Question

Why does aqueous Fe (III) ion develop intense red colour when it reacts with \[SC{N^ - }\] ion while Fe (II) ion does not?
A.Fe (III) ion forms a charge transfer complex with \[SC{N^ - }\] ions
B.Fe (III) is reduced to Fe (I) which is deep red in colour
C. \[SN{C^ - }\] is oxidised to \[C{N^ - }\] which forms red coloured complex with Fe (III) ion
D. \[SC{N^ - }\] does not form any complex with Fe (III) ion

Answer 1

Hint: To solve this question we need to first understand the type of product that the reaction of Fe (III) ion and a thiocyanate ion would form. Then we relate the properties of the product formed to the properties that generally cause chromaticity in compounds, to get our answer.

Complete Step-by-Step Answer:
Before we move forward with the solution of the given question, let us first understand some important basic concepts.
When we react this Fe (III) ion heavily with the thiocyanate ions, the chemical reaction observed is as follows:
\[F{e^{3 + }} + 6SC{N^ - } \to {[Fe{(SCN)_6}]^{3 - }}\]
The product formed is a complex ion which has iron as the central atom and has the thiocyanate ions acting as the ligands. This complex has an octahedral geometrical structure. This molecule can also be explained as a charge transfer complex. A charge transfer complex is formed when there is an electron – donor – acceptor complex between two or more molecules. In this particular situation, the Fe (III) ion is the electron acceptor and the thiocyanate ions are the electron donors. The unpaired electrons in the thiocyanate ions cause the formation of a charge transfer complex.
The colour of coordination complexes arises from electronic transitions between levels who’s spacing corresponds to the wavelengths available in the visible light. In complexes, these transitions are frequently referred to as \[d - d\] transitions because they involve the orbitals that are mainly d in character. Because of this, the charge transfer complex between iron and thiocyanate ions exhibits a dark red colour.

Hence, Option A is the correct option

Note: The atomic number of irons is 26, hence the electronic configuration of iron can be given as \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^6}\]. Hence, we can see that Fe has 2 electrons in its valence shell. In the case of Fe (III) ion, three out of those eight electrons are lost from the 4s and 3d orbitals.