Assertion Hybridisation of `[AuCI_(4)]^(Θ)` is `sp^(3)`
Reason Hybridisation of Au in above complex compound does not depend upon the nature of ligand .

Options

A. If both `(A)` and ` (R )` are correct and `(R )` is the correct explanation of `(A ) ` .
B. If both `(A)` and ` (R )` are correct and `(R )` is the correct explanation of `(A ) ` .
C. If`(A)` is correct, but `( R )` is incorrect .
D. Both `(A)` and `(R ) ` are incorrect .

Explanation

To solve the question, we need to analyze both the assertion and the reason provided regarding the hybridization of the complex ion \([AuCl_4]^{-}\).

### Step 1: Determine the oxidation state of Gold in \([AuCl_4]^{-}\)

- The complex ion \([AuCl_4]^{-}\) contains four chloride ions, each with a charge of -1.
- Let the oxidation state of Au be \(x\).
- The overall charge of the complex is -1, so we can set up the equation:
\[ x + 4(-1) = -1 \]
\[ x - 4 = -1 \implies x = +3 \]
- Thus, the oxidation state of Au in \([AuCl_4]^{-}\) is +3.

### Step 2: Determine the electronic configuration of Au in the +3 oxidation state

- The atomic number of gold (Au) is 79.
- The ground state electronic configuration of Au is:
\[ [Xe] 4f^{14} 5d^{10} 6s^{1} \]
- For Au in the +3 oxidation state, we remove three electrons:
- The 6s electron is removed first, followed by two electrons from the 5d subshell.
- Therefore, the electronic configuration of \([Au]^{3+}\) is:
\[ [Xe] 4f^{14} 5d^{8} \]

### Step 3: Analyze the hybridization

- The coordination number of \([AuCl_4]^{-}\) is 4, which typically suggests a hybridization of \(sp^3\) or \(dsp^2\).
- However, the nature of the ligands can influence the hybridization. Chloride is a weak field ligand and does not cause pairing of electrons in most cases.
- However, due to the high effective nuclear charge of gold, it can cause pairing of the d-electrons even in the presence of weak field ligands.

### Step 4: Determine the hybridization based on the electronic configuration

- In the case of \([AuCl_4]^{-}\), with the configuration of \(5d^8\), we can expect the following:
- The 5d orbitals can hybridize with the 4s and 4p orbitals to form \(dsp^2\) hybridization.
- Therefore, the correct hybridization for \([AuCl_4]^{-}\) is \(dsp^2\), not \(sp^3\).

### Conclusion

- **Assertion**: The hybridization of \([AuCl_4]^{-}\) is \(sp^3\) - **False**.
- **Reason**: The hybridization of Au in the above complex does not depend upon the nature of the ligand - **True**.

Thus, the assertion is incorrect, while the reason is correct.

### Final Answer
- Assertion: False
- Reason: True