Among the triatomic molecules/ions `BeCl_(2),N_(3)^(-),N_(2)O, NO_(2)^(+), O_(3), SCl_(2), lCl_(2)^(-),l_(3)^(-)` and `XeF_(2)`, the total number of linear molecules (s)/ion(s) where the hybridisation of the central atom does not have contribution from the `d`- orbitals (s) is [atomic number of `S=16, Cl=17, I=53` and `Xe=54`]

Explanation

To determine the total number of linear molecules or ions among the given list, where the hybridization of the central atom does not involve d-orbitals, we will analyze each molecule/ion step by step.

### Step 1: Analyze BeCl₂
- **Valence Electrons Calculation**:
- Beryllium (Be) has 2 valence electrons.
- Chlorine (Cl) has 7 valence electrons, and there are 2 Cl atoms.
- Total = 2 + 2(7) = 16 valence electrons.

- **Determine Hybridization**:
- V = 16, V/8 = 2 (quotient), remainder = 0.
- Bond pairs = 2, Lone pairs = 0.
- Hybridization = sp.

- **Shape**: Linear.

### Step 2: Analyze N₃⁻
- **Valence Electrons Calculation**:
- Nitrogen (N) has 5 valence electrons, and there are 3 N atoms.
- Extra negative charge adds 1 electron.
- Total = 3(5) + 1 = 16 valence electrons.

- **Determine Hybridization**:
- V = 16, V/8 = 2 (quotient), remainder = 0.
- Bond pairs = 2, Lone pairs = 0.
- Hybridization = sp.

- **Shape**: Linear.

### Step 3: Analyze N₂O
- **Valence Electrons Calculation**:
- Nitrogen (N) has 5 valence electrons, and there are 2 N atoms and 1 O atom.
- Total = 2(5) + 6 = 16 valence electrons.

- **Determine Hybridization**:
- V = 16, V/8 = 2 (quotient), remainder = 0.
- Bond pairs = 2, Lone pairs = 0.
- Hybridization = sp.

- **Shape**: Linear.

### Step 4: Analyze NO₂⁺
- **Valence Electrons Calculation**:
- Oxygen (O) has 6 valence electrons, and there are 2 O atoms.
- Nitrogen (N) has 5 valence electrons, and we subtract 1 for the positive charge.
- Total = 2(6) + 5 - 1 = 16 valence electrons.

- **Determine Hybridization**:
- V = 16, V/8 = 2 (quotient), remainder = 0.
- Bond pairs = 2, Lone pairs = 0.
- Hybridization = sp.

- **Shape**: Linear.

### Step 5: Analyze O₃
- **Valence Electrons Calculation**:
- Oxygen (O) has 6 valence electrons, and there are 3 O atoms.
- Total = 3(6) = 18 valence electrons.

- **Determine Hybridization**:
- V = 18, V/8 = 2 (quotient), remainder = 2.
- Bond pairs = 2, Lone pairs = 1.
- Hybridization = sp².

- **Shape**: Bent (not linear).

### Step 6: Analyze SCl₂
- **Valence Electrons Calculation**:
- Sulfur (S) has 6 valence electrons, and there are 2 Cl atoms.
- Total = 6 + 2(7) = 20 valence electrons.

- **Determine Hybridization**:
- V = 20, V/8 = 2 (quotient), remainder = 4.
- Bond pairs = 2, Lone pairs = 2.
- Hybridization = sp³.

- **Shape**: Bent (not linear).

### Step 7: Analyze ICl₂⁻
- **Valence Electrons Calculation**:
- Iodine (I) has 7 valence electrons, and there are 2 Cl atoms plus an extra electron for the negative charge.
- Total = 7 + 2(7) + 1 = 22 valence electrons.

- **Determine Hybridization**:
- V = 22, V/8 = 2 (quotient), remainder = 6.
- Bond pairs = 2, Lone pairs = 3.
- Hybridization = sp³d.

- **Shape**: Linear, but hybridization includes d-orbitals (not required).

### Step 8: Analyze I₃⁻
- **Valence Electrons Calculation**:
- Total = 3(7) + 1 = 22 valence electrons.

- **Determine Hybridization**:
- V = 22, V/8 = 2 (quotient), remainder = 6.
- Bond pairs = 2, Lone pairs = 3.
- Hybridization = sp³d.

- **Shape**: Linear, but hybridization includes d-orbitals (not required).

### Step 9: Analyze XeF₂
- **Valence Electrons Calculation**:
- Xenon (Xe) has 8 valence electrons, and there are 2 F atoms.
- Total = 8 + 2(7) = 22 valence electrons.

- **Determine Hybridization**:
- V = 22, V/8 = 2 (quotient), remainder = 6.
- Bond pairs = 2, Lone pairs = 3.
- Hybridization = sp³d.

- **Shape**: Linear, but hybridization includes d-orbitals (not required).

### Summary of Linear Molecules/Ions
The linear molecules/ions that do not involve d-orbitals in hybridization are:
1. BeCl₂
2. N₃⁻
3. N₂O
4. NO₂⁺

**Total Count**: 4

### Final Answer
The total number of linear molecules/ions where the hybridization of the central atom does not have contribution from the d-orbitals is **4**.