A) ferri cyanide
B) ferrous ferri cyanide
C) ferrous cyanide
D) ferri ferro cyanide
Correct Answer: B
Solution :
A blue colour is produced when \[F{{e}^{2+}}\] reacts with \[{{K}_{3}}[Fe{{(CN)}_{6}}]\] potassium ferricyanide. This blue colour is called Tumbulls blue. \[\overset{II}{\mathop{F}}\,eS{{O}_{4}}+{{K}_{3}}[\overset{III}{\mathop{Fe}}\,{{(CN)}_{6}}]\xrightarrow{{}}\underset{TumbulPs\text{ }blue}{\mathop{K\overset{II}{\mathop{Fe}}\,[\overset{III}{\mathop{Fe}}\,{{(CN)}_{6}}]}}\,\] So, Tumbulls blue is ferrous ferricyanide. Note- \[F{{e}^{3+}}\] solution also gives deep blue ppt. of Prussian blue with \[{{K}_{4}}[Fe{{(CN)}_{6}}]\] potassium ferrocyanide. \[\overset{III}{\mathop{Fe}}\,C{{l}_{3}}+{{K}_{4}}[\overset{II}{\mathop{Fe}}\,{{(CN)}_{6}}]\xrightarrow{{}}\] \[\underset{\Pr ussian\,\,blue}{\mathop{K\,.\,\overset{III}{\mathop{Fe}}\,\,[\overset{II}{\mathop{Fe}}\,{{(CN)}_{6}}]}}\,+3KCl\] Recent X-ray work, IR and other methods have proved that Turnbulls blue is identical to Prussian blue. The intense colour arises from electron transfer between Fe (+II) and Fe (+III). \[F{{e}^{3+}}\] partly oxidises \[[\overset{II}{\mathop{Fe}}\,\,{{(CN)}_{6}}{{[}^{4-}}\] to \[[\overset{III}{\mathop{Fe}}\,\,{{(CN)}_{6}}{{[}^{3-}}\] and itself is reduced to \[F{{e}^{2+}}\], and also \[F{{e}^{2+}}\] partly reduced \[{{[\overset{III}{\mathop{Fe}}\,{{(CN)}_{6}}]}^{3-}}\] to \[\overset{III}{\mathop{Fe}}\,{{[\overset{II}{\mathop{Fe}}\,{{(CN)}_{6}}]}^{-}}\]are identical.
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