Consider the transition metal ions Mn^{3+}, C | vedclass

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Question

(A) For low spin octahedral complexes, electrons fill the $t_{2g}$ orbitals first.$Co^{3+} (3d^6): t_{2g}^6 e_g^0$, unpaired electrons $= 0$.$Fe^{3+}

Vedclass · Accepted Answer

$Cr^{3+} > Mn^{3+} > Fe^{3+} > Co^{3+}$

Vedclass · Answer

(A) For low spin octahedral complexes, electrons fill the $t_{2g}$ orbitals first.$Co^{3+} (3d^6): t_{2g}^6 e_g^0$, unpaired electrons $= 0$.$Fe^{3+} (3d^5): t_{2g}^5 e_g^0$, unpaired electrons $= 1$.$Mn^{3+} (3d^4): t_{2g}^4 e_g^0$, unpaired electrons $= 2$.$Cr^{3+} (3d^3): t_{2g}^3 e_g^0$, unpaired electrons $= 3$.Thus, the decreasing order is $Cr^{3+} > Mn^{3+} > Fe^{3+} > Co^{3+}$.

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

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