Consider the following complex ions:
P = \([FeF_6]^{3-}\)
Q = \([V(H_2O)_6]^{2+}\)
R = \([Fe(H_2O)_6]^{2+}\)
The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is:
P = \([FeF_6]^{3-}\)
Q = \([V(H_2O)_6]^{2+}\)
R = \([Fe(H_2O)_6]^{2+}\)
The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is:
- R < Q < P
- R < P < Q
- Q < R < P
- Q < P < R
The Correct Option is C
Approach Solution - 1
To determine the order of complex ions according to their spin-only magnetic moment values, we need to consider the electronic configuration and the number of unpaired electrons in each complex ion.
The spin-only magnetic moment (\( \mu_s \)) is calculated using the formula:
\(\mu_s = \sqrt{n(n+2)}\) B.M.
where \( n \) is the number of unpaired electrons.
- Complex ion \( P = [FeF_6]^{3-} \):
Iron is in the \( +3 \) oxidation state. Electronic configuration of Fe\(^{3+}\) is 3d5. Fluoride ion (F\(^-\)) is a weak field ligand, so it will not cause pairing of electrons. Hence, all the 5 d-electrons remain unpaired.
Number of unpaired electrons in \( P \) = 5.
Magnetic moment, \(\mu_s = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\) B.M.
- Complex ion \( Q = [V(H_2O)_6]^{2+} \):
Vanadium is in the \( +2 \) oxidation state. Electronic configuration of V\(^{2+}\) is 3d3. Water (H\(_2\)O) is a weak field ligand and does not cause pairing of electrons. Therefore, all 3 d-electrons remain unpaired.
Number of unpaired electrons in \( Q \) = 3.
Magnetic moment, \(\mu_s = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\) B.M.
- Complex ion \( R = [Fe(H_2O)_6]^{2+} \):
Iron is in the \( +2 \) oxidation state. The electronic configuration of Fe\(^{2+}\) is 3d6. Water is a weak field ligand and does not cause pairing. Thus, 4 d-electrons remain unpaired.
Number of unpaired electrons in \( R \) = 4.
Magnetic moment, \(\mu_s = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\) B.M.
Based on the magnetic moment values calculated, the correct order is:
Q < R < P
Approach Solution -2
- [FeF6]3−: Fe3+ has the electron configuration [Ar] 3d5. Fluoride (F−) is a weak field ligand, so all 5 d-electrons remain unpaired.
μ = √5(5 + 2) = √35 BM
- [V(H2O)6]2+: V2+ has the electron configuration [Ar] 3d3. With three unpaired electrons, the spin-only magnetic moment is:
μ = √3(3 + 2) = √15 BM
- [Fe(H2O)6]2+: Fe2+ has the electron configuration [Ar] 3d6. Water is a weak field ligand, resulting in four unpaired electrons.
μ = √4(4 + 2) = √24 BM
Thus, the correct order of magnetic moments is Q < R < P.
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