Match List - I with List - II. List-I (a) $[\mathrm{Fe}(\mathrm{CN}){}_6]{}^{3-}$ (b) $[\mathrm{Fe}(\mathrm{H}{}_2\mathrm{O}){}_6]{}^{3+}$ (c) $[\mathrm{Fe}(\mathrm{CN}){}_6]{}^{4-}$ (d) $[\mathrm{Fe}(\mathrm{H}{}_2\mathrm{O}){}_6]{}^{2+}$ List-II (i) $5.92$ BM (ii) $0$ BM (iii) $4.90$ BM (iv) $1.73$ BM Choose the correct answer from the options given below.

Solution & Explanation

\textbf{Answer:} (B) (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii) \textbf{Solution:} Using $\mu =\sqrt{n(n+2)}$ BM with the spectrochemical series ($\mathrm{CN}{}^{-}$ strong field $\to$ low-spin; $\mathrm{H}{}_2\mathrm{O}$ weak field $\to$ high-spin): (a) $[\mathrm{Fe}(\mathrm{CN}){}_6]{}^{3-}$: $\mathrm{Fe}{}^{3+}$ $d^5$, low-spin $\Rightarrow n=1\Rightarrow \mu =\sqrt{3}=1.73$ BM (iv). (b) $[\mathrm{Fe}(\mathrm{H}{}_2\mathrm{O}){}_6]{}^{3+}$: $\mathrm{Fe}{}^{3+}$ $d^5$, high-spin $\Rightarrow n=5\Rightarrow \mu =\sqrt{35}=5.92$ BM (i). (c) $[\mathrm{Fe}(\mathrm{CN}){}_6]{}^{4-}$: $\mathrm{Fe}{}^{2+}$ $d^6$, low-spin $\Rightarrow n=0\Rightarrow \mu =0$ BM (ii). (d) $[\mathrm{Fe}(\mathrm{H}{}_2\mathrm{O}){}_6]{}^{2+}$: $\mathrm{Fe}{}^{2+}$ $d^6$, high-spin $\Rightarrow n=4\Rightarrow \mu =\sqrt{24}=4.90$ BM (iii). Thus (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii), which is option B.