The magnetic moment of [Ru(H2O)6]2+ corresponds to the presence of ......
Ru2+ =[Kr] 4d6. This forms outer complex. Hence, unpaired electrons are 4.
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The magnetic moment of [Ru(H2O)6]2+ corresponds to the presence of ......
The magnetic moment of a complex is a measure of its ability to be influenced by an external magnetic field. It is directly related to the presence of unpaired electrons in the complex. In the case of [Ru(H2O)6]2+, the magnetic moment corresponds to the presence of 4 unpaired electrons.
Explanation:
1. Magnetic Moment:
The magnetic moment of a complex can be defined as the total magnetic field generated by the complex due to the motion of electrons. It is a vector quantity and is usually measured in units of Bohr magneton (µB).
2. Complex: [Ru(H2O)6]2+
[Ru(H2O)6]2+ is a coordination complex in which ruthenium (Ru) is surrounded by six water (H2O) ligands. The complex is in its +2 oxidation state.
3. Ligand Field Theory:
According to Ligand Field Theory, the metal-ligand interaction in a complex gives rise to splitting of the metal's d orbitals into different energy levels. These energy levels are referred to as the eg and t2g sets.
4. Electronic Configuration of Ru2+:
The electronic configuration of Ru2+ in its ground state is [Kr]4d^6. In the presence of ligands, the d orbitals split, and the electrons occupy the lower energy levels first following Hund's rule, which states that electrons fill degenerate orbitals singly before pairing up.
5. Splitting of d Orbitals:
In an octahedral field, the d orbitals split into two sets of energy levels: eg (higher energy) and t2g (lower energy). The three t2g orbitals are dxz, dyz, and dxy, while the two eg orbitals are dx^2-y^2 and dz^2.
6. Filling of Electrons:
Since there are 4 unpaired electrons in the ground state electronic configuration of Ru2+, these electrons will occupy the t2g orbitals singly. Therefore, the [Ru(H2O)6]2+ complex will have 4 unpaired electrons.
7. Magnetic Moment Calculation:
The magnetic moment (µ) of a complex can be calculated using the formula µ = √(n(n+2)) where n is the number of unpaired electrons. In this case, n = 4, so the magnetic moment is √(4(4+2)) = √(4*6) = √24.
Conclusion:
Therefore, the magnetic moment of [Ru(H2O)6]2+ corresponds to the presence of 4 unpaired electrons.
The magnetic moment of [Ru(H2O)6]2+ corresponds to the presence of ......
But Google search the answer is zero so correct 4 or 0 unpaired electron