Identify the two species and oxidation states

The question refers to:

• anionic species of $[Fe(H_2O)_6]^{2+}$? This wording is awkward, but clearly the two complexes intended are:
  • $[Fe(H_2O)_6]^{2+}$ (cationic)
  • $[Fe(CN)_6]^{4-}$ or $[Fe(CN)_6]^{3-}$ depending on notation

From the answer choices, the intended pair is the common comparison:

• $[Fe(H_2O)_6]^{2+}$
• $[Fe(CN)_6]^{4-}$

Let us determine the magnetic moments.

For $[Fe(H_2O)_6]^{2+}$

• Oxidation state of Fe: $x + 6(0) = +2 \Rightarrow x = +2$
• So, iron is $Fe^{2+}$.
• Atomic number of Fe = 26
• Electronic configuration of $Fe$: $[Ar]3d^6 4s^2$
• Therefore, $Fe^{2+} = [Ar]3d^6$

$H_2O$ is a weak field ligand, so this is a high-spin octahedral complex.

For high-spin $d^6$ in octahedral field: $t_{2g}^4 e_g^2$ Number of unpaired electrons, $n = 4$.

Spin-only magnetic moment: $\mu = \sqrt{n(n+2)}$ $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\,BM$

For $[Fe(CN)_6]^{4-}$

• Oxidation state of Fe: $x + 6(-1) = -4$ $x - 6 = -4$ $x = +2$
• So again iron is $Fe^{2+} = 3d^6$.

$CN^-$ is a strong field ligand, so this is a low-spin octahedral complex.

For low-spin $d^6$ in octahedral field: $t_{2g}^6 e_g^0$ Number of unpaired electrons, $n = 0$.

Hence, $\mu = \sqrt{0(0+2)} = 0\,BM$

Match with the options

The two magnetic moments are:

• $[Fe(H_2O)_6]^{2+}$ : $4.9\,BM$
• $[Fe(CN)_6]^{4-}$ : $0\,BM$

So the pair is: $4.9 \text{ and } 0$

This corresponds to Option B.

Comparison with stored answer

Stored correct answer = B.

Our derived answer also = B.

Hence, they agree.