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The correct option(s) to distinguish nitrate salts of Mn+2 and Cu+2 taken separately is/are:
This question has multiple correct options
(A) Mn+2 shows the characteristic green colour in the flame test.
(B) Only Cu+2 shows the formation of precipitate by passing H2S in an acidic medium.
(C) Only Mn+2 shows the formation of precipitate by passing H2S in a faintly basic medium.
(D) Cu+2/Cu has higher reduction potential than Mn+2/Mn (measured under similar conditions).

Answer
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Hint: Cu+2 and Mn+2 both give green colour in flame test. Cu+2 belongs to group-II of cationic radical will give ppt. of Cus in acidic medium. Both ions form ppt. in the basic medium. Cu+2/Cu=+0.34V (SRP) and Mn+2/Mn=1.18V (SRP).

Complete step by step solution:
A flame test is an analytical procedure in chemistry to detect the presence of certain elements, primarily metal ions, based on each element’s characteristic emission spectrum. Cu+2 and Mn+2 both give green colour in flame tests so we cannot distinguish between both ions.
H2S is passed through the solution in an acidic medium to precipitate the sulphides of group-II cations. Sparingly soluble sulphides Cu+2 ions by the passage of hydrogen sulphide through the acidic medium. So, statement (B) is correct.
Mn+2 does not show the formation of precipitate by passing H2S in a faintly basic medium. So, the statement (C) is not correct.
Cu+2/Cu=+0.34V and Mn+2/Mn=1.18V. Hence, Cu+2/Cu has higher reduction potential than Mn+2/Mn. So, the statement (D) is correct.

Hence, the correct options are (B) and (D).

Note: The Ksp value of group-II sulphide is low in comparison to Ksp value of group-IV sulphides. Hence, they do not get precipitated in an acidic medium, where S2 ion concentration is quite low. The main purpose of flame test is to observe the characteristic colour produced by certain metallic ions when vaporized in a flame and then to identify an unknown metallic ion by means of its flame test.
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