The correct statement about `ICl_(5)` and `ICl_(4)^(-)` is

Options

A. both are isostructural.
B. `ICl_(5)` is trigonal bipyramidal and `ICl_(4)^(-)` is tetrahedral.
C. `ICl_(5)` is square pyramidal and `ICl_(4)^(-)` is tetrahedral.
D. `ICl_(5)` is square pyramidal and `ICl_(4)^(-)` is square planar.

Answer

To determine the correct statement about \( \text{ICl}_5 \) and \( \text{ICl}_4^- \), we need to analyze the structures and geometries of both compounds.

### Step 1: Determine the Valence Electrons
- **Iodine (I)** has 7 valence electrons.
- **Chlorine (Cl)** has 7 valence electrons, and there are 5 Cl atoms in \( \text{ICl}_5 \) and 4 Cl atoms in \( \text{ICl}_4^- \).

### Step 2: Calculate Total Valence Electrons
For \( \text{ICl}_5 \):
- Total = 7 (I) + 5 × 7 (Cl) = 7 + 35 = 42 electrons.

For \( \text{ICl}_4^- \):
- Total = 7 (I) + 4 × 7 (Cl) + 1 (extra electron due to charge) = 7 + 28 + 1 = 36 electrons.

### Step 3: Determine the Steric Number
The steric number is calculated as:
\[ \text{Steric Number} = \text{Number of Bond Pairs} + \text{Number of Lone Pairs} \]

**For \( \text{ICl}_5 \)**:
- Bond pairs: 5 (one for each Cl)
- Lone pairs: 1 (remaining electrons)
- Steric Number = 5 + 1 = 6.

**For \( \text{ICl}_4^- \)**:
- Bond pairs: 4 (one for each Cl)
- Lone pairs: 2 (remaining electrons)
- Steric Number = 4 + 2 = 6.

### Step 4: Determine Hybridization and Geometry
- **For \( \text{ICl}_5 \)**:
- Steric Number = 6 → Hybridization = \( \text{sp}^3\text{d}^2 \)
- Geometry = Octahedral.
- Shape (considering lone pairs) = Square Pyramidal.

- **For \( \text{ICl}_4^- \)**:
- Steric Number = 6 → Hybridization = \( \text{sp}^3\text{d}^2 \)
- Geometry = Octahedral.
- Shape (considering lone pairs) = Square Planar.

### Step 5: Compare Structures
- \( \text{ICl}_5 \) has a **Square Pyramidal** shape.
- \( \text{ICl}_4^- \) has a **Square Planar** shape.

### Conclusion
The correct statement about \( \text{ICl}_5 \) and \( \text{ICl}_4^- \) is:
- **\( \text{ICl}_5 \) is Square Pyramidal and \( \text{ICl}_4^- \) is Square Planar.**

### Final Answer
The correct option is: **D. \( \text{ICl}_5 \) is Square Pyramidal and \( \text{ICl}_4^- \) is Square Planar.**