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

When considering the magnetic properties of compounds, we primarily focus on whether they contain unpaired electrons. Substances with unpaired electrons are usually attracted to a magnetic field, while those without unpaired electrons are typically unaffected. This property arises from the alignment of the magnetic moments associated with the unpaired electrons.

Now, let us analyze each compound:

1. \(Na_2[CuCl_4]\) - Copper (Cu) can exist in various oxidation states, but in this compound, it's in the +2 oxidation state (since the overall charge of the complex is 2-). Cu(II) ions have \(d^9\) configuration, with no unpaired electrons. So, this compound is diamagnetic and will not be repelled by a magnetic field.

2. \(K_3[CoF_6]\) - Cobalt (Co) is typically found in several oxidation states, and in this compound, it's in the +3 oxidation state (since the overall charge of the complex is \(3^-\)). Co(III) ions have d6 configuration, and depending on the geometry of the complex, it may or may not have unpaired electrons. If the complex is octahedral, then it will have no unpaired electrons (high spin configuration). So, this compound will not be repelled by a magnetic field.

3. \(K_4[Fe(CN)_6]\) - In this compound, iron (Fe) is in the +2 oxidation state (Fe(II)). Fe(II) ions typically have a d6 electron configuration. The cyanide ligands (\(CN^-\)) are strong-field ligands, which tend to cause the d electrons to pair up before filling the next orbital. This results in a low spin complex, where all the d orbitals are filled with electrons paired up. Since all the d orbitals are fully occupied with electron pairs, there are no unpaired electrons in the complex.

In terms of magnetic behavior, compounds with no unpaired electrons are termed diamagnetic. Therefore, \(K_4[Fe(CN)_6]\) will  be repelled when placed in an external magnetic field.

4. \(K_3[Fe(CN)_6]\) - Iron (Fe) in this compound is in the +3 oxidation state (since the overall charge of the complex is \(3^-\)). Fe(III) ions typically have \(d^5\) configuration. This compound, will also form a high spin complex with one unpaired electron. So, this compound will also be attracted to a magnetic field (paramagnetic).

Therefore, compounds 3  (\(K_4[Fe(CN)_6]\) will be repelled when placed in an external magnetic field.