Read the passage given below and answer the following questions:
In spite of the predictions of stable noble gas compounds since at least 1902, unsuccessful attempts at their synthesis gave rise to the widely held opinion that noble gases are not only noble but also inert. It was not until 1962 that this dogma was shattered when Bartlett in Canada published the first stable noble gas compound `XePtF_(6)` . This discovery triggered a worldwide frenzy in this area, and within a short time span many new xenon, radon, and krypton compounds were prepared and characterized. The recent discoveries show the ability of xenon to act as a ligand . The discovery by Seppelt’s group that more than one xenon atom can attach itself to a metal center which in the case of gold leads to surprisingly stable Au- Xe bonds. The bonding in `[AuXe_(4)]^(2+)` involves 4 Xe ligands attached by relatively strong bonds to a single Au(II) center in a square planar arrangement with a Xe-Au bond length of about 274 pm This discovery provides not only the first example of multiple xenon ligands but also represents the first strong metal - xenon bond.
(Source: Christe, K. O. (2001). A renaissance in noble gas chemistry. Angewandte Chemie International Edition, 40(8), 1419-1421.)
Hybridisation shown by Au in `[AuXe_(4)]^(2+)` is :

Options

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

Answer

To determine the hybridization of the gold (Au) in the complex \([AuXe_4]^{2+}\), we can follow these steps:

### Step 1: Identify the oxidation state of gold in the complex
In the complex \([AuXe_4]^{2+}\), the overall charge is +2. Since xenon (Xe) is a neutral ligand, the oxidation state of gold must be +2.

### Step 2: Determine the electron configuration of Au in the +2 oxidation state
The electron configuration of neutral gold (Au) is:
\[ \text{[Xe]} 4f^{14} 5d^{10} 6s^1 \]

When gold is in the +2 oxidation state, it loses two electrons. The electron configuration becomes:
\[ \text{[Xe]} 4f^{14} 5d^9 \]

### Step 3: Determine the number of ligands and their arrangement
In \([AuXe_4]^{2+}\), there are four xenon ligands coordinated to the gold center. The arrangement of these ligands is square planar.

### Step 4: Determine the hybridization
For a square planar arrangement, the hybridization can be determined as follows:
- The central atom (Au) uses one s orbital, two p orbitals, and two d orbitals to form bonds with the four ligands.
- Therefore, the hybridization is \(dsp^2\).

### Step 5: Conclusion
The hybridization of Au in \([AuXe_4]^{2+}\) is \(dsp^2\).

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