The set which was all the species planar is : -

Options

A. `I_(3)^(-), CO_(2) , XeF_(4) , I_(2)Cl_(6)`
B. `CO_(2), SF_(4), ClF_(3), BrF_(5)`
C. `XeF_(4), H_(2)O, XeO_(4), PF_(3)`
D. `Icl_(2)^(+), I Cl_(2)^(-), CO_(2), XeO_(3)`

Answer

To determine which set contains all planar species, we need to analyze each species given in the options. We will calculate the hybridization and molecular geometry of each species to identify if they are planar.

### Step 1: Analyze I3⁻
- **Valence Electrons**: Iodine (I) has 7 valence electrons. For I3⁻, we have:
- 3 Iodine atoms: 3 × 7 = 21 electrons
- 1 extra electron for the negative charge: +1
- Total = 21 + 1 = 22 electrons
- **Hybridization Calculation**:
\[ \text{Hybridization} = \frac{\text{Valence Electrons} + \text{Monovalent Atoms} - \text{Charge}}{2} = \frac{21 + 0 + 1}{2} = 11 \]
- This indicates 5 bond pairs and 3 lone pairs.
- **Geometry**: The structure is linear due to the presence of 3 lone pairs on the central iodine atom.

### Step 2: Analyze CO2
- **Valence Electrons**: Carbon (C) has 4 valence electrons, and each oxygen (O) has 6:
- 1 C + 2 O = 4 + 2 × 6 = 16 electrons
- **Hybridization Calculation**:
\[ \text{Hybridization} = \frac{4 + 2 \times 6}{2} = \frac{16}{2} = 8 \]
- This indicates 2 bond pairs and 0 lone pairs.
- **Geometry**: The structure is linear (O=C=O).

### Step 3: Analyze XeF4
- **Valence Electrons**: Xenon (Xe) has 8 valence electrons, and each fluorine (F) has 7:
- 1 Xe + 4 F = 8 + 4 × 7 = 36 electrons
- **Hybridization Calculation**:
\[ \text{Hybridization} = \frac{8 + 4}{2} = 6 \]
- This indicates 4 bond pairs and 2 lone pairs.
- **Geometry**: The structure is square planar.

### Step 4: Analyze I2Cl6
- **Valence Electrons**: Each iodine has 7, and each chlorine has 7:
- 2 I + 6 Cl = 2 × 7 + 6 × 7 = 56 electrons
- **Hybridization Calculation**:
\[ \text{Hybridization} = \frac{14 + 6}{2} = 10 \]
- This indicates 6 bond pairs and 0 lone pairs.
- **Geometry**: The structure is planar due to its dimeric form.

### Conclusion
All species analyzed (I3⁻, CO2, XeF4, and I2Cl6) are planar. Therefore, the correct answer is:

**Option 1: I3⁻, CO2, XeF4, I2Cl6**

---