The set in which all the species are 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 the set in which all the species are planar, we will analyze each species given in the sets by calculating their hybridization and drawing their molecular geometries. Here’s a step-by-step solution:

### Step 1: Analyze the first set of species.

1. **I3- (Iodide ion)**:
- Valence electrons: Iodine has 7 valence electrons, and there are 2 monovalent atoms (I).
- Hybridization calculation:
\[ \text{Hybridization} = \frac{1}{2} \left( V + M - C + A \right) = \frac{1}{2} \left( 7 + 2 + 1 \right) = \frac{10}{2} = 5 \quad \text{(sp}^3\text{d)} \]
- Geometry: The structure is trigonal bipyramidal with three iodine atoms in a plane, making it planar.

2. **CO2 (Carbon dioxide)**:
- Valence electrons: Carbon has 4, and there are 2 oxygen atoms (not counted as monovalent).
- Hybridization calculation:
\[ \text{Hybridization} = \frac{1}{2} \left( 4 + 0 \right) = \frac{4}{2} = 2 \quad \text{(sp)} \]
- Geometry: Linear structure, hence planar.

3. **XeF4 (Xenon tetrafluoride)**:
- Valence electrons: Xenon has 8, and there are 4 fluorine atoms.
- Hybridization calculation:
\[ \text{Hybridization} = \frac{1}{2} \left( 8 + 4 \right) = \frac{12}{2} = 6 \quad \text{(sp}^3\text{d}^2) \]
- Geometry: Square planar, hence planar.

4. **I2Cl6 (Dimer of ICl3)**:
- Valence electrons: Each iodine has 7, and there are 6 chlorine atoms.
- Hybridization calculation:
\[ \text{Hybridization} = \frac{1}{2} \left( 7 + 6 \right) = \frac{13}{2} = 6.5 \quad \text{(complex structure)} \]
- Geometry: Planar due to its dimeric structure.

### Conclusion for the first set:
All species in the first set (I3-, CO2, XeF4, I2Cl6) are planar.

### Step 2: Analyze the second set of species.

1. **CO2**: Already analyzed as planar.
2. **SF4 (Sulfur tetrafluoride)**:
- Valence electrons: Sulfur has 6, and there are 4 fluorine atoms.
- Hybridization calculation:
\[ \text{Hybridization} = \frac{1}{2} \left( 6 + 4 \right) = \frac{10}{2} = 5 \quad \text{(sp}^3\text{d)} \]
- Geometry: Seesaw shape, not planar.

3. **ClF3 (Chlorine trifluoride)**:
- Valence electrons: Chlorine has 7, and there are 3 fluorine atoms.
- Hybridization calculation:
\[ \text{Hybridization} = \frac{1}{2} \left( 7 + 3 \right) = \frac{10}{2} = 5 \quad \text{(sp}^3\text{d)} \]
- Geometry: T-shaped, not planar.

4. **BrF5 (Bromine pentafluoride)**:
- Valence electrons: Bromine has 7, and there are 5 fluorine atoms.
- Hybridization calculation:
\[ \text{Hybridization} = \frac{1}{2} \left( 7 + 5 \right) = \frac{12}{2} = 6 \quad \text{(sp}^3\text{d}^2) \]
- Geometry: Square pyramidal, not planar.

### Conclusion for the second set:
Not all species are planar.

### Step 3: Analyze the third set of species.

1. **XeF4**: Already analyzed as planar.
2. **H2O (Water)**:
- Valence electrons: Oxygen has 6, and there are 2 hydrogen atoms.
- Hybridization calculation:
\[ \text{Hybridization} = \frac{1}{2} \left( 6 + 2 \right) = \frac{8}{2} = 4 \quad \text{(sp}^3\text{)} \]
- Geometry: Bent shape, not planar.

3. **XeO4 (Xenon tetraoxide)**:
- Valence electrons: Xenon has 8, and there are 4 oxygen atoms.
- Hybridization calculation:
\[ \text{Hybridization} = \frac{1}{2} \left( 8 + 4 \right) = \frac{12}{2} = 6 \quad \text{(sp}^3\text{d}^2) \]
- Geometry: Tetrahedral, not planar.

4. **PF3 (Phosphorus trifluoride)**:
- Valence electrons: Phosphorus has 5, and there are 3 fluorine atoms.
- Hybridization calculation:
\[ \text{Hybridization} = \frac{1}{2} \left( 5 + 3 \right) = \frac{8}{2} = 4 \quad \text{(sp}^3\text{)} \]
- Geometry: Trigonal pyramidal, not planar.

### Conclusion for the third set:
Not all species are planar.

### Step 4: Analyze the fourth set of species.

1. **ICl2+**:
- Valence electrons: Iodine has 7, and there are 2 chlorine atoms, minus 1 for the cationic charge.
- Hybridization calculation:
\[ \text{Hybridization} = \frac{1}{2} \left( 7 + 2 - 1 \right) = \frac{8}{2} = 4 \quad \text{(sp}^3\text{)} \]
- Geometry: Linear, hence planar.

2. **ICl2-**:
- Valence electrons: Iodine has 7, and there are 2 chlorine atoms, plus 1 for the anionic charge.
- Hybridization calculation:
\[ \text{Hybridization} = \frac{1}{2} \left( 7 + 2 + 1 \right) = \frac{10}{2} = 5 \quad \text{(sp}^3\text{d)} \]
- Geometry: Linear, hence planar.

3. **CO2**: Already analyzed as planar.
4. **XeO3 (Xenon trioxide)**:
- Valence electrons: Xenon has 8, and there are 3 oxygen atoms.
- Hybridization calculation:
\[ \text{Hybridization} = \frac{1}{2} \left( 8 + 3 \right) = \frac{11}{2} = 5.5 \quad \text{(complex structure)} \]
- Geometry: Not planar.

### Conclusion for the fourth set:
Not all species are planar.

### Final Answer:
The only set in which all the species are planar is the **first set**: {I3-, CO2, XeF4, I2Cl6}.

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