Question

Select correct statement(s):
I) When excess $FeC{l}_3$ solution is added to $K_4\left[Fe\right(CN{)}_5]$ solution, in addition to
$F{e}^{III}\left[F{e}^{II}\right(CN{)}_6{]}^{-},F{e}^{II}\left[F{e}^{III}\right(CN{)}_6{]}^{-}$is also formed due to side redox reaction
(II) When $FeC{l}_2$ is added to $K_3\left[Fe\right(CN{)}_6]$ solution, in addition to $F{e}^{II}\left[F{e}^{III}\right(CN{)}_6{]}^{-},F{e}^{III}\left[F{e}^{II}\right(CN{)}_6{]}^{-}$ is also formed due to side redox reaction
(III) $F{e}^{III}\left[F{e}^{II}\right(CN{)}_6{]}^{-}$ is paramagnetic while $F{e}^{II}\left[F{e}^{III}\right(CN{)}_6{]}^{-}$ is diamagnetic
(IV) $F{e}^{III}\left[F{e}^{II}\right(CN{)}_6{]}^{-}$ is diamagnetic while $F{e}^{II}\left[F{e}^{III}\right(CN{)}_6{]}^{-},$ is paramagnetic

• A. I, II
• B. III, IV
• C. both (a) and (b)
• D. None of these

Answer 1

The correct option is A I, II

$F{e}^{III}\left[F{e}^{II}\right(CN{)}_6{]}^{-}$ and $F{e}^{II}\left[F{e}^{III}\right(CN{)}_6{]}^{-}$ both are paramagnetic due to the presence of an unpaired electron.