Which of the following is correct about `[ICl_(4)]^(-)` and `ICl_(5)` compound

Options

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

Answer

To determine the correct statements regarding the structures of \([ICl_4]^-\) and \(ICl_5\), we will analyze the molecular geometry of both compounds step by step.

### Step 1: Determine the Valence Electrons
- **For \([ICl_4]^-\)**:
- Iodine (I) has 7 valence electrons.
- Each chlorine (Cl) has 7 valence electrons, and there are 4 chlorine atoms contributing a total of \(4 \times 7 = 28\) electrons.
- The negative charge adds 1 additional electron.
- Total = \(7 + 28 + 1 = 36\) valence electrons.

- **For \(ICl_5\)**:
- Iodine has 7 valence electrons.
- Each chlorine has 7 valence electrons, and there are 5 chlorine atoms contributing a total of \(5 \times 7 = 35\) electrons.
- Total = \(7 + 35 = 42\) valence electrons.

### Step 2: Determine the Electron Pair Geometry
- **For \([ICl_4]^-\)**:
- Total valence electrons = 36.
- The central iodine atom will be surrounded by 4 bonding pairs (from the 4 Cl atoms) and 2 lone pairs.
- This gives a total of \(4 + 2 = 6\) electron pairs.

- **For \(ICl_5\)**:
- Total valence electrons = 42.
- The central iodine atom will be surrounded by 5 bonding pairs (from the 5 Cl atoms) and 1 lone pair.
- This gives a total of \(5 + 1 = 6\) electron pairs.

### Step 3: Determine the Molecular Geometry
- **For \([ICl_4]^-\)**:
- With 4 bonding pairs and 2 lone pairs, the molecular geometry is square planar.

- **For \(ICl_5\)**:
- With 5 bonding pairs and 1 lone pair, the molecular geometry is square pyramidal.

### Conclusion
- \([ICl_4]^-\) is square planar.
- \(ICl_5\) is square pyramidal.

Thus, the correct statement is: **ICl4- is square planar and ICl5 is square pyramidal.** This corresponds to option 2.