The calculated magnetic moments (spin only value) for species $$[FeCl_4]^{2-}$$, $$[Co(C_2O_4)_3]^{3-}$$ and $$MnO_4^{2-}$$ respectively are:

We need to find the spin-only magnetic moment for each species using the formula $$\mu = \sqrt{n(n+2)}$$ BM, where $$n$$ is the number of unpaired electrons.

For $$[FeCl_4]^{2-}$$: Iron is in the +2 oxidation state ($$Fe^{2+}$$) with configuration $$3d^6$$. Since $$Cl^-$$ is a weak field ligand and the complex is tetrahedral, there is no crystal field splitting large enough to cause pairing. In a tetrahedral field, $$d^6$$ gives 4 unpaired electrons. Thus $$\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.90$$ BM.

For $$[Co(C_2O_4)_3]^{3-}$$: Cobalt is in the +3 oxidation state ($$Co^{3+}$$) with configuration $$3d^6$$. Oxalate ($$C_2O_4^{2-}$$) is a moderate-to-strong field ligand, and in an octahedral field, $$d^6$$ with a strong field results in all electrons being paired in the $$t_{2g}$$ orbitals. This gives 0 unpaired electrons. Thus $$\mu = 0$$ BM.

For $$MnO_4^{2-}$$: Manganese is in the +6 oxidation state ($$Mn^{6+}$$) with configuration $$3d^1$$. This gives 1 unpaired electron. Thus $$\mu = \sqrt{1(1+2)} = \sqrt{3} = 1.73$$ BM.

The magnetic moments are 4.90, 0 and 1.73 BM respectively.

The correct answer is Option (1).