Statement I: K_3[Fe(CN)_6] is a low-spin complex Statement II: In the…

The correct answer is option 1. Both Statement I and statement II are correct and statement II is the correct explanation of statement I

Let us look into each of the given statements:

Statement I: \(K_3[Fe(CN)_6]\) is a low-spin complex

\(K_3[Fe(CN)_6]\) consists of three potassium ions \((K^+)\) and one hexacyanoferrate(III) ion \(([Fe(CN)_6]^{3-})\). The iron ion in this complex is in the \(+3\) oxidation state \((Fe^{3+})\).  Iron in the +3 oxidation state has an electronic configuration of \([Ar]3d^5\). This means it has five electrons in its 3d orbitals. when the six \(CN^-\) ligand approaches the central metal ion it causes pairing of electrons in the d-orbital as \(CN^-\) is a strong field ligand and thus leads to the formation of a low spin complex.

Statement II: In the presence of a strong \(CN^−\) ligand, the d-electrons are rearranged

Cyanide \((CN^−)\) is a strong field ligand, which means it causes a large splitting (High \(\Delta _0\) value) of the d-orbitals in an octahedral crystal field. This splitting is much larger compared to weak field ligands like water \((H_2O)\) or chloride \((Cl^−)\).

In an octahedral field, the five degenerate d-orbitals split into two sets namely \(t_{2g}\) (lower energy, consisting of \(d_{xy}, d_{xz}, d_{yz}\)) and \(e_g\) (higher energy, consisting of \(d_{z^2}\) and \(d_{x^2-y^2}\)). Whether a complex is low-spin or high-spin depends on the relative magnitudes of the splitting energy \((\Delta _0)\) and the pairing energy \((P)\). In the presence of a strong field ligand like \(CN^−\), the splitting energy is large enough to overcome the pairing energy. As a result, electrons prefer to pair up in the lower-energy \(t_{2g}\) orbitals rather than occupy the higher-energy \(e_g\) orbitals.

For \([Fe(CN)_6]^{3-}\), the splitting energy is significant due to the strong field created by the cyanide ligands. \(Fe^{3+}\) has five \(3d\) electrons. In a strong field environment, these electrons will fill the lower-energy \(t_{2g}\) orbitals first. Therefore, the electronic configuration in the octahedral field with a strong ligand like \(CN^−\) is \(t_{2g}^5\) (with all electrons paired except one).

Because the electrons are paired up as much as possible in the lower energy orbitals \((t_{2g})\), \([Fe(CN)_6]^{3-}\) is a low-spin complex.

Conclusion:

Statement I is correct because \(K_3[Fe(CN)_6]\) indeed forms a low-spin complex due to the strong field effect of the cyanide ligands.

Statement II is correct because the presence of the strong field ligand |(CN^−\) leads to a large splitting of the d-orbitals, causing the electrons to rearrange in such a way that they occupy the lower-energy \(t_{2g}\) orbitals first, resulting in a low-spin configuration. Therefore, Statement II is the correct explanation for Statement I.