The hybridization of Cr in `[Cr(en)_(3)]^(3+)` is

Options

A. `d^(2)sp^(3)`
B. `sp^(3)d`
C. `sp^(3)d^(2)`
D. `dsp^(3)`

Explanation

To determine the hybridization of chromium in the complex \([Cr(en)_{3}]^{3+}\), we can follow these steps:

### Step 1: Determine the Oxidation State of Chromium
The complex has an overall charge of +3. Since ethylenediamine (en) is a neutral ligand, it does not contribute to the oxidation state. Therefore, we can set up the equation:

\[ \text{Oxidation state of Cr} + \text{(number of en ligands)} \times \text{(oxidation state of en)} = \text{overall charge} \]

This simplifies to:

\[ X + 0 = +3 \implies X = +3 \]

### Step 2: Write the Electron Configuration of Chromium
The atomic number of chromium (Cr) is 24. The electron configuration of neutral chromium is:

\[ [Ar] 3d^5 4s^1 \]

For chromium in the +3 oxidation state, we remove three electrons. The two electrons from the 4s orbital and one electron from the 3d orbital are removed, leading to:

\[ \text{Cr}^{3+} : 3d^3 4s^0 \]

### Step 3: Determine the Coordination Number
In the complex \([Cr(en)_{3}]^{3+}\), there are three ethylenediamine (en) ligands. Each en ligand is bidentate, meaning it can coordinate through two donor atoms. Therefore, the total coordination number is:

\[ \text{Coordination number} = 3 \times 2 = 6 \]

### Step 4: Identify the Hybridization
To accommodate six coordination sites, we need to determine the hybridization of the chromium ion. The hybridization can be determined by the number of orbitals involved:

- For a coordination number of 6, the hybridization is typically \(d^2sp^3\).
- The \(d\) orbitals come from the 3d subshell, and we will also use one s orbital and three p orbitals.

### Step 5: Conclusion
Thus, the hybridization of chromium in the complex \([Cr(en)_{3}]^{3+}\) is:

\[ \text{Hybridization} = d^2sp^3 \]

### Final Answer
The hybridization of Cr in \([Cr(en)_{3}]^{3+}\) is \(d^2sp^3\).

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