Animated Solution for Chemistry - Coordination Compounds: The correct order of the calculated spin only magnetic moments of complexes (A) to (D) is (A) ${\text{Ni(CO)}}_4$ (B) ${\text{[Ni(H}}_2{\text{O)}}_6{\text{]Cl}}_2$ (C) ${\text{Na}}_2{\text{[Ni(CN)}}_4\text{]}$ (D) ${\text{PdCl}}_2{\text{(PPh}}_3{\text{)}}_2$

Finding the spin-only magnetic moment of coordination complexes is like solving a microscopic puzzle. The key to this puzzle lies in a simple yet powerful formula:

Here, $n$ represents the number of unpaired electrons in the central metal ion. Our mission is to determine $n$ for four distinct complexes and arrange them in order of their magnetic moments.

Complex A and B

Let's start with Complex A: ${\text{Ni(CO)}}_4$. In this complex, nickel is in a zero oxidation state, giving it an outer electronic configuration of $3d^{8}4s^2$. However, carbon monoxide ($\text{CO}$) is a notoriously strong field ligand. It exerts such a strong influence that it forces the two $4s$ electrons to pair up inside the $3d$ orbitals. This results in a completely filled $3d^{10}$ configuration. With zero unpaired electrons ($n=0$), the magnetic moment $\mu$ is exactly $0$ BM.

Moving on to Complex B: ${\text{[Ni(H}}_2{\text{O)}}_6{\text{]Cl}}_2$. Here, nickel is in a $+2$ oxidation state, leaving it with a $3d^8$ configuration. Water (${\text{H}}_2\text{O}$) is a weak field ligand, meaning it doesn't have the energy to force electrons to pair up against their will. Following Hund's rule, we fill the $d$-orbitals and find exactly $2$ unpaired electrons ($n=2$). Plugging this into our formula gives:

Complex C and D

Next up is Complex C: ${\text{Na}}_2{\text{[Ni(CN)}}_4\text{]}$. Nickel is again in the $+2$ state ($3d^8$). But this time, it's surrounded by cyanide ions (${\text{CN}}^{-}$), which are very strong field ligands. They force the unpaired electrons to pair up, vacating one $d$-orbital to allow for $ds{p}^2$ hybridization (forming a square planar geometry). Since all electrons are paired ($n=0$), the magnetic moment $\mu$ is $0$ BM.

Finally, we encounter Complex D: ${\text{PdCl}}_2{\text{(PPh}}_3{\text{)}}_2$. Palladium is in a $+2$ oxidation state, giving a $4d^8$ configuration. Here lies a crucial exception in coordination chemistry! Palladium belongs to the $4d$ transition series. Elements in the $4d$ and $5d$ series have a significantly higher crystal field splitting energy (${\Delta }_0$) compared to their $3d$ counterparts. Because of this massive splitting, they always form low-spin, square planar complexes when the coordination number is $4$, regardless of whether the ligand is strong or weak. Thus, pairing occurs, leaving zero unpaired electrons ($n=0$), and $\mu =0$ BM.

Let's summarize our findings: - $\mu (A)=0\text{ BM}$ - $\mu (B)=2.82\text{ BM}$ - $\mu (C)=0\text{ BM}$ - $\mu (D)=0\text{ BM}$

Comparing these values, it is clear that complexes A, C, and D have identical magnetic moments of zero, while complex B has a higher value. Therefore, the correct order is:

This perfectly matches option (d). Always remember to watch out for those sneaky $4d$ and $5d$ metals!