Animated Solution for Chemistry - Coordination Compounds: Among the species given below, the total number of diamagnetic species is____. $\text{H atom},{\text{NO}}_2\text{ monomer},{\text{O}}_2^{-}\text{ (superoxide)},\text{dimeric sulphur in vapour phase},{\text{Mn}}_3{\text{O}}_4,({\text{NH}}_4{)}_2[{\text{FeCl}}_4],({\text{NH}}_4{)}_2[{\text{NiCl}}_4],{\text{K}}_2{\text{MnO}}_4,{\text{K}}_2{\text{CrO}}_4$

Welcome to a thrilling exploration of magnetic properties in chemistry! In this JEE Advanced 2018 problem, we are tasked with identifying the diamagnetic species from a diverse list of nine chemical entities. The fundamental rule of magnetism in chemistry is beautifully simple: if a species possesses even a single unpaired electron, it is paramagnetic. It will be weakly attracted to an external magnetic field. Conversely, if every single electron is perfectly paired up with a partner of opposite spin, the species is diamagnetic and will be weakly repelled by a magnetic field. Our mission is to hunt down the species with zero unpaired electrons.

Let's begin with the simplest entities on our list. The hydrogen atom ($extH$) has an atomic number of 1. Its electronic configuration is simply $1s^1$. That single electron has no partner, making the hydrogen atom inherently paramagnetic.

Next, we examine the nitrogen dioxide ($ext{NO}_2$) monomer. To determine its magnetic nature, we can simply count its valence electrons. Nitrogen brings 5 valence electrons, and each of the two oxygen atoms brings 6. This gives us a total of $5+2(6)=17$ valence electrons. Here is a golden rule: any molecule with an odd number of total electrons must have at least one unpaired electron. It is mathematically impossible to pair up an odd number of items! Therefore, $ext{NO}_2$ is paramagnetic.

Moving forward, we encounter diatomic species: the superoxide ion ($ext{O}_2^{-}$) and dimeric sulphur in the vapour phase ($ext{S}_2$). Simple Lewis structures fail us here; we must invoke Molecular Orbital Theory (MOT).

The superoxide ion, $ext{O}_2^{-}$, has 17 electrons (16 from the two oxygen atoms plus 1 extra for the negative charge). When we fill the molecular orbitals according to the Aufbau principle, the highest occupied energy level consists of two degenerate ${\pi }^{\ast }$ antibonding orbitals. Placing the last three electrons into these orbitals results in one pair and one unpaired electron. Thus, $ext{O}_2^{-}$ is paramagnetic.

What about $ext{S}_2$? Sulphur is located directly below oxygen in the periodic table. In the vapour phase, diatomic sulphur behaves analogously to $ext{O}_2$. It has 16 valence electrons, and the last two electrons enter the degenerate ${\pi }^{\ast }$ orbitals. Following Hund's rule of maximum multiplicity, they occupy separate orbitals with parallel spins. These two unpaired electrons make $ext{S}_2$ paramagnetic.

Now we face $ext{Mn}_{3}ext{O}_4$. This compound is a classic trap! It is not a simple oxide but rather a mixed oxide, structurally formulated as $extMnO\cdot ext{Mn}_{2}ext{O}_3$. This means manganese exists simultaneously in two different oxidation states within the crystal lattice: $+2$ and $+3$.

Let's look at their electronic configurations. $ext{Mn}^{2+}$ has a $3d^5$ configuration, meaning it has 5 unpaired electrons. $ext{Mn}^{3+}$ has a $3d^4$ configuration, possessing 4 unpaired electrons. Since both constituent ions are loaded with unpaired d-electrons, $ext{Mn}_{3}ext{O}_4$ is highly paramagnetic.

The Weak Field Dilemma

Our journey takes us into the realm of coordination compounds with $(ext{NH}_4{)}_2[ext{FeCl}_4]$ and $(ext{NH}_4{)}_2[ext{NiCl}_4]$.

In the complex ion $[ext{FeCl}_4{]}^{2-}$, iron is in the $+2$ oxidation state. A neutral iron atom is $[extAr]4s^{2}3d^6$, so $ext{Fe}^{2+}$ is $[extAr]3d^6$. The chloride ion ($ext{Cl}^{-}$) is a weak field ligand. According to Crystal Field Theory, it produces a small crystal field splitting energy (${\Delta }_t$). Because ${\Delta }_t$ is less than the pairing energy, the electrons prefer to occupy higher energy orbitals rather than pair up. Distributing 6 electrons across the split d-orbitals leaves us with 4 unpaired electrons. It is paramagnetic.

Similarly, in $[ext{NiCl}_4{]}^{2-}$, nickel is in the $+2$ oxidation state, giving it a $3d^8$ configuration. Again, the weak field chloride ligands do not force pairing. Filling 8 electrons into the tetrahedral d-orbital splitting pattern leaves exactly 2 unpaired electrons. This complex is also paramagnetic.

A Tale of Empty Orbitals

We are down to our last two candidates: potassium manganate ($ext{K}_{2}ext{MnO}_4$) and potassium chromate ($ext{K}_{2}ext{CrO}_4$).

In the manganate ion ($ext{MnO}_4^{2-}$), let's determine the oxidation state of manganese. Four oxygen atoms contribute a charge of $-8$. To result in an overall $-2$ charge, manganese must be in the $+6$ oxidation state. A neutral manganese atom is $4s^{2}3d^5$. Removing 6 electrons leaves a single electron in the d-orbitals ($3d^1$). This one unpaired electron makes the manganate ion paramagnetic.

Finally, we analyze the chromate ion ($ext{CrO}_4^{2-}$). Using the same logic, chromium is in the $+6$ oxidation state. A neutral chromium atom has an anomalous configuration of $4s^{1}3d^5$. If we strip away all 6 of these valence electrons to form $ext{Cr}^{6+}$, we are left with a pristine $3d^0$ configuration!

There are absolutely zero d-electrons left. Every single electron in the core shells is perfectly paired. Therefore, the chromate ion is perfectly diamagnetic.

Out of the nine species provided, only $ext{K}_{2}ext{CrO}_4$ lacks unpaired electrons. The total number of diamagnetic species is exactly 1.