Which of the following compounds will be repelled when placed in an external magnetic field?

\(K_4[Fe(CN)_6]\)

The correct answer is option 3. \(K_4[Fe(CN)_6]\).

Here is an explanation of why \(K_4[\text{Fe(CN)}_6]\) is diamagnetic and will be repelled by an external magnetic field:

Oxidation State of Iron in \(K_4[\text{Fe(CN)}_6]\)

In \(K_4[\text{Fe(CN)}_6]\) iron is in the +2 oxidation state (\(\text{Fe}^{2+}\)). Electronic Configuration of \(\text{Fe}^{2+}\). The atomic number of iron (Fe) is 26. The electronic configuration of neutral iron is: \([Ar] 3d^6 4s^2\). When iron loses two electrons to become \(\text{Fe}^{2+}\), the electronic configuration becomes: \([Ar] 3d^6\).

Ligand Field and Crystal Field Splitting

In \(K_4[\text{Fe(CN)}_6]\), iron is surrounded by six cyanide (\(\text{CN}^-\)) ligands in an octahedral geometry. Cyanide is a strong field ligand. Strong field ligands cause a large crystal field splitting (\(\Delta_0\)) in the octahedral field.

Low-Spin Configuration

In the presence of strong field ligands like \(\text{CN}^-\), \(\text{Fe}^{2+}\) undergoes low-spin pairing of electrons. For \(\text{Fe}^{2+}\) (\([Ar] 3d^6\)), in a low-spin octahedral complex with strong field ligands, the \(3d\) orbitals split into two sets: \(t_{2g}\) (lower energy) and \(e_g\) (higher energy).

The six \(3d\) electrons will pair up in the \(t_{2g}\) orbitals as follows:

\(t_{2g}^6 e_g^0\): All six \(3d\) electrons are paired in the lower energy \(t_{2g}\) orbitals.

Magnetic Properties

Diamagnetic Substances: Have all their electrons paired. There are no unpaired electrons, and they are repelled by an external magnetic field.

Paramagnetic Substances: Have one or more unpaired electrons and are attracted by an external magnetic field.

Since in \(K_4[\text{Fe(CN)}_6]\), the \(3d\) electrons are paired due to the strong field created by the cyanide ligands, this complex is diamagnetic.

Summary

Electron Pairing: The strong field ligands cause pairing of all \(3d\) electrons in \(\text{Fe}^{2+}\), resulting in a diamagnetic complex.

Magnetic Behavior: Because \(K_4[\text{Fe(CN)}_6]\) does not have unpaired electrons, it will be repelled by an external magnetic field, as diamagnetic substances are.

Therefore, the compound \(K_4[\text{Fe(CN)}_6]\) is correctly identified as diamagnetic and will be repelled by an external magnetic field.