Question

The number of paramagnetic complexes among $[FeF_6]^{3-}, [Fe(CN)_6]^{3-}, [Mn(CN)_6]^{3-}, [Co(C_2O_4)_3]^{3-}, [MnCl_6]^{3-}$ and $[CoF_6]^{3-}$, which involved $d^2sp^3$ hybridization is ...............

Hint:

To determine paramagnetism, look for unpaired electrons in the complex. Strong field ligands usually pair up electrons, while weak field ligands leave electrons unpaired.

Answer 1

Correct Answer: 2

To determine the number of paramagnetic complexes, we need to check the number of unpaired electrons in each complex. Paramagnetic complexes have at least one unpaired electron, while diamagnetic complexes have all paired electrons. 

Step 1: Identification
1. \([FeF_6]^{3-}\): Iron in \(Fe^{3+}\) has the electron configuration \([Ar] 3d^5\). Fluoride (\(F^{-}\)) is a weak field ligand and does not cause electron pairing. \(Fe^{3+}\) undergoes \(d^2sp^3\) hybridization, leaving 5 unpaired electrons in the \(d\)-orbitals. Paramagnetic.
2. \([Fe(CN)_6]^{3-}\): Iron in \(Fe^{3+}\) has the electron configuration \([Ar] 3d^5\). Cyanide (\(CN^{-}\)) is a strong field ligand that causes electron pairing. \(Fe^{3+}\) undergoes \(d^2sp^3\) hybridization, resulting in no unpaired electrons. Diamagnetic.
3. \([Mn(CN)_6]^{3-}\): Manganese in \(Mn^{3+}\) has the electron configuration \([Ar] 3d^4\). Cyanide (\(CN^{-}\)) is a strong field ligand and causes pairing of electrons. \(Mn^{3+}\) undergoes \(d^2sp^3\) hybridization, resulting in 2 unpaired electrons. Paramagnetic.
4. \([Co(C_2O_4)_3]^{3-}\): Cobalt in \(Co^{3+}\) has the electron configuration \([Ar] 3d^6\). Oxalate (\(C_2O_4^{2-}\)) is a weak field ligand. \(Co^{3+}\) undergoes \(d^2sp^3\) hybridization, resulting in 3 unpaired electrons. Paramagnetic.
5. \([MnCl_6]^{3-}\): Manganese in \(Mn^{3+}\) has the electron configuration \([Ar] 3d^4\). Chloride (\(Cl^{-}\)) is a weak field ligand and does not cause electron pairing. \(Mn^{3+}\) undergoes \(d^2sp^3\) hybridization, leaving 4 unpaired electrons. Paramagnetic.
6. \([CoF_6]^{3-}\): Cobalt in \(Co^{3+}\) has the electron configuration \([Ar] 3d^6\). Fluoride (\(F^{-}\)) is a weak field ligand and does not cause electron pairing. \(Co^{3+}\) undergoes \(d^2sp^3\) hybridization, leaving 2 unpaired electrons. Paramagnetic.

Step 2: Conclusion.
The number of paramagnetic complexes is 2. These are:
\([FeF_6]^{3-}\) \([Mn(CN)_6]^{3-}\)
Thus, the number of paramagnetic complexes is 2.

Answer 2

Step 1: 
For d²sp³ hybridization, the complex must be inner orbital (low spin), which occurs when the ligand is strong field and causes pairing of electrons.
Let’s analyze each complex one by one:
 

(1) [FeF₆]³⁻
- Metal ion: Fe³⁺ (3d⁵).
- Ligand: F⁻ (weak field).
- Weak field → no pairing → outer orbital hybridization (sp³d²).
- Complex is paramagnetic (5 unpaired electrons).
Not d²sp³.
 

(2) [Fe(CN)₆]³⁻
- Metal ion: Fe³⁺ (3d⁵).
- Ligand: CN⁻ (strong field).
- Strong field → electron pairing → inner orbital hybridization (d²sp³).
- One unpaired electron remains → paramagnetic.
✔ d²sp³ and paramagnetic.
 

(3) [Mn(CN)₆]³⁻
- Metal ion: Mn³⁺ (3d⁴).
- Ligand: CN⁻ (strong field).
- Strong field → pairing occurs → d²sp³ hybridization.
- Two unpaired electrons remain → paramagnetic.
✔ d²sp³ and paramagnetic.
 

(4) [Co(C₂O₄)₃]³⁻
- Metal ion: Co³⁺ (3d⁶).
- Ligand: C₂O₄²⁻ (strong field).
- Strong field → complete pairing → d²sp³ hybridization.
- All electrons paired → diamagnetic.
✖ d²sp³ but diamagnetic.
 

(5) [MnCl₆]³⁻
- Metal ion: Mn³⁺ (3d⁴).
- Ligand: Cl⁻ (weak field).
- Weak field → no pairing → sp³d² hybridization.
- Paramagnetic.
✖ Not d²sp³.
 

(6) [CoF₆]³⁻
- Metal ion: Co³⁺ (3d⁶).
- Ligand: F⁻ (weak field).
- Weak field → no pairing → sp³d² hybridization.
- Paramagnetic.
✖ Not d²sp³.
 

Step 2:
Hence, the complexes that are both paramagnetic and involve d²sp³ hybridization are:
[Fe(CN)₆]³⁻ and [Mn(CN)₆]³⁻.

Final Answer:
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