Question

The octahedral complex ion $[FeC{l}_6{]}^{3-}$ and $\left[Fe\right(CN{)}_6{]}^{3-}$ are both paramagnetic.

The number of unpaired electrons and correct configuration of $[FeC{l}_6{]}^{3-}$ is :

• A. Five unpaired electrons $\left(t_{2g}^3{e}_g^2\right)$
• B. One unpaired electron $\left(t_{2g}^1{e}_g^3\right)$
• C. Three unpaired electrons $\left(t_{2g}^1{e}_g^2\right)$
• D. All the above electrons paired.

Answer 1

The correct option is A Five unpaired electrons $\left(t_{2g}^3{e}_g^2\right)$

High spin complex $[FeC{l}_6{]}^{3-}$ has the d-configuration as $t_{2g}^3{e}_g^2$.

$F{e}^{3+}$ configuration is $\left[Ar\right]3d^5$ hence according to the CFSE for octahedral geometry (high spin) first 3 electrons will go to $t_{2g}$ level and then next two to $e_g$ level.

No pairing takes place.

Option A is correct.