In `[Pt(NH_(3))_(2)Cl_(2)]`, pt-Cl bond length is 2 Å and Cl-Cl distance is 2.88 Å then the compound is:

Options

A. tetrahedral
B. square pyramidal
C. cis-square planar
D. trans-square planar

Answer

To determine the nature of the compound `[Pt(NH₃)₂Cl₂]`, we need to analyze the given bond lengths and the coordination environment of the platinum ion. Here’s a step-by-step solution:

### Step 1: Identify the oxidation state of platinum
In the complex `[Pt(NH₃)₂Cl₂]`, ammonia (NH₃) is a neutral ligand and chlorine (Cl) is a negatively charged ligand. Since there are two Cl ligands, we can set up the equation for the oxidation state of platinum (Pt):

Let the oxidation state of Pt be \( x \).
\[ x + 2(0) + 2(-1) = 0 \]
\[ x - 2 = 0 \implies x = +2 \]

**Hint:** Remember that the charge of neutral ligands is zero, while negatively charged ligands contribute negatively to the overall charge.

### Step 2: Determine the electronic configuration of Pt²⁺
The electronic configuration of neutral platinum (Pt) is \([Xe] 4f^{14} 5d^9 6s^1\). For Pt²⁺, we remove two electrons (typically from the outermost shell):

\[ \text{Pt}^{2+}: [Xe] 4f^{14} 5d^8 \]

**Hint:** The oxidation state affects the electronic configuration by removing electrons from the outermost orbitals.

### Step 3: Analyze the coordination number and geometry
The coordination number of platinum in this complex is 4, as it is coordinated by two ammonia ligands and two chloride ligands. The geometry for a coordination number of 4 can be either tetrahedral or square planar. However, for d⁸ metal ions like Pt²⁺, the square planar geometry is more common.

**Hint:** The coordination number helps determine the geometry of the complex.

### Step 4: Determine the hybridization
In a square planar geometry, the hybridization is typically \( dsp^2 \) or \( dsp^3 \) depending on the ligands. For Pt²⁺ with 4 ligands, it undergoes \( dsp^2 \) hybridization.

**Hint:** The hybridization can be determined based on the geometry and the number of ligands.

### Step 5: Analyze bond lengths and distances
We are given the Pt-Cl bond length as 2 Å and the Cl-Cl distance as 2.88 Å. In a square planar arrangement, if we assume a cis configuration, the Cl-Cl distance should be less than the sum of the bond lengths (2 Å + 2 Å = 4 Å). Since the Cl-Cl distance is 2.88 Å, which is less than 4 Å, this supports the cis arrangement.

**Hint:** Compare the bond lengths and distances to determine the configuration (cis or trans).

### Conclusion
Based on the analysis, the compound `[Pt(NH₃)₂Cl₂]` is a **cis** isomer due to the bond lengths and the geometry of the complex.

**Final Answer:** The compound is a **cis isomer**.