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JEE Main
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Chemistry
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Practical Chemistry

JEE Main · Chemistry · Practical Chemistry

The correct option(s) to distinguish nitrate salts of Mn^ 2 + and...

The correct option(s) to distinguish nitrate salts of Mn2+Mn^{2 +} and Cu2+Cu^{2 +} taken separately is (are)

(1) Mn2+Mn^{2 +} shows the characteristic green colour in the flame test

(2) only Cu2+Cu^{2 +} shows the formation of precipitate by passing H2S{H}_{2}S in acidic medium

(3) only Mn2+Mn^{2 +} shows the formation of precipitate by passing H2S{H}_{2}S in faintly basic medium

(4) Cu2+∣Cu{\text{Cu}}^{2 +} | \text{Cu} has higher reduction potential than Mn2+∣Mn{\text{Mn}}^{2 +} | \text{Mn} (measured under similar conditions)

Options

  1. A1, 2
  2. B1, 3
  3. C2, 4
  4. D1, 2, 4

Correct answer

C. 2, 4

Step-by-step solution

(A)\left(A\right) Manganese show pale purple colour in flame test.

(B)Cu2+→H2SHClCuS↓Black ppt.\left(B\right)Cu^{2 +}\overset{{H}_{2}SHCl}{→}\mathop{CuS ↓}_{\text{Black ppt.}}

\left(C\right)BothBothCu^{2 +}andandMn^{2 +}formprecipitatewithform precipitate with{H}_{2}Sinbasicmedium.<br/><br/>(D)ECu2+∣Cuo=+0.34V,EMn2+∣Mno=−1.18Vin basic medium.<br/><br/>\left(\text{D}\right){E}_{{\text{Cu}}^{2 +} | \text{Cu}}^{o} = + 0.34\text{V}, {E}_{{\text{Mn}}^{2 +} | \text{Mn}}^{o} = - 1.18\text{V}
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