Question:
In which of the following given sets, complexes are correctly arranged in the increasing order of their spin only magnetic moment values?
align*
I. & [Fe(CN)_6]^4- < [Fe(CN)_6]^3- < [Fe(H_2O)_6]^3+
II. & [Co(NH_3)_6]^3+ < [Ni(H_2O)_6]^2+ < [Cr(H_2O)_6]^3+
III. & [V(H_2O)_6]^3+ < [Cr(CN)_6]^3- < [Fe(H_2O)_6]^2+
align*
In which of the following given sets, complexes are correctly arranged in the increasing order of their spin only magnetic moment values?
align*
I. & [Fe(CN)_6]^4- < [Fe(CN)_6]^3- < [Fe(H_2O)_6]^3+
II. & [Co(NH_3)_6]^3+ < [Ni(H_2O)_6]^2+ < [Cr(H_2O)_6]^3+
III. & [V(H_2O)_6]^3+ < [Cr(CN)_6]^3- < [Fe(H_2O)_6]^2+ align*
II. & [Co(NH_3)_6]^3+ < [Ni(H_2O)_6]^2+ < [Cr(H_2O)_6]^3+
III. & [V(H_2O)_6]^3+ < [Cr(CN)_6]^3- < [Fe(H_2O)_6]^2+ align*
Show Hint
Strong field ligands like \(CN^-\) and CO produce low-spin complexes, while weak field ligands like \(H_2O\) and \(F^-\) generally produce high-spin complexes.
Updated On: Jun 8, 2026
- I, II only
- I, II, III
- II, III only
- I, III only
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Concept:
The spin-only magnetic moment is given by:
\[
\mu = \sqrt{n(n+2)} \text{ B.M.}
\]
where \(n\) is the number of unpaired electrons.
Greater the number of unpaired electrons, greater is the magnetic moment.
Step 1: Analyze the first set of complexes.
\[ [Fe(CN)_6]^{4-} \] Here: \[ Fe^{2+} : 3d^6 \] Since \(CN^-\) is a strong field ligand, pairing occurs: \[ n = 0 \] \[ [Fe(CN)_6]^{3-} \] \[ Fe^{3+} : 3d^5 \] Low-spin configuration gives: \[ n = 1 \] \[ [Fe(H_2O)_6]^{3+} \] \(H_2O\) is a weak field ligand, so no pairing: \[ n = 5 \] Thus: \[ 0 < 1 < 5 \] Hence the increasing order is correct.
Step 2: Apply the same logic to sets II and III.
For the remaining sets, the complexes are arranged according to increasing number of unpaired electrons, so their magnetic moments also increase correctly. Therefore, all three arrangements are correct. Hence: \[ \boxed{(B)\ I,\ II,\ III} \]
Step 1: Analyze the first set of complexes.
\[ [Fe(CN)_6]^{4-} \] Here: \[ Fe^{2+} : 3d^6 \] Since \(CN^-\) is a strong field ligand, pairing occurs: \[ n = 0 \] \[ [Fe(CN)_6]^{3-} \] \[ Fe^{3+} : 3d^5 \] Low-spin configuration gives: \[ n = 1 \] \[ [Fe(H_2O)_6]^{3+} \] \(H_2O\) is a weak field ligand, so no pairing: \[ n = 5 \] Thus: \[ 0 < 1 < 5 \] Hence the increasing order is correct.
Step 2: Apply the same logic to sets II and III.
For the remaining sets, the complexes are arranged according to increasing number of unpaired electrons, so their magnetic moments also increase correctly. Therefore, all three arrangements are correct. Hence: \[ \boxed{(B)\ I,\ II,\ III} \]
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