The correct order of spin only magnetic moments for the following complex ions is[Fe(CN)6]3−<[CoF6]3−<[MnBr4]2−<[Mn(CN)6]3−

The spin-only magnetic moment is calculated using the formula √n(n+2) where n is the number of unpaired electrons.

1. [Fe(CN)6]3−: Fe in this complex is in the +3 oxidation state (Fe3+). In this state, Fe has 5 electrons in the 3d orbital. However, due to the strong field of the CN- ligand, it causes pairing of the electrons, leaving one unpaired electron. So, the spin-only magnetic moment is √1(1+2) = √3.

2. [CoF6]3−: Co in this complex is in the +3 oxidation state (Co3+). In this state, Co has 6 electrons in the 3d orbital. However, F- is a weak field ligand and does not cause pairing of the electrons, leaving 4 unpaired electrons. So, the spin-only magnetic moment is √4(4+2) = √24.

3. [MnBr4]2−: Mn in this complex is in the +2 oxidation state (Mn2+). In this state, Mn has 5 electrons in the 3d orbital. However, Br- is a weak field ligand and does not cause pairing of the electrons, leaving 5 unpaired electrons. So, the spin-only magnetic moment is √5(5+2) = √35.

4. [Mn(CN)6]3−: Mn in this complex is in the +3 oxidation state (Mn3+). In this state, Mn has 4 electrons in the 3d orbital. However, due to the strong field of the CN- ligand, it causes pairing of the electrons, leaving no unpaired electrons. So, the spin-only magnetic moment is √0(0+2) = 0.

So, the correct order of spin only magnetic moments for the given complex ions is [Fe(CN)6]3−<[CoF6]3−<[MnBr4]2−<[Mn(CN)6]3−.