Coordination Compounds Part 4

Concept:In low spin octahedral complexes, strong field ligands cause pairing of electrons in the lower energy $t_{2g}$ orbitals before occupying $e_g$ orbitals.Explanation:First, determine the $3d$ electron count for each $M^{3+}$ ion after removing three electrons.$Mn^{3+}$: $Mn$ ($Z=25$) has $[Ar]3d^5 4s^2$, so $Mn^{3+}$ has $3d^4$ electrons.$Cr^{3+}$: $Cr$ ($Z=24$) has $[Ar]3d^5 4s^1$, so $Cr^{3+}$ has $3d^3$ electrons.$Fe^{3+}$: $Fe$ ($Z=26$) has $[Ar]3d^6 4s^2$, so $Fe^{3+}$ has $3d^5$ electrons.$Co^{3+}$: $Co$ ($Z=27$) has $[Ar]3d^7 4s^2$, so $Co^{3+}$ has $3d^6$ electrons.Now fill these $d$ electrons in low spin (strong field) condition: electrons first pair up in the three $t_{2g}$ orbitals, then occupy $e_g$ orbitals.• $Cr^{3+}$ $d^3$: All three $t_{2g}$ orbitals singly occupied → 3 unpaired electrons.• $Mn^{3+}$ $d^4$: $t_{2g}$ gets one pair and two singles → 2 unpaired electrons.• $Fe^{3+}$ $d^5$: $t_{2g}$ gets two pairs and one single → 1 unpaired electron.• $Co^{3+}$ $d^6$: $t_{2g}$ completely filled with three pairs → 0 unpaired electrons.Thus, the decreasing order of unpaired electrons is: $Cr^{3+} (3) > Mn^{3+} (2) > Fe^{3+} (1) > Co^{3+} (0)$.Answer:Option A: $Cr^{3+} > Mn^{3+} > Fe^{3+} > Co^{3+}$ is correct.