Given below are two statements: Statement I: Both $[\mathrm{Co}(\mathrm{NH}{}_3){}_6]{}^{3+}$ and $[\mathrm{CoF}{}_6]{}^{3-}$ complexes are octahedral but differ in their magnetic behaviour. Statement II: $[\mathrm{Co}(\mathrm{NH}{}_3){}_6]{}^{3+}$ is diamagnetic whereas $[\mathrm{CoF}{}_6]{}^{3-}$ is paramagnetic. In the light of the above statements, choose the correct answer from the options given below:

Solution & Explanation

\textbf{Answer:} (D) Both Statement I and Statement II are true. \textbf{Solution:} In both complexes cobalt is $\mathrm{Co}{}^{3+}$, a $d^6$ ion, and both are octahedral. $[\mathrm{Co}(\mathrm{NH}{}_3){}_6]{}^{3+}$: $\mathrm{NH}{}_3$ is a strong-field ligand, so ${\Delta }_o>$ pairing energy and all six electrons pair up: $t_{2g}^6{e}_g^0$, $n=0$ unpaired $\Rightarrow$ low-spin, diamagnetic. $[\mathrm{CoF}{}_6]{}^{3-}$: $\mathrm{F}{}^{-}$ is a weak-field ligand, so the electrons remain unpaired as far as possible: $t_{2g}^4{e}_g^2$, $n=4$ unpaired $\Rightarrow$ high-spin, paramagnetic. Thus both complexes are octahedral but differ in magnetic behaviour (Statement I true), and the specific assignment (Statement II) is also correct. Hence option D.