Question

The d - electron configurations of $M{n}^{2+},F{e}^{2+},C{o}^{3+},andN{i}^{2+}are3d^5,3d^6,3d^6,3d^8,$ respectively. Which of the following aqua complexes will exhibit the minimum paramagnetic behavior?

• A. $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$
• B. $\left[Co\right(H_{2}O{)}_6{]}^{3+}$
• C. $\left[Ni\right(H_{2}O{)}_6{]}^{2+}$
• D. $\left[Mn\right(H_{2}O{)}_6{]}^{2+}$

Answer 1

The correct option is B

$\left[Co\right(H_{2}O{)}_6{]}^{3+}$

For minimum paramagnetism, the number of unpaired electrons should be the least. Already the electron configuration for the ions have been given. Now all we have to do is figure out is how the ligands function in this case:

It is simple to figure out the oxidation state of the central metal in $\left[Co\right(H_{2}O{)}_6{]}^{3+}$. What is surprising though is that $H_{2}O$ - like $OX$ (oxalate) and $N{H}_3$, acts as a strong field ligand. Thus in the $d^6$ configuration, CFSE dictates that the 3 orbitals in $t_{2g}$ are all paired and fully-filled with 6 electrons!

$\left[Co\right(H_{2}O{)}_6{]}^{3+}=d^{2}s{p}^3$ low spin, diamagnetic, zero B.M.