To determine the degenerate orbitals of the complex ion \([Cr(H_2O)_6]^{3+}\), we will follow these steps:
### Step 1: Identify the oxidation state of chromium in the complex.
- Chromium (Cr) has an atomic number of 24. In the complex \([Cr(H_2O)_6]^{3+}\), the overall charge is +3. Since water (H2O) is a neutral ligand, the oxidation state of Cr in this complex is +3.
### Step 2: Determine the electronic configuration of Cr in the +3 oxidation state.
- The ground state electronic configuration of Cr is \([Ar] 4s^1 3d^5\).
- For Cr in the +3 oxidation state, we remove 3 electrons: 1 from the 4s orbital and 2 from the 3d orbital.
- Therefore, the electronic configuration for \([Cr]^{3+}\) is \([Ar] 3d^3\).
### Step 3: Understand the splitting of d-orbitals in an octahedral field.
- In an octahedral field, the five d-orbitals split into two sets of energy levels: \(t_{2g}\) (lower energy) and \(e_g\) (higher energy).
- The \(t_{2g}\) set consists of three orbitals: \(d_{xy}\), \(d_{xz}\), and \(d_{yz}\).
- The \(e_g\) set consists of two orbitals: \(d_{x^2-y^2}\) and \(d_{z^2}\).
### Step 4: Fill the d-orbitals according to Hund's rule and the Aufbau principle.
- With 3 electrons in the \(3d\) orbitals, they will fill the \(t_{2g}\) orbitals first before occupying the \(e_g\) orbitals.
- Thus, the 3 electrons will occupy the \(d_{xy}\), \(d_{xz}\), and \(d_{yz}\) orbitals.
### Step 5: Identify the degenerate orbitals.
- The \(t_{2g}\) orbitals (\(d_{xy}\), \(d_{xz}\), and \(d_{yz}\)) are degenerate, meaning they have the same energy level.
- The \(e_g\) orbitals (\(d_{x^2-y^2}\) and \(d_{z^2}\)) are also degenerate among themselves but are at a higher energy level than the \(t_{2g}\) orbitals.
### Conclusion:
The degenerate orbitals of \([Cr(H_2O)_6]^{3+}\) are:
- \(d_{xy}\), \(d_{xz}\), and \(d_{yz}\) (which belong to the \(t_{2g}\) set).
### Final Answer:
The degenerate orbitals of \([Cr(H_2O)_6]^{3+}\) are \(d_{xy}\), \(d_{xz}\), and \(d_{yz}\).
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