Question

The formula of turnbulls and prussian blue respectively are

• A. $K_3[\stackrel{(III)}{\mathrm{Fe}}{(CN)}_6],K_4[\stackrel{(II)}{\mathrm{Fe}}{(CN)}_6]$
• B. $K\stackrel{(II)}{\mathrm{Fe}}[\stackrel{(III)}{\mathrm{Fe}}{(CN)}_6],K\stackrel{(III)}{\mathrm{Fe}}[\stackrel{(II)}{\mathrm{Fe}}{(CN)}_6]$
• C. $\stackrel{(III)}{\mathrm{KFe}}[\stackrel{(II)}{\mathrm{Fe}}{(CN)}_6],\stackrel{\text{(II)}}{\mathrm{KFe}}[\stackrel{(III)}{\mathrm{Fe}}{(CN)}_6]$
• D. $K_4[\stackrel{(II)}{\mathrm{Fe}}{(CN)}_6],K_3[\stackrel{(III)}{\mathrm{Fe}}{(CN)}_6]$

Answer 1

Answer: (B)

Turn bulls blue colour is obtained when $F{e}^{2+}$ salt react with $K_3[\stackrel{\text{III}}{\mathrm{Fe}}{(CN)}_6]$ and Prussians blue colour is obtained when $F{e}^{3+}$ salts react with $K_4[\stackrel{\text{III}}{\mathrm{Fe}}{(CN)}_6]$ $\stackrel{\text{II}}{\mathrm{Fe}}C{l}_2+K_3[Fe{(CN)}_6]\to$ $\underset{Turnbullsblue}{K\stackrel{\text{II}}{\mathrm{Fe}}[\stackrel{\text{III}}{\mathrm{Fe}}{(CN)}_6]}+2KCl$ $\stackrel{\text{III}}{\mathrm{Fe}}C{l}_3+K_4[\stackrel{\text{II}}{\mathrm{Fe}}{(CN)}_6]\to$ $\underset{Prussainsblue}{K\stackrel{\text{III}}{\mathrm{Fe}}[\stackrel{\text{II}}{\mathrm{Fe}}{(CN)}_6]}+3KCl$