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.
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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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
