Animated Solution for Chemistry - Coordination Compounds: The pair having the same magnetic moment is [at. no. Cr = 24, Mn = 25, Fe = 26 and Co = 27]

Welcome to a fascinating journey into the world of coordination chemistry! In this problem, we are tasked with finding a pair of complexes that share the exact same magnetic moment. But what exactly dictates this magnetic property?

The answer lies in the spin-only magnetic moment formula:

Here, $n$ represents the number of unpaired electrons in the central metal ion. This elegant equation tells us a simple truth: if two complexes have the same number of unpaired electrons, their magnetic moments will be identical. Our mission, therefore, is to act as electron detectives and count the unpaired electrons for each metal ion.

Let's begin by examining the complexes that feature water ($H_{2}O$) as a ligand. Water is a neutral molecule, which means the oxidation state of the central metal ion is simply the overall charge of the complex. In all our water-based complexes, the metal is in a $+2$ oxidation state.

Furthermore, according to the spectrochemical series, water is a weak field ligand. This means it produces a relatively small crystal field splitting energy (${\Delta }_o$). When the splitting energy is small, it's easier for electrons to jump to the higher energy $e_g$ orbitals rather than pairing up in the lower $t_{2g}$ orbitals. This results in what we call a high-spin complex.

1. The Chromium Complex: $[Cr(H_{2}O{)}_6{]}^{2+}$ Chromium ($Cr$) has an atomic number of 24. In its $+2$ state, its electronic configuration is $[Ar]3d^4$. When we distribute these 4 electrons in a high-spin octahedral field, the first three go into the $t_{2g}$ level, and the fourth jumps up to the $e_g$ level. Configuration: $t_{2g}^3{e}_g^1$. Unpaired electrons ($n$) = 4.

2. The Iron Complex: $[Fe(H_{2}O{)}_6{]}^{2+}$ Iron ($Fe$) has an atomic number of 26. In its $+2$ state, its electronic configuration is $[Ar]3d^6$. Distributing 6 electrons in a high-spin field means we first place 5 electrons singly across all orbitals ($t_{2g}^3{e}_g^2$), and the 6th electron must pair up in the lower $t_{2g}$ level. Configuration: $t_{2g}^4{e}_g^2$. Unpaired electrons ($n$) = 4.

3. The Manganese Complex: $[Mn(H_{2}O{)}_6{]}^{2+}$ Manganese ($Mn$) has an atomic number of 25. In its $+2$ state, its electronic configuration is $[Ar]3d^5$. With 5 electrons in a high-spin field, each of the five d-orbitals gets exactly one electron. Configuration: $t_{2g}^3{e}_g^2$. Unpaired electrons ($n$) = 5.

Now, let's shift our focus to the cobalt complex, $[CoC{l}_4{]}^{2-}$. Here, we have four chloride ($C{l}^{-}$) ligands. Since each chloride has a $-1$ charge, the cobalt must be in a $+2$ oxidation state to give an overall $-2$ charge.

Cobalt ($Co$) has an atomic number of 27, so $C{o}^{2+}$ has a $[Ar]3d^7$ configuration.

Because there are only 4 ligands, this complex adopts a tetrahedral geometry. In a tetrahedral crystal field, the splitting pattern is inverted compared to an octahedral field: the $e$ set is lower in energy, and the $t_2$ set is higher. Moreover, tetrahedral splitting (${\Delta }_t$) is always smaller than pairing energy, so tetrahedral complexes are almost always high-spin.

Let's distribute the 7 electrons. The first two go into the lower $e$ level, the next three go into the higher $t_2$ level, and the remaining two pair up in the lower $e$ level. Configuration: $e^4{t}_2^3$. Unpaired electrons ($n$) = 3.

Let's review our findings: - $[Cr(H_{2}O{)}_6{]}^{2+}$ has 4 unpaired electrons. - $[Fe(H_{2}O{)}_6{]}^{2+}$ has 4 unpaired electrons. - $[Mn(H_{2}O{)}_6{]}^{2+}$ has 5 unpaired electrons. - $[CoC{l}_4{]}^{2-}$ has 3 unpaired electrons.

It is crystal clear that both the chromium and iron complexes possess exactly 4 unpaired electrons. Consequently, they will exhibit the exact same magnetic moment. The mystery is solved, and the correct pair is identified!