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D and F Block Elements question

2018 · 16 Apr · Shift 1 · Q22
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D and F Block Elements question

2018 · 16 Apr · Shift 1 · Q22

JEE MainChemistryD and F Block ElementsMCQ+4 / −1
The incorrect statement is :
  1. A
    Cu2+Cu^{2+}Cu2+ salts give red coloured borax bead test in reducing flame.
  2. B
    Cu2+Cu^{2+}Cu2+ and Ni2+Ni^{2+}Ni2+ ions give black precipitate with H2SH_2SH2​S in presence of HClHClHCl solution.
  3. C
    Ferric ion gives blood red color with potasium thiocyanate.
  4. D
    Cu2+Cu^{2+}Cu2+ ion gives chocolate coloured preciitate with potassium ferrocyanide solution.
View written solutionFree

Correct answer: B

  1. Check option A: Borax bead test for copper

    In the borax bead test, copper salts give:

    • blue-green bead in oxidising flame
    • red/opaque bead in reducing flame due to formation of metallic copper or cuprous oxide.

    So statement A is correct.

  2. Check option B: Reaction of Cu2+Cu^{2+}Cu2+ and Ni2+Ni^{2+}Ni2+ with H2SH_2SH2​S in presence of HClHClHCl

    In qualitative analysis:

    • In acidic medium (HClHClHCl present), only group II cations precipitate with H2SH_2SH2​S because sulfide ion concentration is low.
    • Cu2+Cu^{2+}Cu2+ belongs to group II and gives black precipitate of CuSCuSCuS.
    • Ni2+Ni^{2+}Ni2+ belongs to group IV, and it does not precipitate with H2SH_2SH2​S in acidic medium. It precipitates as NiSNiSNiS only in alkaline medium.

    Therefore the statement that both Cu2+Cu^{2+}Cu2+ and Ni2+Ni^{2+}Ni2+ give black precipitate with H2SH_2SH2​S in presence of HClHClHCl is incorrect.

  3. Check option C: Ferric ion with potassium thiocyanate

    Fe3+Fe^{3+}Fe3+ gives a blood-red complex with thiocyanate ion: Fe3++SCN−→[FeSCN]2+Fe^{3+} + SCN^- \rightarrow [FeSCN]^{2+}Fe3++SCN−→[FeSCN]2+

    Hence statement C is correct.

  4. Check option D: Cu2+Cu^{2+}Cu2+ with potassium ferrocyanide

    Cu2+Cu^{2+}Cu2+ reacts with potassium ferrocyanide, K4[Fe(CN)6]K_4[Fe(CN)_6]K4​[Fe(CN)6​], to give a chocolate-brown precipitate of copper ferrocyanide.

    Therefore statement D is correct.

  5. Conclusion

    The only incorrect statement is: B\boxed{B}B​

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