Question

Atomic number of Mn. Fe and Co are 25, 26 and 27 respectively. Which of the following inner orbital octahedral complex ions are diamangetic?
(a) $\left[Co\right(N{H}_3{)}_6{]}^{3+}$
(b) $\left[Mn\right(CN{)}_6{]}^{3-}$
(c) $\left[Fe\right(CN{)}_6{]}^{4-}$
(d) $\left[Fe\right(CN{)}_6{]}^{3-}$

Answer 1

Molecular orbital electronic configuration of $C{o}^{3+}$ in $\left[Co\right(N{H}_3{)}_6{]}^{3+}$ is



Number of unpaired electron = 0
Magnetic property = Diamagnetic
Molecular orbital electronic configuration of $M{n}^{3+}$ in $\left[Mn\right(CN{)}_6{]}^{3-}$



Number of unpaired electron = 2
Magnetic property = Paramagnetic
Molecular orbital electronic configuration of $F{e}^{2+}$ in $\left[Fe\right(CN{)}_6{]}^{4-}$ is



Number of unpaired electron = 0
Magnetic property = Diamagnetic
Molecular orbital electronic configuration of $F{e}^{2+}$ in $\left[Fe\right(CN{)}_6{]}^{4-}$ is



Number of unpaired electron = 1
Magnetic property = Paramagnetic
Thus, $\left[Co\right(N{H}_3{)}_6{]}^{3+}$ and $\left[Fe\right(CN{)}_6{]}^{4-}$ are diamagnetic.
Hence, correct choices are options (a) and (c).