Give the number of `3d` electrons occupied in `t_(2g)` orbitals of hydrated `Cr^(3+)` ion (octahedral) .

Answer

To determine the number of `3d` electrons occupied in the `t_(2g)` orbitals of the hydrated `Cr^(3+)` ion in an octahedral complex, we can follow these steps:

### Step 1: Identify the oxidation state and electron configuration of Chromium
Chromium (Cr) has an atomic number of 24. In its neutral state, the electron configuration is:
\[ \text{Cr: } [\text{Ar}] 4s^1 3d^5 \]

When chromium is in the +3 oxidation state (as in `Cr^(3+)`), it loses three electrons. The electrons are removed first from the 4s orbital and then from the 3d orbital.

### Step 2: Determine the electron configuration of `Cr^(3+)`
Removing three electrons from the neutral chromium configuration:
1. Remove the 1 electron from the 4s orbital.
2. Remove 2 electrons from the 3d orbital.

Thus, the electron configuration for `Cr^(3+)` becomes:
\[ \text{Cr}^{3+}: [\text{Ar}] 3d^3 \]

### Step 3: Understand the splitting of d-orbitals in an octahedral field
In an octahedral field, the five d-orbitals split into two sets:
- `t_(2g)` (lower energy): consists of three orbitals
- `e_g` (higher energy): consists of two orbitals

### Step 4: Fill the d-electrons into the split orbitals
For `Cr^(3+)`, we have 3 electrons in the 3d subshell. According to Hund's rule, we fill the lower energy `t_(2g)` orbitals first before moving to the higher energy `e_g` orbitals.

- The three electrons will occupy the three `t_(2g)` orbitals:
- 1 electron in each of the three `t_(2g)` orbitals.

### Step 5: Conclusion
The number of `3d` electrons in the `t_(2g)` orbitals of the hydrated `Cr^(3+)` ion is:
\[ \text{Number of electrons in } t_{2g} = 3 \]

### Final Answer
The number of `3d` electrons occupied in `t_(2g)` orbitals of hydrated `Cr^(3+)` ion is **3**.

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