Question

Turnbull's blue is $F{e}_3{\left[Fe{\left(CN\right)}_6\right]}_2$.

• A. True
• B. False

Answer 1

The correct option is A True

$\left[Fe\right(CN{)}_6{]}^{4-}+F{e}^{3+}$ ---- (Prussian blue)

$\left[Fe\right(CN{)}_6{]}^{3-}+F{e}^{2+}\to F{e}_3{\left[Fe{\left(CN\right)}_6\right]}_2$ ---- (Turnbull's Blue)

Hence, the given statement is true.