Which of the following is/are correct about `[Cu(NH_(3))_(4)]SO_(4)` :

Options

A. It is a Tetrahedral complex
B. It is paramagentic with one unpaired electron in the d-subshell
C. It gives white precipitate with `BaCl_(2)`
D. Its aqueous solution does not conduct electricity.

Explanation

To determine the correct statements about the complex \([Cu(NH_3)_4]SO_4\), we will analyze the geometry, magnetic properties, precipitation reactions, and electrical conductivity of the complex step by step.

### Step 1: Determine the oxidation state of copper in the complex
The complex is \([Cu(NH_3)_4]SO_4\). The sulfate ion \((SO_4)^{2-}\) has a charge of -2. Since the complex is neutral overall, the charge of the copper complex \([Cu(NH_3)_4]^{2+}\) must be +2.

**Hint:** Remember that the charge of the complex must balance the charge of the sulfate ion to maintain overall neutrality.

### Step 2: Find the electronic configuration of \(Cu^{2+}\)
Copper has an atomic number of 29, and its electronic configuration is \(Ar \, 3d^{10} \, 4s^1\). When copper loses two electrons to form \(Cu^{2+}\), the electronic configuration becomes \(Ar \, 3d^9\).

**Hint:** When determining the oxidation state, always adjust the electronic configuration accordingly by removing the appropriate number of electrons.

### Step 3: Analyze the ligand and its effect on electron pairing
The ligand \(NH_3\) is a strong field ligand, which causes electron pairing. In \(Cu^{2+}\), there are 9 electrons in the \(3d\) subshell. The strong field nature of \(NH_3\) will pair one of the \(3d\) electrons, leading to the configuration \(3d^{10}\) and leaving the \(4s\) empty.

**Hint:** Strong field ligands can cause pairing of electrons in the \(d\) orbitals, which is crucial for determining the geometry and magnetic properties.

### Step 4: Determine the hybridization and geometry of the complex
With four \(NH_3\) ligands and the \(Cu^{2+}\) ion, the hybridization is \(sp^3\), leading to a square planar geometry. This is typical for \(d^8\) metal complexes.

**Hint:** The number of ligands and their arrangement can provide insights into the hybridization and geometry of the complex.

### Step 5: Assess the magnetic properties of the complex
Since all \(3d\) electrons are paired in the \(3d\) subshell, the complex is diamagnetic. However, if we consider the \(4p\) orbital, there is one unpaired electron in the \(4p\) orbital, making the complex paramagnetic.

**Hint:** Check the electron configuration after considering ligand effects to determine if there are any unpaired electrons.

### Step 6: Examine the reaction with barium chloride
When \([Cu(NH_3)_4]^{2+}\) dissociates in solution, it releases \(SO_4^{2-}\). The sulfate ion reacts with barium chloride \((BaCl_2)\) to form barium sulfate \((BaSO_4)\), which is a white precipitate.

**Hint:** Look for ions that can form precipitates when mixed with other solutions.

### Step 7: Evaluate the electrical conductivity of the aqueous solution
Since the complex dissociates into ions in solution, it will conduct electricity. Therefore, the aqueous solution of the complex does conduct electricity.

**Hint:** The presence of free ions in solution is key to determining if a solution can conduct electricity.

### Conclusion
Based on the analysis:
1. The statement that it is a tetrahedral complex is incorrect; it is square planar.
2. The statement that it is paramagnetic with one unpaired electron in the \(d\) sub-shell is incorrect; the unpaired electron is in the \(p\) orbital.
3. The statement that it gives a white precipitate with \(BaCl_2\) is correct.
4. The statement that its aqueous solution does not conduct electricity is incorrect; it does conduct electricity.

**Correct Answer:** The only correct statement is that it gives a white precipitate with \(BaCl_2\).