Question

The formula of turnbulls and prussian blue respectively are

• A. $ {{K}_{3}}[\overset{(III)}{\mathop{Fe}}\,\,{{(CN)}_{6}}],\,\,{{K}_{4}}[\overset{(II)}{\mathop{Fe}}\,\,{{(CN)}_{6}}] $
• B. $ K\overset{(II)}{\mathop{Fe}}\,\,[\overset{(III)}{\mathop{Fe}}\,\,{{(CN)}_{6}}],\,\,K\overset{(III)}{\mathop{Fe}}\,\,[\overset{(II)}{\mathop{Fe}}\,\,{{(CN)}_{6}}] $
• C. $ \overset{\,\,\,\,\,\,\,\,(III)}{\mathop{KFe}}\,\,[\overset{(II)}{\mathop{Fe}}\,\,{{(CN)}_{6}}],\,\,\overset{\,\,\,\,\,\,\text{(II)}}{\mathop{KFe}}\,\,[\overset{(III)}{\mathop{Fe}}\,\,\,{{(CN)}_{6}}] $
• D. $ {{K}_{4}}\,[\overset{(II)}{\mathop{Fe}}\,\,{{(CN)}_{6}}],\,\,{{K}_{3}}\,[\overset{(III)}{\mathop{Fe}}\,\,\,{{(CN)}_{6}}] $

Answer 1

Answer: (B)

Turn bulls blue colour is obtained when $ F{{e}^{2+}} $ salt react with $ {{K}_{3}}[\overset{\text{III}}{\mathop{Fe}}\,\,{{(CN)}_{6}}] $ and Prussians blue colour is obtained when $ F{{e}^{3+}} $ salts react with $ {{K}_{4}}[\overset{\text{III}}{\mathop{Fe}}\,\,{{(CN)}_{6}}] $ $ \overset{\text{II}}{\mathop{Fe}}\,C{{l}_{2}}+{{K}_{3}}[Fe{{(CN)}_{6}}]\xrightarrow{{}} $ $ \underset{Turn\,bulls\,blue}{\mathop{K\overset{\text{II}}{\mathop{Fe}}\,\,[\overset{\text{III}}{\mathop{Fe}}\,{{(CN)}_{6}}]}}\,+2KCl $ $ \overset{\text{III}}{\mathop{Fe}}\,C{{l}_{3}}+{{K}_{4}}[\overset{\text{II}}{\mathop{Fe}}\,\,{{(CN)}_{6}}]\xrightarrow{{}} $ $ \underset{Prussains\,blue}{\mathop{K\overset{\text{III}}{\mathop{Fe}}\,\,[\overset{\text{II}}{\mathop{Fe}}\,\,{{(CN)}_{6}}]}}\,+3KCl $