`[Pt(NH_3)_4]CL_2` is

Options

A. pyramidal
B. pentagonal
C. tetrahedral
D. square planar

Answer

To determine the shape of the coordination compound `[Pt(NH₃)₄]Cl₂`, we will follow these steps:

### Step 1: Determine the oxidation state of platinum (Pt).
- Ammonia (NH₃) is a neutral ligand, contributing 0 to the overall charge.
- Each chloride ion (Cl) has a charge of -1, and there are 2 chloride ions, contributing a total of -2.
- Let the oxidation state of platinum be \( x \). The overall charge of the complex is neutral, so we can set up the equation:
\[ x + 0 + (-2) = 0 \]
This simplifies to:
\[ x - 2 = 0 \implies x = +2 \]
- Therefore, the oxidation state of platinum in this complex is +2.

### Step 2: Identify the electron configuration of platinum.
- Platinum (Pt) has an atomic number of 78. Its electron configuration is:
\[ [Xe] 4f^{14} 5d^9 6s^1 \]
- In the +2 oxidation state, platinum loses two electrons, typically from the 6s and one from the 5d subshell:
\[ [Xe] 4f^{14} 5d^8 \]

### Step 3: Determine the hybridization of the complex.
- Since platinum is in the +2 oxidation state and is surrounded by 4 ammonia ligands (which are strong field ligands), we can use the Valence Bond Theory (VBT) to determine the hybridization.
- The presence of 4 ligands suggests that we will have \( dsp^2 \) hybridization, which involves one d orbital, one s orbital, and two p orbitals.

### Step 4: Determine the geometry of the complex.
- The \( dsp^2 \) hybridization corresponds to a square planar geometry.
- Therefore, the shape of the coordination compound `[Pt(NH₃)₄]Cl₂` is square planar.

### Final Answer:
The coordination compound `[Pt(NH₃)₄]Cl₂` has a square planar geometry.

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