Select the correct order of following property .

Options

A. % s -character : `sp^(3) gt sp^(2) gt sp`
B. `Ohat(N)O` bond angle : `NO_(3)^(-) gt overset(o+)(N)O_(2)`
C. All angles in `CH_(2)F_(2)` are not identical
D. C-F bond length , `CF_(4) gt CH_(3)F gt CH_(2)F_(2) gt CF_(3)H`

Answer

To solve the question of selecting the correct order of the given properties, we will analyze each property mentioned step by step.

### Step 1: Calculate Percentage S Character
1. **Identify Hybridization Types**: We have three types of hybridization: sp3, sp2, and sp.
2. **Calculate Percentage S Character**:
- For **sp3 hybridization**:
- Number of s orbitals = 1
- Total hybrid orbitals = 4
- Percentage s character = (1/4) * 100 = 25%
- For **sp2 hybridization**:
- Number of s orbitals = 1
- Total hybrid orbitals = 3
- Percentage s character = (1/3) * 100 = 33.33%
- For **sp hybridization**:
- Number of s orbitals = 1
- Total hybrid orbitals = 2
- Percentage s character = (1/2) * 100 = 50%

3. **Order of Percentage S Character**:
- sp < sp2 < sp3
- Therefore, the correct order is: 50% (sp) > 33.33% (sp2) > 25% (sp3)

### Step 2: Analyze Bond Angles
1. **Nitrate Ion (NO3-)**:
- Geometry: Trigonal planar
- Bond angle: Approximately 120 degrees
2. **Nitronium Ion (NO2+)**:
- Geometry: Linear
- Bond angle: 180 degrees
3. **Comparison**:
- The bond angle in nitrate is less than that in the nitronium ion.

### Step 3: Analyze Bond Angles in CH2F2
1. **Geometry of CH2F2**:
- Hybridization: sp3
- Bond angles: All angles are approximately 109.5 degrees.
2. **Conclusion**: All bond angles in CH2F2 are identical.

### Step 4: Analyze CF Bond Lengths
1. **CF4, CF3H, and other compounds**:
- All these compounds have sp3 hybridization.
- Therefore, all carbon-fluorine bond lengths will be equal due to the same hybridization and tetrahedral structure.

### Final Conclusion
- All the statements provided in the question are incorrect based on our analysis.
- Thus, the answer is: **None of these**.