Correct order of spin-only magnetic moment of the following compl… — Solution | JEE Chemistry

MCQMediumJEE 2023Crystal Field Theory

Correct order of spin-only magnetic moment of the following complex ions is:

(Given At. No. Fe: $26$ , Co: $27$ )

Answer

Correct answer:A

Standard Method

Given: The complexes are $[FeF_6]^{3-}$ , $[CoF_6]^{3-}$ and $[Co(C_2O_4)_3]^{3-}$ .

Find: The correct order of spin-only magnetic moment.

The spin-only magnetic moment depends on the number of unpaired electrons.

For $[FeF_6]^{3-}$ , $Fe^{3+}$ has $5$ unpaired electrons, resulting in the highest magnetic moment.

For $[CoF_6]^{3-}$ , $Co^{3+}$ in a weak field ligand (fluoride) has $4$ unpaired electrons.

For $[Co(C_2O_4)_3]^{3-}$ , $Co^{3+}$ in a strong field ligand (oxalate) has $0$ unpaired electrons.

Thus, the order of magnetic moment is

[FeF_6]^{3-} > [CoF_6]^{3-} > [Co(C_2O_4)_3]^{3-}

Therefore, the correct option is A. The solution lists B, but the worked order matches option A.

Assuming the listed option letter is always correct without checking the worked order. Here, the solution text gives an order that matches option A, not the displayed letter B. Always verify the actual sequence written in the explanation.
Ignoring that the spin-only magnetic moment depends on the number of unpaired electrons. Comparing ligands directly without counting unpaired electrons can lead to the wrong order.
Treating all $Co^{3+}$ complexes as having the same spin state. The ligand field matters: fluoride is weak field, while oxalate is taken as stronger field in the given solution, so the number of unpaired electrons changes.

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