Question

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:

• A. R < Q < P
• B. R < P < Q
• C. Q < R < P
• D. Q < P < R

Answer 1

The Correct Option is C

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.

1. 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.

1. 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.

1. 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

Answer 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.