Coordination CompoundHard

Question

Consider the transition metal ions ${Mn}^{3 +},{Cr}^{3 +},{Fe}^{3 +}$ and ${Co}^{3 +}$ and all form low spin octahedral complexes. The correct decreasing order of unpaired electrons in their respective d-orbitals of the complexes is

Options

A.${Cr}^{3 +} > {Fe}^{3 +} > {Co}^{3 +} > {Mn}^{3 +}$
B.${Mn}^{3 +} > {Fe}^{3 +} > {Co}^{3 +} > {Cr}^{3 +}$
C.${Fe}^{3 +} > {Co}^{3 +} > {Mn}^{3 +} > {Cr}^{3 +}$
D.${Cr}^{3 +} > {Mn}^{3 +} > {Fe}^{3 +} > {Co}^{3 +}$

Solution

$\ {Co}^{3 +} \longrightarrow 3{\text{ }d}^{6} \Rightarrow t_{2\text{ }g}^{2,2,2}e_{g}^{0,0}$, unpaired electron $= 0$

$$\begin{matrix} & {Fe}^{3 +} \rightarrow 3{\text{ }d}^{5} \Rightarrow t_{2\text{ }g}^{2,2,1}e_{g}^{0,0}\text{~}\text{unpaired electron}\text{~} = 1 \\ & {Cr}^{3 +} \rightarrow 3{\text{ }d}^{3} \Rightarrow t_{2\text{ }g}^{1,1,1}e_{g}^{0,0}\text{~}\text{unpaired electron}\text{~} = 3 \\ & {Mn}^{3 +} \rightarrow 3{\text{ }d}^{4} \Rightarrow t_{2\text{ }g}^{2,1,1}e_{g}^{0,0}\text{~}\text{unpaired electron}\text{~} = 2 \end{matrix}$$

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