Discussion Forum



1)

The correct options (s) to distinguish nitrate salts to Mn2+ and Cu2+ taken separately is (are)

   

 


A) $Mn^{2+}$ shows the characteristic green colour in the flame test

B) Only $Cu^{2+}$ shows the formation of precipitate by passing $H_{2}S$ in acidic medium.

C) Only $Mn^{2+}$ shows the formation of precipitate by passing $H_{2}S$ in faintly basic medium.

D) $Cu^{2+}$/Cu has higher reduction potential than $Mn^{2+}$/Mn ( measured under similar conditions)

Answer:

Option B,D

Explanation:

Statement wise explanation is

Statement   (a)    Mn2+    produces yellow-green colour in the flame test while Cu2+ produce bluish-green colour in a flame test. Thus, due to the presence of green colour in both the cases, flame test is not a suitable method to distinguish between nitrate salts of Cu2+ and Mn2+. Hence this statement is wrong.

Statement  (b)  Cu2+ belong to group II of cationic or basic radicals. It gives black ppt. of CuS if H2S  is passed through it in the presence of acid (eg HCl). Mn2+ does not show this property hence this can be considered as a suitable method to distinguish between Mn2+ and Cu2+. Hence, this statement is correct.

Statement (c)     In faintly basic medium when H2S is passed both Cu2+ and Mn2+ forms precipitates. Thus, it is not a suitable method to distinguish between them. Hence, this statement is incorrect.

Statement (d)   The standard reduction potential of Cu2+ /Cu is +0.34 V while that of Mn2+/Mn is -1.18V. This can be used to distinguish between Cu2+  and Mn2+. In general, less electropostive metals have higher SRP.

               Hence, this statement is correct.