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*

• A. I, II only
• B. I, II, III
• C. II, III only
• D. I, III only

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.

Answer 1

The Correct Option is B

Step 1: Recall the magnetic moment formula.
Spin-only magnetic moment is $\mu=\sqrt{n(n+2)}$ B.M., where $n$ is the number of unpaired electrons. More unpaired electrons means a bigger moment.
Step 2: Use ligand strength.
Strong field ligands like $CN^-$ pair up electrons (low spin, fewer unpaired). Weak field ligands like $H_2O$ keep electrons unpaired (high spin, more unpaired).
Step 3: Work out Set I.
$[Fe(CN)_6]^{4-}$ is $Fe^{2+}$, $d^6$ low spin, $n=0$. $[Fe(CN)_6]^{3-}$ is $Fe^{3+}$, $d^5$ low spin, $n=1$. $[Fe(H_2O)_6]^{3+}$ is $d^5$ high spin, $n=5$. So $0<1<5$, increasing, correct.
Step 4: Work out Set II.
Counting unpaired electrons for $[Co(NH_3)_6]^{3+}$, $[Ni(H_2O)_6]^{2+}$ and $[Cr(H_2O)_6]^{3+}$ gives a rising order, so Set II is correct.
Step 5: Work out Set III.
Likewise the three complexes in Set III line up in increasing unpaired electrons, so Set III is correct.
Step 6: Combine.
All three sets are correctly ordered. \[ \boxed{I,\ II,\ III} \]