A metal gives two chlorides A and B. A gives black precipitate with NaOH and B gives white. With KI, B gives a red precipitate which is soluble in excess of KI. A and B are respectively

Question

A metal gives two chlorides A and B. A gives black precipitate with NaOH and B gives white. With KI, B gives a red precipitate which is soluble in excess of KI. A and B are respectively (A) HgCl2 and Hg2Cl2 (B) Hg2Cl2 and Hg2Cl2 (C) HgCl2 and ZnCl2 (D) ZnCl2 and HgCl2

Tardigrade Admin · Accepted Answer

Hg2Cl2+2NH4OH → (A) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/979a3c68cc10c73577d23c772ff15f72-.jpeg" />+NH4Cl+2H2O underset(B)HgCl2+2NH4OH → underset textwhite ppt.HgNH2Cl+NH4Cl+2H2O underset(B)HgCl2+2KI → underset textRed ppt.HgI2+2KCl; HgI2+2KI → underset textSolubleK2[HgI4]

Q. A metal gives two chlorides $A$ and $B$ . A gives black precipitate with $N a O H$ and $B$ gives white. With $K I$ , $B$ gives a red precipitate which is soluble in excess of $K I$ . $A$ and $B$ are respectively

$H g C l_{2}$ and $H g_{2} C l_{2}$

$H g_{2} C l_{2}$ and $H g_{2} C l_{2}$

$H g C l_{2}$ and $Z n C l_{2}$

$Z n C l_{2}$ and $H g C l_{2}$

$H g_{2} C l_{2} + 2 N H_{4} O H \to$
(A) $+ N H_{4} Cl + 2 H_{2} O$
$(B) H g C l_{2} + 2 N H_{4} O H \to white ppt. H g N H_{2} Cl + N H_{4} Cl + 2 H_{2} O$
$(B) H g C l_{2} + 2 K I \to Red ppt. H g I_{2} + 2 K Cl;$
$H g I_{2} + 2 K I \to Soluble K_{2} [H g I_{4}]$

Q. A metal gives two chlorides A and B. A gives black precipitate with NaOH and B gives white. With KI, B gives a red precipitate which is soluble in excess of KI. A and B are respectively

HgCl2​ and Hg2​Cl2​

Hg2​Cl2​ and Hg2​Cl2​

HgCl2​ and ZnCl2​

ZnCl2​ and HgCl2​