Q.$\text{Fe}^{2+}$ ($d^6$) forms both $[\text{Fe}(\text{CN})_6]^{4-}$ and $[\text{Fe}(\text{H}_2\text{O})_6]^{2+}$. State the electronic configuration, spin state (high-spin or low-spin), and magnetic behaviour of each, given that $\text{CN}^-$ is a strong-field ligand ($\Delta_o > P$) and $\text{H}_2\text{O}$ is a weak-field ligand ($\Delta_o < P$).

[!TLDR] Same Fe$^{2+}$ $d^6$ ion, different ligand field: CN$^-$ strong ($\Delta_o>P$) gives low-spin diamagnetic; H$_2$O weak ($\Delta_o<P$) gives high-spin paramagnetic. [!ANSWER] $[\text{Fe}(\text{CN})_6]^{4-}$: $t_{2g}^6e_g^0$, low-spin, dia …

Both complexes contain $\text{Fe}^{2+}$, a $d^6$ ion, so the only variable deciding their electron configuration is the field strength of the ligand present, compared against the pairing energy $P$. In $[\text{Fe}(\text{CN})_6]^{4-}$, $\text{CN}^-$ is a strong-field ligand, so $\Delta_o > P$: it is energetically cheaper for electrons to pair up in the lower $t_{2g}$ set than to occupy the higher $e_g$ set, so all six electrons fill $t_{2g}$ (pairing as needed), giving $t_{2g}^6 e_g^0$ — the low-spin configuration, with zero unpaired electrons, so the complex is diamagnetic. In $[\text{Fe}(\text{H}_2\text{O})_6]^{2+}$, $\text{H}_2\text{O}$ is a weak-field ligand, so $\Delta_o < P$: pairing is more costly than occupying $e_g$, so the electrons spread out by Hund's rule across all five orbitals first, giving $t_{2g}^4 e_g^2$ — the high-spin configuration, with four unpaired electrons (2 in $t_{2g}$, 2 in $e_g$, exactly as worked out for $[\text{CoF}_6]^{3-}$ earlier), so the complex is paramagnetic. This pair of complexes mirrors the $\text{Co}^{ …

Since both complexes involve the same $d^6$ metal ion, compare only the ligand field strength ($\Delta_o$) against the fixed pairing energy $P$ for each ligand separately to determine high-sp …

Assuming a metal ion has one single, fixed magnetic behaviour regardless of ligand — as with the earlier $\text{Co}^{3+}$ examples, the SAME $d^6$ ion gives opposi …