Animated Solution for Chemistry - Coordination Compounds: In which of the following order the given complex ions are arranged correctly with respect to their decreasing spin only magnetic moment? (i) $[Fe{F}_6{]}^{3-}$ (ii) $[Co(N{H}_3{)}_6{]}^{3+}$ (iii) $[NiC{l}_4{]}^{2-}$ (iv) $[Cu(N{H}_3{)}_4{]}^{2+}$

Spin-Only Magnetic Moment

To conquer this problem, we first need to arm ourselves with the fundamental equation for the spin-only magnetic moment. The magnetic moment, denoted by $\mu$, is given by the formula:

Here, $n$ represents the number of unpaired electrons in the central metal ion. The key takeaway from this formula is beautifully simple: the magnetic moment is directly proportional to the number of unpaired electrons. The more unpaired electrons a complex has, the higher its magnetic moment will be. Our mission is to find $n$ for each of the given complexes.

Let's break down each complex one by one, examining the oxidation state of the central metal, its $d$-electron configuration, and the nature of the ligands attached to it.

1. The Hexafluoroferrate(III) Ion: $[Fe{F}_6{]}^{3-}$ First up, we have iron surrounded by six fluoride ions. The oxidation state of iron here is $+3$, which leaves it with a $3d^5$ configuration. Now, we must consult the spectrochemical series. Fluoride ($F^{-}$) is a classic weak field ligand. It doesn't have the strength to force electrons to pair up against their will (the pairing energy is greater than the crystal field splitting energy, ${\Delta }_o<P$). Consequently, all five electrons occupy separate orbitals, remaining unpaired. Result: $n=5$.

2. The Hexaamminecobalt(III) Ion: $[Co(N{H}_3{)}_6{]}^{3+}$ Next, we examine cobalt coordinated to six ammonia molecules. Cobalt is in a $+3$ oxidation state, giving it a $3d^6$ configuration. Ammonia ($N{H}_3$) is generally a strong field ligand, especially with $+3$ metal ions. It creates a large crystal field splitting (${\Delta }_o>P$), forcing the six electrons to pair up in the lower energy $t_{2g}$ orbitals. Result: All electrons are paired, so $n=0$.

3. The Tetrachloridonickelate(II) Ion: $[NiC{l}_4{]}^{2-}$ Moving on to nickel bonded to four chloride ions. Nickel is in a $+2$ oxidation state, which corresponds to a $3d^8$ configuration. Chloride ($C{l}^{-}$) is a weak field ligand. When we distribute eight electrons across the five $d$-orbitals, we fill them singly first, then start pairing. This results in three pairs and exactly two unpaired electrons. Result: $n=2$.

4. The Tetraamminecopper(II) Ion: $[Cu(N{H}_3{)}_4{]}^{2+}$ Finally, we have copper with four ammonia ligands. Copper is in a $+2$ oxidation state, leaving it with a $3d^9$ configuration. With nine electrons, the arrangement is straightforward regardless of the ligand strength: four orbitals will have paired electrons, and one orbital will have a single unpaired electron. Result: $n=1$.

Now that we have the number of unpaired electrons for each complex, we can easily determine the decreasing order of their magnetic moments.

Let's list our findings: (i) $[Fe{F}_6{]}^{3-}$ has $n=5$ (ii) $[Co(N{H}_3{)}_6{]}^{3+}$ has $n=0$ (iii) $[NiC{l}_4{]}^{2-}$ has $n=2$ (iv) $[Cu(N{H}_3{)}_4{]}^{2+}$ has $n=1$

Arranging these in decreasing order of $n$ gives us: $5>2>1>0$. Translating this back to the complexes, we get: ${\mu }_i>{\mu }_{iii}>{\mu }_{iv}>{\mu }_{ii}$

This perfectly matches option (a). By systematically analyzing the electron configurations and ligand strengths, we've successfully cracked the problem!