Out off `N_(2)O,SO_(2),I_(3)^(+),I_(3)^(-),H_(2)O,NO_(2)^(-),N_(3)^(-)`, the linear species are:

Options

A. `NO_(2)^(-),I_(3)^(+),H_(2)O`
B. `N_(2)O,I_(3)^(+),N_(3)^(-)`
C. `N_(2)O,I_(3)^(-),N_(3)^(-)`
D. `N_(3)^(-),I_(3)^(-),NO_(2)^(-)`

Answer

To determine which of the given species are linear, we will analyze each one step by step based on their valence electrons and molecular geometry.

### Step 1: Analyze N₂O (Dinitrogen monoxide)
1. **Count the Valence Electrons**:
- Nitrogen (N) has 5 valence electrons, and there are 2 nitrogen atoms: \(5 \times 2 = 10\).
- Oxygen (O) has 6 valence electrons: \(6\).
- Total = \(10 + 6 = 16\) valence electrons.

2. **Determine Electron Pairs**:
- Since 16 is greater than 8, we divide by 2 to find the number of electron pairs: \(16 / 2 = 8\) electron pairs.

3. **Hybridization**:
- 8 electron pairs correspond to sp hybridization, which indicates a linear shape.

### Step 2: Analyze SO₂ (Sulfur dioxide)
1. **Count the Valence Electrons**:
- Sulfur (S) has 6 valence electrons, and there are 2 oxygen atoms: \(6 + (6 \times 2) = 18\) valence electrons.

2. **Determine Electron Pairs**:
- Since 18 is greater than 8, we divide by 2: \(18 / 2 = 9\) electron pairs.

3. **Hybridization**:
- 9 electron pairs correspond to sp³d hybridization, which does not indicate a linear shape.

### Step 3: Analyze I₃⁻ (Triiodide ion)
1. **Count the Valence Electrons**:
- Iodine (I) has 7 valence electrons, and there are 3 iodine atoms: \(7 \times 3 = 21\).
- Add 1 for the negative charge: \(21 + 1 = 22\) valence electrons.

2. **Determine Electron Pairs**:
- Since 22 is greater than 8, we divide by 8: \(22 / 8 = 2\) with a remainder of 6.
- Divide the remainder by 2: \(6 / 2 = 3\).
- Total electron pairs = \(2 + 3 = 5\).

3. **Hybridization**:
- 5 electron pairs correspond to sp³d hybridization, which indicates a linear shape.

### Step 4: Analyze I₃⁻ (Triiodide ion) again
1. **Count the Valence Electrons**:
- As calculated before, I₃⁻ has 22 valence electrons.

2. **Determine Electron Pairs**:
- Total electron pairs = 5.

3. **Hybridization**:
- The shape is linear due to the presence of 3 lone pairs and 2 bond pairs.

### Step 5: Analyze H₂O (Water)
1. **Count the Valence Electrons**:
- Oxygen has 6 valence electrons, and there are 2 hydrogen atoms: \(6 + (1 \times 2) = 8\) valence electrons.

2. **Determine Electron Pairs**:
- Total electron pairs = \(8 / 2 = 4\).

3. **Hybridization**:
- 4 electron pairs correspond to sp³ hybridization, which does not indicate a linear shape.

### Step 6: Analyze NO₂⁻ (Nitrite ion)
1. **Count the Valence Electrons**:
- Nitrogen has 5 valence electrons, and there are 2 oxygen atoms: \(5 + (6 \times 2) = 17\).
- Add 1 for the negative charge: \(17 + 1 = 18\) valence electrons.

2. **Determine Electron Pairs**:
- Total electron pairs = \(18 / 2 = 9\).

3. **Hybridization**:
- 9 electron pairs correspond to sp³d hybridization, which does not indicate a linear shape.

### Step 7: Analyze N₃⁻ (Azide ion)
1. **Count the Valence Electrons**:
- Nitrogen has 5 valence electrons, and there are 3 nitrogen atoms: \(5 \times 3 = 15\).
- Add 1 for the negative charge: \(15 + 1 = 16\) valence electrons.

2. **Determine Electron Pairs**:
- Total electron pairs = \(16 / 2 = 8\).

3. **Hybridization**:
- 8 electron pairs correspond to sp hybridization, which indicates a linear shape.

### Summary of Linear Species
- The linear species identified are:
- N₂O (linear)
- I₃⁻ (linear)
- N₃⁻ (linear)

### Final Answer
The linear species from the given options are: **N₂O, I₃⁻, and N₃⁻.**