Question
Question: The correct statement about the magnetic properties of \({{\left[ Fe{{(CN)}_{6}} \right]}^{3-}}\) an...
The correct statement about the magnetic properties of [Fe(CN)6]3− and [FeF6]3−is (Z = 26):
(A)- Both are paramagnetic
(B)- Both are diamagnetic
(C)- [Fe(CN)6]3− is diamagnetic, [FeF6]3− is paramagnetic.
(D)- [Fe(CN)6]3− is paramagnetic, [FeF6]3−is diamagnetic
Solution
The magnetic behaviour of the complex is determined by the number of unpaired electrons present in it. That is, if unpaired electrons are present, then it is paramagnetic, whereas, in the absence or paired electrons, it is diamagnetic in nature.
Complete step by step solution:
In the two given complexes of iron, we will determine the magnetic nature of the complex on the basis of the number of unpaired electrons present in the valence shell.
The iron, Fe with atomic number = 26, has the electronic configuration as Fe=[Ar]3d64s2.
So, in [Fe(CN)6]3−, the cyanide ligand has (-1) charge, so, the iron has oxidation state (x) given as follows:
x+(6×−1)=(−3)
x=(+3), that is, in Fe3+ state.
Then, for Fe3+ state, the configuration will be Fe3+=[Ar]3d54s0. Thus, having five unpaired electrons in its d-orbital.
In the presence of the strong field ligand, CN− , it causes pairing of the d-orbital electrons. Due to which only one unpaired electron is left.
Similarly, in the [FeF6]3− complex, the fluoride ligand having (-1) charge, the iron in (+3) state.
The fluoride ion being a weak field ligand, is insufficient to cause the pairing, due to which the complex has five unpaired electrons in it.
Therefore, both the complexes [Fe(CN)6]3− and [FeF6]3− are paramagnetic in nature, due to the presence of one unpaired and five unpaired electrons present in the d-orbital respectively.
So, the correct statement regarding the magnetic properties of the two complexes is option (A)- Both are paramagnetic.
Note: It is the presence of the ligands, which helps in determining the magnetic nature of the complex. As seen, the strong field ligand causes the pairing, so the complex formed is a low- spin or inner-orbital complex. Whereas, the weak field ligand is unable to cause the pairing, so the complex formed is a high- spin or outer-orbital complex.