent is K₂HgI₄, option (d).

The idea to keep. Nessler's reagent is alkaline K₂[HgI₄]: the red HgI₂ dissolves in excess KI. If it turns brown with a sample, think NH₄⁺.

Answer from the NTA final answer key. Solution by PrepWiser.