Question

Identify the diamagnetic octahedral complex ions from below ; 
A. [Mn(CN)\(_6\)]\(^{3-}\) 
B. [Co(NH\(_3\))\(_6\)]\(^{3+}\) 
C. [Fe(CN)\(_6\)]\(^{4-}\) 
D. [Co(H\(_2\)O)\(_3\)F\(_3\)] 
Choose the correct answer from the options given below :

• A. B and D Only
• B. A and D Only
• C. A and C Only
• D. B and C Only

Hint:

To determine the magnetic properties of a coordination complex, find the oxidation state of the central metal ion and its d electron configuration. Then, consider the strength of the ligands to determine the electron pairing in the octahedral field splitting. Strong field ligands (like CN\(^-\), NH\( _3 \)) cause pairing, while weak field ligands (like H\( _2 \)O, F\(^-\), Cl\(^-\)) generally do not. Diamagnetic complexes have all paired electrons, while paramagnetic complexes have unpaired electrons.

Answer 1

The Correct Option is D

Let's analyze each complex ion to determine its magnetic properties. A diamagnetic complex has no unpaired electrons. diamagnetic complex ion has all its electrons paired. Let's analyze each complex:

A. [Mn(CN)₆]³⁻:

• Mn is in the +3 oxidation state: Mn³⁺ has a d⁴ configuration.
• CN⁻ is a strong field ligand, so it will cause pairing in the +3 oxidation state. Mn³⁺ has an electronic configuration of d⁴.
• CN⁻ is a strong field ligand, so the complex is low spin.
• In an octahedral low-spin complex, the d⁴ configuration will have the electrons fill the t₂g orbitals first: ( of electrons (low spin).
• The d⁴ configuration in a strong field octahedral complex will have the configuration t₂g⁴ eg⁰, meaning all four electrons aret₂g)⁴ (eg)⁰. Thus, there will be 4 unpaired electrons in the t2g orbitals. The electronic configuration is \(t_{2g}^4e_g^0\).
• Since there are unpaired in the t₂g orbitals.
• Therefore, there are 2 unpaired electrons after filling in, hence paramagnetic
• t2g⁴ eg⁰ (↑↓ ↑↓ __) --> two unpaired
• electrons, this complex is paramagnetic.

B. [Co(NH₃)₆]³⁺:

• Co is in the +3 oxidation state. Co³⁺ has an electronic configuration of d⁶.
• Co is in the +3 oxidation state: Co³⁺ has a d⁶ configuration.
• NH₃ is a strong field ligand, so it will cause pairing of electrons (low spin).
• NH₃ is a strong field ligand, so the complex is low spin.
• In an octahedral low-spin complex, the d⁶ configuration will have the electrons fill the t₂g orbitals first: (t₂g)⁶ (eg)⁰. Thus, there will be no unpaired electrons. The electronic The d⁶ configuration in a strong field octahedral complex will have the configuration t₂g⁶ eg⁰, meaning all six electrons are in the t₂g orbitals.
• Therefore, there are no unpaired electrons, hence diamagnetic.
• t2g⁶ eg⁰ (↑↓ ↑↓ ↑
• The d⁶ configuration in a strong field octahedral complex will have the configuration t₂g⁶ eg⁰, meaning all six electrons are in the t₂g orbitals.
• Therefore, there are no unpaired electrons, hence diamagnetic.
• t2g⁶ eg⁰ (↑↓ ↑↓ ↑↓ )--> zero unpaired
• Since there are no unpaired electrons, this complex is diamagnetic.

C. [Fe(CN)₆]⁴⁻:

• Fe is in the +2 oxidation state. Fe²⁺ has an electronic configuration of d⁶.
• Fe is in the +2 oxidation state: Fe²⁺ has a d⁶ configuration.
• CN⁻ is a strong field ligand, so it will cause pairing of electrons (low spin).
• The d⁶ configuration of d⁶.
• CN⁻ is a strong field ligand, so the complex is low spin.
• In an octahedral low-spin complex, the d⁶ configuration will have the electrons fill the t₂g orbitals first: (t₂g)⁶ (eg)⁰. Thus, there configuration in a strong field octahedral complex will have the configuration t₂g⁶ eg⁰, meaning all six electrons are in the t₂g orbitals.
• Therefore, there are no unpaired electrons, hence diamagnetic.
• t2g⁶ eg⁰ (↑↓ ↑↓ ↑↓ )--> zero
• strong field ligand, so it will cause pairing of electrons (low spin).
• In an octahedral low-spin complex, the d⁶ configuration will have the electrons fill the t₂g orbitals first: (t₂g)⁶ (eg)⁰. Thus, there will be no unpaired electrons. The electronic configuration is \(t_{2g}^6e_g^0\).
• Since there are no unpaired electrons, this complex is diamagnetic.

D. [Co(H₂O)₃F₃]:

• The complex is neutral. unpaired
• Co is in the +3 oxidation state: Co³⁺ has a d⁶ configuration.
• H₂O is a weak field ligand and F is also a weak field ligand, so it will NOT cause pairing of electrons Co is in +3 oxidation state.
• H₂O is a weak field ligand, and F is also a weak field ligand. So, this is likely a high-spin complex.
• Co³⁺ is d⁶, so in a high-spin complex, the configuration would be (t₂g) (high spin).
• Therefore, there are 4 unpaired electrons, hence paramagnetic
• The d⁶ configuration in a weak field octahedral complex will have the configuration t₂g⁴ eg²
• t2g⁴ eg² (↑↓ ↑ ↑ ↑) --> Four unpaired electrons

Thus, the diamagnetic complexes are B and C.

Final Answer: The final answer is B and C.