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 :
Show Hint
- B and D Only
- A and D Only
- A and C Only
- B and C Only
The Correct Option is D
Solution and Explanation
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.
Top Questions on Coordination chemistry
- The complex ions \([Co(NH_3)_5(NO_2)]^{2+}\) and \([Co(NH_3)_5(ONO)]^{2+}\) are called
- CBSE CLASS XII - 2025
- Chemistry
- Coordination chemistry
Werner’s coordination theory in 1893 was the first attempt to explain the bonding in coordination complexes. It must be remembered that this theory was put forward before the electron had been discovered by J.J. Thomson in 1897, and before the electronic theory of valency. Werner did not have any of the modern instrumental techniques and all his studies were made using simple experimental techniques. Werner was able to explain the nature of bonding in complexes and he concluded that in complexes, the metal shows two different sorts of valency: primary and secondary. Primary valences are normally ionisable whereas secondary valences are non-ionisable.
- CBSE CLASS XII - 2025
- Chemistry
- Coordination chemistry
- Write the name and structure of a hexadentate ligand.
- CBSE CLASS XII - 2025
- Chemistry
- Coordination chemistry
- Why is [Ni(CN)4]2- square planar while [Ni(CO)4] is tetrahedral? [Atomic number : Ni = 28]
- CBSE CLASS XII - 2025
- Chemistry
- Coordination chemistry
- Based on Valence Bond Theory, explain the geometry and magnetic character of [NiCl$_4$]$^{2-$. [Atomic number: Ni = 28]}
- CBSE CLASS XII - 2025
- Chemistry
- Coordination chemistry
Questions Asked in JEE Main exam
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
- JEE Main - 2025
- Integration
- Given below are two statements: Statement I: One mole of propyne reacts with excess of sodium to liberate half a mole of H₂ gas. Statement II: Four g of propyne reacts with NaNH₂ to liberate NH₃ gas which occupies 224 mL at STP. In the light of the above statements, choose the most appropriate answer from the options given below:
- JEE Main - 2025
- Organic Chemistry
- Given below are two statements about X-ray spectra of elements:
Statement (I):A plot of \(\sqrt {v}\) (v=frequency of X-rays emitted) vs atomic mass is a straight line.
Statement (II): A plot of v (v=frequency of X-rays emitted)} vs atomic number is a straight line.- JEE Main - 2025
- X-ray Spectra and Moseley's Law
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
- JEE Main - 2025
- Organic Chemistry
- The ratio of the power of a light source \( S_1 \) to that of the light source \( S_2 \) is 2. \( S_1 \) is emitting \( 2 \times 10^{15} \) photons per second at 600 nm. If the wavelength of the source \( S_2 \) is 300 nm, then the number of photons per second emitted by \( S_2 \) is \_\_\_\_\_ \( \times 10^{14} \).
- JEE Main - 2025
- Units and measurement