Animated Solution for Chemistry - Coordination Compounds: Which one of the following species responds to an external magnetic field

When a question asks which species "responds to an external magnetic field," it is essentially asking you to identify the paramagnetic substance. Paramagnetism arises from the presence of unpaired electrons in the atomic or molecular orbitals. These unpaired electrons act like tiny bar magnets that align with an external magnetic field, causing an attraction. Conversely, if all electrons are paired, their magnetic moments cancel out, resulting in a diamagnetic substance that weakly repels magnetic fields.

To solve this, we must determine the electronic configuration of the central metal ion in each complex and apply the principles of Valence Bond Theory (VBT) or Crystal Field Theory (CFT) to see how the ligands affect electron pairing.

Let's break down each option systematically:

Option (a): $[Fe(H_{2}O{)}_6{]}^{3+}$ Here, the central metal is Iron (Fe). The oxidation state of Fe is $+3$. The ground state electronic configuration of Fe is $[Ar]3d^{6}4s^2$. Removing three electrons gives us $F{e}^{3+}=[Ar]3d^5$. Now, we look at the ligand. Water ($H_{2}O$) is a weak field ligand according to the spectrochemical series. It does not provide enough crystal field splitting energy (${\Delta }_o$) to overcome the pairing energy ($P$). Therefore, the electrons will not pair up, and they will singly occupy the five $3d$ orbitals. This leaves us with $n=5$ unpaired electrons, making the complex highly paramagnetic.

Option (b): $[Ni(CN{)}_4{]}^{2-}$ Nickel (Ni) is in a $+2$ oxidation state. The configuration of $N{i}^{2+}$ is $[Ar]3d^8$. The ligand here is Cyanide ($C{N}^{-}$), which is a very strong field ligand. It forces the electrons to pair up against Hund's rule. The 8 electrons will completely fill four of the $3d$ orbitals, leaving one $3d$ orbital empty for $ds{p}^2$ hybridization. Since all electrons are paired ($n=0$), this complex is diamagnetic.

Option (c): $[Co(CN{)}_6{]}^{3-}$ Cobalt (Co) is in a $+3$ oxidation state, giving a configuration of $C{o}^{3+}=[Ar]3d^6$. Again, we have the strong field $C{N}^{-}$ ligand. It forces all 6 electrons to pair up into just three $3d$ orbitals ($t_{2g}^6$ in CFT terms). With zero unpaired electrons ($n=0$), this complex is also diamagnetic.

The Case of Nickel Tetracarbonyl

Option (d): $[Ni(CO{)}_4]$ This is a classic and often tricky example. Nickel is in a zero oxidation state, so its full valence configuration is intact: $N{i}^0=[Ar]3d^{8}4s^2$. Carbon monoxide ($CO$) is an exceptionally strong field ligand (a strong $\pi$-acceptor). The intense ligand field forces the two $4s$ electrons to demote into the $3d$ subshell, pairing up with the existing $3d$ electrons. This results in a pseudo-noble gas configuration of $3d^{10}$. Every single electron is paired ($n=0$), rendering the complex completely diamagnetic.

After evaluating all options, it is clear that only $[Fe(H_{2}O{)}_6{]}^{3+}$ possesses unpaired electrons. Therefore, it is the only paramagnetic species and the only one that will respond to an external magnetic field.

As a bonus, we can calculate the exact spin-only magnetic moment ($\mu$) for this complex using the formula:

Substituting $n=5$:

This high magnetic moment confirms its strong response to magnetic fields.