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How do we differentiate between Fe^{3+}  and Cr^{3+} in group III?

  • Option 1)

    By taking excess of NH_4OH solution

  • Option 2)

    By increasing NH_4^{+} ion concentration

  • Option 3)

    By decreasing OH^{-} ion concentration

  • Option 4)

    Both (b) and (c).

 

Answers (1)

best_answer

As we learned

 

Test of Chromium(3+) -

Cr(OH)_3  is oxidised to CrO^{2-}_4 in presence of NaOH & H_2O_2

2Cr(OH)_3 + H_2O_2 +4OH^{-}\rightarrow 2CrO^{2-}_4 + 8H_2O

CrO^{2-}_4gives yellow ppt with Pb^{2+}

Pb^{2+} +CrO^{2-}_4 \rightarrow PbCrO_4 Yellow

CrO^{2-}_4 in acidic medium with H_2O_2 \: gives\: CrO_5 \: (Blue)

CrO^{2-}_4 +2H^{+} + 2H_2O_2 \rightarrow CrO_5(Blue) +3H_2O_2 \:

- wherein

Yellow PbCrO_4 \: and \: blue \: CrO_3\:confirmed \:the \:presence \:of\:Cr^{3+}

 

 

Test of Ferric -

Fe(OH)_3 + 3H^{+} \rightarrow Fe^{3+} +3H_2O

Fe^{3+} +6SCN^{-} \rightarrow [Fe(SCN)_6]^{3-} (deep\ red) with K4[Fe(CN)6] .Fe3+ gives Prussian blue color

- wherein

Fe^{3+} + K_{4}[Fe(CN)_6]\rightarrow KFe[Fe(CN)_6] as \:Fe_4[Fe(CN)_6]_3

 

 

 Group reagent of Group III is NH4OH in presence of NH4Cl.

Upon increasing the concentration of NH_{4}^+{} ions it will suppress the concentration of -OH ions by common ion effect. Hence, the solubility prooduct of Fe\left ( OH \right )_{3} will be attained and colour of participate will be different.

 

 


Option 1)

By taking excess of NH_4OH solution

Option 2)

By increasing NH_4^{+} ion concentration

Option 3)

By decreasing OH^{-} ion concentration

Option 4)

Both (b) and (c).

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Plabita

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