The hybridization of the complex `[CrCl_(2)(NO_(2))_(2)(NH_(3))_(2)]^(-)` is:

Options

A. `sp^(3)d^(2)`
B. `d^(2)sp^(3)`
C. `sp^(3)d`
D. cannot be predicted

Explanation

To determine the hybridization of the complex \([CrCl_2(NO_2)_2(NH_3)_2]^-\), we can follow these steps:

### Step 1: Determine the Oxidation State of Chromium
Let the oxidation state of chromium be \(X\). The charges of the ligands are as follows:
- Chloride (\(Cl^-\)): -1 (2 ligands contribute -2)
- Nitrite (\(NO_2^-\)): -1 (2 ligands contribute -2)
- Ammonia (\(NH_3\)): 0 (2 ligands contribute 0)

The overall charge of the complex is -1. Therefore, we can set up the equation:
\[ X - 2 - 2 = -1 \]
Solving for \(X\):
\[ X - 4 = -1 \implies X = +3 \]
Thus, the oxidation state of chromium is +3.

### Step 2: Write the Electron Configuration of Chromium
The ground state electron configuration of chromium (Cr) is:
\[ [Ar] 4s^1 3d^5 \]
For \(Cr^{3+}\), we remove 3 electrons (1 from 4s and 2 from 3d):
\[ Cr^{3+}: [Ar] 3d^3 \]

### Step 3: Determine the Hybridization
In an octahedral complex, the central metal ion forms six coordinate bonds with ligands. The hybridization can be determined by the number of orbitals involved in bonding.

For \(Cr^{3+}\) with a \(3d^3\) configuration, we will use the \(d\), \(s\), and \(p\) orbitals to form hybrid orbitals. The hybridization for an octahedral complex is typically \(d^2sp^3\).

### Step 4: Count the Number of Ligands
The complex has:
- 2 \(Cl^-\) ligands
- 2 \(NO_2^-\) ligands
- 2 \(NH_3\) ligands

This gives a total of 6 ligands, which confirms the octahedral geometry.

### Step 5: Conclusion
The hybridization of the complex \([CrCl_2(NO_2)_2(NH_3)_2]^-\) is \(d^2sp^3\).

### Summary
- Oxidation state of Cr: +3
- Electron configuration of \(Cr^{3+}\): \([Ar] 3d^3\)
- Hybridization: \(d^2sp^3\)