
The correct option(s) to distinguish nitrate salts of $M{n^{ + 2}}$ and $C{u^{ + 2}}$ taken separately is/are:
This question has multiple correct options
(A) $M{n^{ + 2}}$ shows the characteristic green colour in the flame test.
(B) Only $C{u^{ + 2}}$ shows the formation of precipitate by passing ${H_2}S$ in an acidic medium.
(C) Only $M{n^{ + 2}}$ shows the formation of precipitate by passing ${H_2}S$ in a faintly basic medium.
(D) $C{u^{ + 2}}/Cu$ has higher reduction potential than $M{n^{ + 2}}/Mn$ (measured under similar conditions).
Answer
153.3k+ views
Hint: $C{u^{ + 2}}$ and $M{n^{ + 2}}$ both give green colour in flame test. $C{u^{ + 2}}$ belongs to group-II of cationic radical will give ppt. of $Cus$ in acidic medium. Both ions form ppt. in the basic medium. $C{u^{ + 2}}/Cu$$ = + 0.34V$ (SRP) and $M{n^{ + 2}}/Mn = - 1.18V$ (SRP).
Complete step by step solution:
A flame test is an analytical procedure in chemistry to detect the presence of certain elements, primarily metal ions, based on each element’s characteristic emission spectrum. $C{u^{ + 2}}$ and $M{n^{ + 2}}$ both give green colour in flame tests so we cannot distinguish between both ions.
${H_2}S$ is passed through the solution in an acidic medium to precipitate the sulphides of group-II cations. Sparingly soluble sulphides $C{u^{ + 2}}$ ions by the passage of hydrogen sulphide through the acidic medium. So, statement (B) is correct.
$M{n^{ + 2}}$ does not show the formation of precipitate by passing ${H_2}S$ in a faintly basic medium. So, the statement (C) is not correct.
$C{u^{ + 2}}/Cu$$ = + 0.34V$ and $M{n^{ + 2}}/Mn = - 1.18V$. Hence, $C{u^{ + 2}}/Cu$ has higher reduction potential than $M{n^{ + 2}}/Mn$. So, the statement (D) is correct.
Hence, the correct options are (B) and (D).
Note: The ${K_{sp}}$ value of group-II sulphide is low in comparison to ${K_{sp}}$ value of group-IV sulphides. Hence, they do not get precipitated in an acidic medium, where ${S^{2 - }}$ 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.
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. $C{u^{ + 2}}$ and $M{n^{ + 2}}$ both give green colour in flame tests so we cannot distinguish between both ions.
${H_2}S$ is passed through the solution in an acidic medium to precipitate the sulphides of group-II cations. Sparingly soluble sulphides $C{u^{ + 2}}$ ions by the passage of hydrogen sulphide through the acidic medium. So, statement (B) is correct.
$M{n^{ + 2}}$ does not show the formation of precipitate by passing ${H_2}S$ in a faintly basic medium. So, the statement (C) is not correct.
$C{u^{ + 2}}/Cu$$ = + 0.34V$ and $M{n^{ + 2}}/Mn = - 1.18V$. Hence, $C{u^{ + 2}}/Cu$ has higher reduction potential than $M{n^{ + 2}}/Mn$. So, the statement (D) is correct.
Hence, the correct options are (B) and (D).
Note: The ${K_{sp}}$ value of group-II sulphide is low in comparison to ${K_{sp}}$ value of group-IV sulphides. Hence, they do not get precipitated in an acidic medium, where ${S^{2 - }}$ 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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