Consider the following complex ions, P, Q and R. P = [FeF6]3–, Q = [V(H2O)6]2+ and R = [Fe(H2O)6]2+, the correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is:

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Question

Consider the following complex ions, P, Q and R.

P = [FeF₆]^3–, Q = [V(H₂O)₆]²⁺ and R = [Fe(H₂O)₆]²⁺,

the correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is:

Q < P < R

R < Q < P

Q < R < P

R < P < Q

(P) [FeF₆]^–3 $\to$ Fe⁺³ $\to$ d5 configuration with WFL

unpaired electron = 5

(Q) [V(H₂O)₆]²⁺ $\to$ V⁺² $\to$ d³ configuration

unpaired electron = 3

(R) [Fe(H₂O)₆]²⁺ $\to$ Fe⁺² $\to$ d⁶ configuration with WFL

unpaired electron = 4

Spin only magnetic moment (µ) = $\sqrt{n (n + 2)}$ B.M. (n = unpaired electron)

So, µ of P > R > Q

To determine the correct order of spin-only magnetic moment values for the given complex ions, we need to analyze the number of unpaired electrons in each complex. The spin-only magnetic moment (μ) is calculated using the formula:

$μ = \sqrt{n (n + 2)}$ B.M., where n is the number of unpaired electrons.

Step 1: Determine the oxidation state of the central metal ion in each complex.

P: [FeF₆]^3–. F is –1, so total charge from ligands is –6. Let oxidation state of Fe be x. Then x + (–6) = –3 ⇒ x = +3. So, Fe³⁺.
Q: [V(H₂O)₆]²⁺. H₂O is neutral, so oxidation state of V is +2.
R: [Fe(H₂O)₆]²⁺. H₂O is neutral, so oxidation state of Fe is +2.

Step 2: Identify the nature of the ligand (weak or strong field) to determine the electron configuration and unpaired electrons.

P: Fe³⁺ (atomic number 26) has electron configuration [Ar] 3d⁵. F^– is a weak field ligand, so it does not pair electrons. Thus, high spin configuration with 5 unpaired electrons (n=5).
Q: V²⁺ (atomic number 23) has electron configuration [Ar] 3d³. H₂O is a weak field ligand, so high spin with 3 unpaired electrons (n=3).
R: Fe²⁺ (atomic number 26) has electron configuration [Ar] 3d⁶. H₂O is a weak field ligand, so high spin with 4 unpaired electrons (n=4).

Step 3: Calculate the spin-only magnetic moment for each.

P: n=5, μ = √[5(5+2)] = √35 ≈ 5.92 B.M.
Q: n=3, μ = √[3(3+2)] = √15 ≈ 3.87 B.M.
R: n=4, μ = √[4(4+2)] = √24 ≈ 4.90 B.M.

Step 4: Compare the values. Q (3.87) < R (4.90) < P (5.92). So the correct order is Q < R < P.

Final Answer: The correct order is Q < R < P.

Formulae

Spin-only magnetic moment: $μ = \sqrt{n (n + 2)}$ B.M., where n is the number of unpaired electrons.

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