Coordination Compounds: Tricky Questions Solved (5)

Question

Determine the IUPAC name and number of unpaired electrons in $[\text{Fe}(\text{CN})_6]^{4-}$.

Solution — Step by Step

The complex has zero unpaired electrons (diamagnetic) and IUPAC name hexacyanidoferrate(II).

Why This Works

Crystal field theory predicts that in an octahedral complex with a strong-field ligand, the splitting energy $\Delta_o$ exceeds the pairing energy. So electrons pair up in the lower $t_{2g}$ set rather than occupy the higher $e_g$ set.

For $d^6$ low-spin: $t_{2g}^6$ — all paired. For $d^6$ high-spin: $t_{2g}^4 e_g^2$ — four unpaired.

Alternative Method

Use the magnetic moment formula $\mu = \sqrt{n(n+2)}\,\text{BM}$. If experimental $\mu = 0$, then $n = 0$ — confirms low-spin. Many JEE questions give $\mu$ and ask you to deduce $n$.

Common Mistake

Writing “ferro” instead of “ferrate” for anionic complexes. The fix: anionic complexes always end in “-ate” with the Latin name when applicable. Iron $\to$ ferrate, copper $\to$ cuprate, lead $\to$ plumbate.