There are four species `CO_(2),N_(3) Θ,NO_(2)^(o+)` and `I_(3)^(Θ)` Which of the following statement is correct about these species ? .

Options

A. All are linear and having sp hybridisation central atoms .
B. All are linear but only `CO_(2),N_(3)^(Θ)` and `I_(3)^(Θ)` have sp hybridisation on their central atom .
C. All are linear but only `CO_(2),N_(3)^(Θ)` and `overset(o+)NO_(2)` have sp hybridisation on their cental atom .
D. `CO_(2),N_(3)^(Θ)` and `NO_(3)^(o+)` are linear but `I_(3)^(Θ)` is not .

Explanation

To determine the correct statement about the species \( CO_2 \), \( N_3^- \), \( NO_2^+ \), and \( I_3^- \), we need to analyze each species in terms of hybridization and molecular shape.

### Step 1: Analyze \( CO_2 \)
- **Central Atom**: Carbon (C)
- **Bonding Pairs**: 2 (double bonds with two oxygen atoms)
- **Lone Pairs**: 0
- **Hybridization**: The total number of electron pairs (bonding + lone) is 2, which corresponds to \( sp \) hybridization.
- **Shape**: Linear

### Step 2: Analyze \( N_3^- \)
- **Central Atom**: The central nitrogen atom in the linear structure.
- **Bonding Pairs**: 2 (triple bond between two nitrogen atoms and a single bond with the third nitrogen)
- **Lone Pairs**: 0
- **Hybridization**: The total number of electron pairs is 2, leading to \( sp \) hybridization.
- **Shape**: Linear

### Step 3: Analyze \( NO_2^+ \)
- **Central Atom**: Nitrogen (N)
- **Bonding Pairs**: 2 (double bond with one oxygen and a single bond with the other)
- **Lone Pairs**: 0
- **Hybridization**: The total number of electron pairs is 2, resulting in \( sp \) hybridization.
- **Shape**: Linear

### Step 4: Analyze \( I_3^- \)
- **Central Atom**: Iodine (I)
- **Bonding Pairs**: 2 (bonded to two other iodine atoms)
- **Lone Pairs**: 3
- **Hybridization**: The total number of electron pairs is 5 (2 bonding + 3 lone), leading to \( sp^3d \) hybridization.
- **Shape**: Linear (due to the arrangement of lone pairs in equatorial positions)

### Step 5: Summary of Findings
- \( CO_2 \): Linear, \( sp \) hybridized
- \( N_3^- \): Linear, \( sp \) hybridized
- \( NO_2^+ \): Linear, \( sp \) hybridized
- \( I_3^- \): Linear, \( sp^3d \) hybridized

### Conclusion
The correct statement is that all species are linear, but only \( CO_2 \), \( N_3^- \), and \( NO_2^+ \) have \( sp \) hybridization on their central atom.