Total number of moles of AgCl precipitated on addition of excess of AgNO 3 to one mole each of the following complexes [ Co ( NH 3)4 Cl 2] Cl, [ Ni ( H 2 O )6] Cl 2,[ Pt ( NH 3)2 Cl 2] and [ Pd ( NH 3)4] Cl 2 is

Question

Total number of moles of AgCl precipitated on addition of excess of AgNO 3 to one mole each of the following complexes [ Co ( NH 3)4 Cl 2] Cl, [ Ni ( H 2 O )6] Cl 2,[ Pt ( NH 3)2 Cl 2] and [ Pd ( NH 3)4] Cl 2 is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

[ Co ( NH 3)4 Cl 2] Cl ⇒ Gives 1 mole AgCl [ Ni ( H 2 O )6] Cl 2 ⇒ Gives 2 moles AgCl [ Pt ( NH 3)2 Cl 2] ⇒ Gives No AgCl [ Pd ( NH 3)4] Cl 2 ⇒ Gives 2 moles AgCl Total number of moles of AgCl =5 mole.

Q. Total number of moles of $A g Cl$ precipitated on addition of excess of $A g N O_{3}$ to one mole each of the following complexes $[C o (N H_{3})_{4} C l_{2}] Cl$ , $[N i (H_{2} O)_{6}] C l_{2}, [Pt (N H_{3})_{2} C l_{2}]$ and $[P d (N H_{3})_{4}] C l_{2}$ is

$[C o (N H_{3})_{4} C l_{2}] Cl \Rightarrow$ Gives 1 mole $A g Cl$
$[N i (H_{2} O)_{6}] C l_{2} \Rightarrow$ Gives 2 moles $A g Cl$
$[Pt (N H_{3})_{2} C l_{2}] \Rightarrow$ Gives No $A g Cl$
$[P d (N H_{3})_{4}] C l_{2} \Rightarrow$ Gives 2 moles $A g Cl$
Total number of moles of $A g Cl = 5$ mole.

Q. Total number of moles of AgCl precipitated on addition of excess of AgNO3​ to one mole each of the following complexes [Co(NH3​)4​Cl2​]Cl, [Ni(H2​O)6​]Cl2​,[Pt(NH3​)2​Cl2​] and [Pd(NH3​)4​]Cl2​ is

[Co(NH3​)4​Cl2​]Cl⇒ Gives 1 mole AgCl [Ni(H2​O)6​]Cl2​⇒ Gives 2 moles AgCl [Pt(NH3​)2​Cl2​]⇒ Gives No AgCl [Pd(NH3​)4​]Cl2​⇒ Gives 2 moles AgCl Total number of moles of AgCl=5 mole.

Q. Total number of moles of $A g Cl$ precipitated on addition of excess of $A g N O_{3}$ to one mole each of the following complexes $[C o (N H_{3})_{4} C l_{2}] Cl$ , $[N i (H_{2} O)_{6}] C l_{2}, [Pt (N H_{3})_{2} C l_{2}]$ and $[P d (N H_{3})_{4}] C l_{2}$ is

$[C o (N H_{3})_{4} C l_{2}] Cl \Rightarrow$ Gives 1 mole $A g Cl$
$[N i (H_{2} O)_{6}] C l_{2} \Rightarrow$ Gives 2 moles $A g Cl$
$[Pt (N H_{3})_{2} C l_{2}] \Rightarrow$ Gives No $A g Cl$
$[P d (N H_{3})_{4}] C l_{2} \Rightarrow$ Gives 2 moles $A g Cl$
Total number of moles of $A g Cl = 5$ mole.