Question

Which of the following inner orbital octahedral complex ions are diamagnetic?

• A. $\left[Co\right(N{H}_3{)}_6{]}^{+3}$
• B. $\left[Mn\right(CN{)}_6{]}^{+3}$
• C. $\left[Fe\right(CN{)}_6{]}^{-4}$
• D. $\left[Fe\right(CN{)}_6{]}^{-3}$

Answer 1

Answer: (A), (C)

The correct options are
A

$\left[Co\right(N{H}_3{)}_6{]}^{+3}$

C

$\left[Fe\right(CN{)}_6{]}^{-4}$

All the ligands are strong and would thus force electron pairing given a chance! But from the given, both $C{o}^{3+}$ and $F{e}^{+2}$ both have the $d^6$ configuration, which means there just won't be any unpaired electrons. Three of the $d$-orbitals are completely filled - paired because of strong field ligands. To accommodate six ligands, with two unfilled $d$-orbitals, the hybridization then becomes $d^{2}s{p}^3$. Hence, these two are diamagnetic!

Let us look at the electronic configuration for the $C{o}^{3+}$ complex:

For the ferrocyanide anion, the lower $t_{2g}$ gets completely filled (low-spin configuration) because of the strong field ligand and high ${\Delta }_{oct}$: