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 in the complex \([Cr(NO_2)_4(NH_3)_2]^-\), we can follow these steps:

### Step 1: Determine the Oxidation State of Chromium
The first step is to find the oxidation state of chromium (Cr) in the complex.

- The nitrito ligand (NO₂) has a charge of -1. Since there are four nitrito ligands, their total contribution to the charge is \(4 \times (-1) = -4\).
- The ammonia (NH₃) ligand is neutral, contributing 0 to the charge.
- The overall charge of the complex is -1.

Let \(x\) be the oxidation state of chromium. We can set up the equation:

\[ x + (-4) + 0 = -1 \]
\[ x - 4 = -1 \]
\[ x = +3 \]

So, the oxidation state of chromium in this complex is +3.

### Step 2: Determine the Electron Configuration of Chromium
The electron configuration of chromium (Cr) in its elemental form (atomic number 24) is:

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

Since we have determined that the oxidation state of chromium in this complex is +3, we need to remove three electrons from the electron configuration:

1. Remove the 4s electron first: \(4s^0\)
2. Remove two 3d electrons: \(3d^3\)

Thus, the electron configuration of Cr in this complex is:

\[ [Ar] 3d^3 \]

### Step 3: Identify the Number of Ligands and Geometry
In the complex \([Cr(NO_2)_4(NH_3)_2]^-\):
- There are 4 nitrito ligands and 2 ammonia ligands, making a total of 6 ligands.

Since there are 6 ligands, the geometry of the complex is octahedral.

### Step 4: Determine the Hybridization
In an octahedral complex, the hybridization can be determined by the number of d, s, and p orbitals involved in hybridization.

For Cr in the +3 oxidation state with a \(3d^3\) configuration:
- The d orbitals involved are \(3d\).
- The s orbital involved is \(4s\).
- The p orbitals involved are \(4p\).

In an octahedral field, the \(d\) orbitals split into two sets: \(t_{2g}\) (lower energy) and \(e_g\) (higher energy). The \(3d\) orbitals will be used for bonding.

The hybridization can be determined as follows:
- 3 \(d\) orbitals + 2 \(p\) orbitals + 1 \(s\) orbital gives us \(d^2sp^3\) hybridization.

Thus, the hybridization of chromium in the complex \([Cr(NO_2)_4(NH_3)_2]^-\) is:

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

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

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