To solve the problem, we need to analyze the two statements regarding the molecular geometries and hybridization of ClF₂⁻ and ClF₂⁺.
### Step-by-Step Solution:
1. **Identify the Valence Electrons of Chlorine:**
- Chlorine (Cl) is in Group 17 of the periodic table, which means it has 7 valence electrons.
2. **Calculate the Hybridization for ClF₂⁻:**
- For ClF₂⁻, we need to calculate the number of hybrid orbitals using the formula:
\[
X = \frac{1}{2} \left( \text{Valence electrons} + \text{Monovalent atoms} - \text{Charge on cation} + \text{Charge on anion} \right)
\]
- Here, the valence electrons = 7 (from Cl), monovalent atoms = 2 (from 2 F), charge on cation = 0 (since it is neutral), and charge on anion = -1.
- Plugging in the values:
\[
X = \frac{1}{2} \left( 7 + 2 - 0 + 1 \right) = \frac{1}{2} \left( 10 \right) = 5
\]
- Since \(X = 5\), the hybridization is sp³d.
3. **Determine the Molecular Geometry of ClF₂⁻:**
- With sp³d hybridization and 2 bond pairs (from the two F atoms) and 3 lone pairs, the molecular geometry is trigonal bipyramidal.
- However, due to the presence of lone pairs, the shape is linear because the lone pairs occupy the equatorial positions.
4. **Calculate the Hybridization for ClF₂⁺:**
- Now, for ClF₂⁺, we use the same formula:
\[
X = \frac{1}{2} \left( \text{Valence electrons} + \text{Monovalent atoms} - \text{Charge on cation} + \text{Charge on anion} \right)
\]
- Here, the valence electrons = 7, monovalent atoms = 2, charge on cation = +1 (since it is a cation), and charge on anion = 0.
- Plugging in the values:
\[
X = \frac{1}{2} \left( 7 + 2 - 1 + 0 \right) = \frac{1}{2} \left( 8 \right) = 4
\]
- Since \(X = 4\), the hybridization is sp³.
5. **Determine the Molecular Geometry of ClF₂⁺:**
- With sp³ hybridization, there are 2 bond pairs and 2 lone pairs. The molecular geometry is tetrahedral, but the shape is bent (or V-shaped) due to the two lone pairs.
6. **Evaluate the Statements:**
- **Statement 1:** ClF₂⁻ is linear while ClF₂⁺ is bent. This statement is **true**.
- **Statement 2:** The Cl atom in ClF₂⁻ and ClF₂⁺ has the same state of hybridization. This statement is **false** (one is sp³d and the other is sp³).
### Final Conclusion:
- Statement 1 is true.
- Statement 2 is false.
