QJEEM22Ch43fa: JEE Main 2022 Chemistry - Class 12 JEEM22Ch43fa Solution: Amongst FeCl 3 .3H 2 O, K 3 [Fe(CN) 6 ] and [Co(NH 3 ) 6 ]Cl 3…

$\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3} \mathrm{Cl}_{3}\right] \rightarrow$ Outer-orbital complex

$\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \rightarrow \text { Inner-orbital complex }$

$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3} \rightarrow$ Inner-orbital complex

Since $\mathrm{CN}^{-}$ is a strong field ligand than $\mathrm{NH}_{3}$ . Hence $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ is the inner-orbital complex that absorbs light at shortest wavelength.

$\mathrm{Fe}(\text{III}) \rightarrow$ valence shell configuration $3 \mathrm{~d}^{5}$

Since $\mathrm{CN}^{-}$ will do pairing, so unpaired electron $=1$

$\mu=\sqrt{1(1+2)}=\sqrt{3} \mathrm{BM} \simeq 2 \mathrm{BM}$