The hybridization of 'Cr' in the complex`[Cr(NO_(2))_(4)(NH_(3))_(2)]^(-)`

Options

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

Answer

To determine the hybridization of chromium (Cr) in the complex \([Cr(NO_2)_4(NH_3)_2]^-\), we will follow these steps:

### Step 1: Identify the oxidation state of chromium
1. The complex has 4 nitro ligands (NO₂⁻) and 2 amine ligands (NH₃).
2. The nitro ligand is a negatively charged ligand (-1 charge), and the amine ligand is neutral (0 charge).
3. The overall charge of the complex is -1.
4. Set up the equation for the oxidation state (x) of chromium:
\[ x + (4 \times -1) + (2 \times 0) = -1 \]
\[ x - 4 = -1 \implies x = +3 \]
5. Therefore, chromium is in the +3 oxidation state.

### Step 2: Determine the electron configuration of chromium
1. The atomic number of chromium (Cr) is 24.
2. The ground state electron configuration of Cr is \([Ar] 4s^1 3d^5\).
3. When chromium is in the +3 oxidation state, it loses 3 electrons (2 from 4s and 1 from 3d):
\[ [Ar] 4s^0 3d^3 \]
4. This means we have 3 electrons in the 3d subshell.

### Step 3: Determine the hybridization
1. The complex has a total of 6 ligands (4 nitro and 2 amine), which means it has an octahedral geometry.
2. In octahedral complexes, the hybridization can be determined by the number of ligands and the type of orbitals involved.
3. The 3d, 4s, and 4p orbitals will be involved in hybridization to accommodate 6 electrons.
4. The hybridization for this configuration is \(d^2sp^3\) because:
- 2 d-orbitals (from 3d)
- 1 s-orbital (from 4s)
- 3 p-orbitals (from 4p)
5. Thus, the hybridization of chromium in this complex is \(d^2sp^3\).

### Conclusion
The hybridization of chromium in the complex \([Cr(NO_2)_4(NH_3)_2]^-\) is \(d^2sp^3\).

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