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The crystal field stabilization energy (CFSE) in [Mn(H2O)6]2+ is
- a)0 ΔO
- b)2.0 ΔO – 2P
- c)0.4 ΔO – 2P
- d)2.0 ΔO
Correct answer is option 'A'. Can you explain this answer?
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The crystal field stabilization energy (CFSE) in [Mn(H2O)6]2+ isa)0 &#...
Crystal Field Stabilization Energy (CFSE) in [Mn(H2O)6]2+
CFSE is the energy that results from the interaction between the electrons of a transition metal ion and the ligands surrounding it. In the case of [Mn(H2O)6]2+, the ligands are water molecules.
The CFSE for [Mn(H2O)6]2+ is determined by calculating the energy difference between the two sets of d-orbitals in the Mn2+ ion, namely the lower energy t2g orbitals and the higher energy eg orbitals.
Calculation of CFSE
The CFSE for [Mn(H2O)6]2+ is given by the formula:
CFSE = -0.4 * n * ∆o
where n is the number of electrons in the d-orbitals and ∆o is the crystal field splitting energy.
For [Mn(H2O)6]2+, n = 5 (since there are five electrons in the d-orbitals) and ∆o is relatively small due to the weak ligand field of the water molecules. Therefore, the CFSE is close to zero.
Conclusion
The correct answer is option 'A', i.e., the CFSE in [Mn(H2O)6]2+ is zero. This is because the ligand field of the water molecules is weak, and the CFSE is small.
CFSE is the energy that results from the interaction between the electrons of a transition metal ion and the ligands surrounding it. In the case of [Mn(H2O)6]2+, the ligands are water molecules.
The CFSE for [Mn(H2O)6]2+ is determined by calculating the energy difference between the two sets of d-orbitals in the Mn2+ ion, namely the lower energy t2g orbitals and the higher energy eg orbitals.
Calculation of CFSE
The CFSE for [Mn(H2O)6]2+ is given by the formula:
CFSE = -0.4 * n * ∆o
where n is the number of electrons in the d-orbitals and ∆o is the crystal field splitting energy.
For [Mn(H2O)6]2+, n = 5 (since there are five electrons in the d-orbitals) and ∆o is relatively small due to the weak ligand field of the water molecules. Therefore, the CFSE is close to zero.
Conclusion
The correct answer is option 'A', i.e., the CFSE in [Mn(H2O)6]2+ is zero. This is because the ligand field of the water molecules is weak, and the CFSE is small.
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