Mercury is not considered as transition metal however it has tendency…

The correct answer is Option (4) → X = HgCl₂ , Y = Hg₂Cl₂

Mercury forms two chlorides: mercury(II) chloride, HgCl₂ (corrosive sublimate, soluble in water) and mercury(I) chloride, Hg₂Cl₂ (calomel, insoluble). In mercury(I) the cation is the dimer \( \ce{Hg2^{2+}} \) with an Hg–Hg bond, so “HgCl” is not a correct formula.

X gives a red precipitate with KI: \( \ce{HgCl2} \) gives scarlet-red mercury(II) iodide, which dissolves in excess KI as \( \ce{K2[HgI4]} \) (Nessler’s reagent):

    \( \displaystyle \ce{HgCl2 + 2KI -> HgI2 v + 2KCl} \)

    \( \displaystyle \ce{HgI2 + 2KI -> K2[HgI4]} \)

Y turns black with NH₄OH: calomel disproportionates; finely divided black mercury and white amidomercury(II) chloride are formed together, so the solid looks black:

    \( \displaystyle \ce{Hg2Cl2 + 2NH3 -> Hg + Hg(NH2)Cl + NH4Cl} \)

So X = HgCl₂ and Y = Hg₂Cl₂.

Why the other options are incorrect

• Option (1) X = Hg₂Cl₂ , Y = HgCl — Incorrect. X and Y are swapped, and HgCl is not a real formula of mercury(I) chloride.

• Option (2) X = Hg₂Cl₂ , Y = HgCl₂ — Incorrect. The two are swapped: it is \( \ce{HgCl2} \) that gives the red \( \ce{HgI2} \), and \( \ce{Hg2Cl2} \) that blackens with ammonia.

• Option (3) X = HgCl , Y = HgCl₂ — Incorrect. HgCl is not a correct formula, and \( \ce{HgCl2} \) does not give a black product with ammonia (it gives a white precipitate).

Note: The question is from beyond the NCERT textbook.