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

We need to find the correct decreasing order of unpaired electrons in low-spin octahedral complexes of $$Mn^{3+}$$, $$Cr^{3+}$$, $$Fe^{3+}$$, and $$Co^{3+}$$.

$$Cr^{3+}$$: $$[Ar]3d^3$$, $$Mn^{3+}$$: $$[Ar]3d^4$$, $$Fe^{3+}$$: $$[Ar]3d^5$$, $$Co^{3+}$$: $$[Ar]3d^6$$

In a low-spin (strong field) octahedral complex, electrons fill $$t_{2g}$$ first before going to $$e_g$$:

$$Cr^{3+}$$ ($$d^3$$): $$t_{2g}^3 e_g^0$$ → 3 unpaired electrons

$$Mn^{3+}$$ ($$d^4$$): $$t_{2g}^4 e_g^0$$ → 2 unpaired electrons (one pair in $$t_{2g}$$)

$$Fe^{3+}$$ ($$d^5$$): $$t_{2g}^5 e_g^0$$ → 1 unpaired electron

$$Co^{3+}$$ ($$d^6$$): $$t_{2g}^6 e_g^0$$ → 0 unpaired electrons

$$Cr^{3+}(3) > Mn^{3+}(2) > Fe^{3+}(1) > Co^{3+}(0)$$

This matches Option C: $$Cr^{3+} > Mn^{3+} > Fe^{3+} > Co^{3+}$$.

Therefore, the answer is Option C.