Two soluble salts of iron A & B,on combining | vedclass

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Question

(A) is $FeCl_{2}$ and $B$ is $FeCl_{3}$. The reagent $(R)$ is $KCN$. $C$ is $K_{4}[Fe(CN)_{6}]$ (pale yellow) and $D$ is $K_{3}[Fe(CN)_{6}]$ (yellow).

Vedclass · Accepted Answer

Blue colour of the compound is due to $d-d$ transition

Vedclass · Answer

(A) is $FeCl_{2}$ and $B$ is $FeCl_{3}$. The reagent $(R)$ is $KCN$. $C$ is $K_{4}[Fe(CN)_{6}]$ (pale yellow) and $D$ is $K_{3}[Fe(CN)_{6}]$ (yellow). Reaction of $A$ $(Fe^{2+})$ with $D$ $([Fe(CN)_{6}]^{3-})$ gives $Fe_{3}[Fe(CN)_{6}]_{2}$ (Turnbull's blue). Reaction of $B$ $(Fe^{3+})$ with $C$ $([Fe(CN)_{6}]^{4-})$ gives $Fe_{4}[Fe(CN)_{6}]_{3}$ (Prussian blue). Both are blue coloured compounds. The blue colour in these complexes is due to metal-to-metal charge transfer $(MMCT)$ between $Fe^{2+}$ and $Fe^{3+}$,not $d-d$ transition. $C$ $([Fe(CN)_{6}]^{4-})$ is diamagnetic ($d^{6}$,low spin),while $D$ $([Fe(CN)_{6}]^{3-})$ is paramagnetic ($d^{5}$,low spin). Thus,the statement that the blue colour is due to $d-d$ transition is incorrect,and the statement about paramagnetism is only true for $D$.

Two soluble salts of iron $A$ & $B$,on combining with excess of a reagent $(R)$ produce compound $C$ (pale yellow) & $D$ (yellow) respectively. Reaction of $A$ with $D$ and that of $B$ with $C$ both produce a blue coloured compound. Which of the following statements is correct?

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