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 :
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 :
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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.
Updated On: Apr 19, 2026
- B and D Only
- A and D Only
- A and C Only
- B and C Only
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The Correct Option is D
Approach Solution - 1
To determine which of the given complexes are diamagnetic, we must first understand the electronic configuration and the nature of the ligands.
- [Mn(CN)6]3-:
- Manganese (Mn) in [Mn(CN)6]3- is in the +3 oxidation state, giving it a configuration of \(d^4\) (Mn: [Ar] 3d5 4s2, so Mn3+: [Ar] 3d4).
- CN- is a strong field ligand and can cause pairing of electrons in the 3d orbitals.
- However, since the d-configuration is \(d^4\), complete pairing leading to diamagnetism will not occur.
- [Co(NH3)6]3+:
- Cobalt (Co) in [Co(NH3)6]3+ is in the +3 oxidation state, resulting in a \(d^6\) configuration (Co: [Ar] 3d7 4s2, so Co3+: [Ar] 3d6).
- NH3 is also a strong field ligand, causing the pairing of d electrons.
- This results in a completely paired electronic configuration, making the complex diamagnetic.
- [Fe(CN)6]4-:
- Iron (Fe) in [Fe(CN)6]4- is in the +2 oxidation state, giving it a configuration of \(d^6\) (Fe: [Ar] 3d6 4s2, so Fe2+: [Ar] 3d6).
- CN-, being a strong field ligand, similarly causes the pairing of the 3d electrons in this case.
- This again leads to a completely paired electronic configuration, making the complex diamagnetic.
- [Co(H2O)3F3]:
- This complex exhibits weak field ligands (such as H2O and F-), so it will not lead to the pairing of electrons.
- For Co3+, it would have a \(d^6\) configuration, and due to weak fields, electrons remain unpaired, resulting in a paramagnetic complex.
Hence, the diamagnetic complexes are [Co(NH3)6]3+ (B) and [Fe(CN)6]4- (C).
Therefore, the correct answer is: B and C Only.
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Approach Solution -2
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.
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