Coordination Compounds Part 4

Concept:Paramagnetism depends on unpaired electrons; ligand field strength determines high spin or low spin in octahedral complexes.Explanation:Statement I: Check each species for unpaired electrons.$[\mathrm{CoF}_6]^{3-}$: Co is $+3$ ($d^6$). $\mathrm{F}^-$ is weak field → high spin → unpaired electrons → paramagnetic.$[\mathrm{TiF}_6]^{3-}$: Ti is $+3$ ($d^1$). One unpaired electron → paramagnetic.$\mathrm{V}_2\mathrm{O}_5$: V is $+5$ ($d^0$). No $d$-electrons → diamagnetic.$[\mathrm{Fe(CN)}_6]^{3-}$: Fe is $+3$ ($d^5$). $\mathrm{CN}^-$ is strong field → low spin → one unpaired electron → paramagnetic.Thus, three paramagnetic species: $[\mathrm{CoF}_6]^{3-}$, $[\mathrm{TiF}_6]^{3-}$, $[\mathrm{Fe(CN)}_6]^{3-}$. Statement I is true.Statement II: Count unpaired electrons in each complex.$\mathrm{K}_4[\mathrm{Fe(CN)}_6]$: Fe$^{2+}$ ($d^6$) + strong field → low spin → $0$ unpaired.$\mathrm{K}_3[\mathrm{Fe(CN)}_6]$: Fe$^{3+}$ ($d^5$) + strong field → low spin → $1$ unpaired.$[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]\mathrm{SO}_4\cdot\mathrm{H}_2\mathrm{O}$: $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{2+}$, Fe$^{2+}$ ($d^6$) + weak field → high spin → $4$ unpaired.$[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]\mathrm{Cl}_3$: $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{3+}$, Fe$^{3+}$ ($d^5$) + weak field → high spin → $5$ unpaired.So order $0 < 1 < 4 < 5$ matches given order. Statement II is true.Answer:Both Statement I and Statement II are true → Option D.