The user wants us to find the total number of linear molecules or ions from the given list, with the condition that the hybridization of the central atom does not involve d-orbitals.

Let's analyze each species based on two criteria:

1. Shape: Is the molecule/ion linear?
2. Hybridization: Does the hybridization of the central atom involve d-orbitals? We are looking for hybridizations like sp, $sp^2$, $sp^3$. A triatomic species can only be linear if the hybridization is sp (Steric Number = 2), or $sp^3d$ (Steric Number = 5, with 3 lone pairs). The condition of no d-orbital contribution restricts us to sp hybridization.

We will determine the Steric Number (SN) for the central atom in each species, where SN = (number of sigma bonds) + (number of lone pairs).

1. BeCl${}_2$

   • Central atom: Beryllium (Be).
   • Valence electrons of Be: 2.
   • Be forms two single bonds with two Cl atoms. It uses both its valence electrons for bonding.
   • Sigma bonds = 2, Lone pairs = 0.
   • SN = 2 + 0 = 2.
   • Hybridization: sp.
   • Shape: Linear.
   • Conclusion: It is linear and its hybridization (sp) does not involve d-orbitals. (This molecule is counted).

2. N${}_3^-$ (Azide ion)

   • Central atom: Nitrogen (N).
   • The structure is $[N=N=N]^-$. The central N atom forms two sigma bonds (one with each adjacent N) and has no lone pairs.
   • Sigma bonds = 2, Lone pairs = 0.
   • SN = 2 + 0 = 2.
   • Hybridization: sp.
   • Shape: Linear.
   • Conclusion: It is linear and its hybridization (sp) does not involve d-orbitals. (This ion is counted).

3. N${}_2$O (Nitrous oxide)

   • Central atom: Nitrogen (N). The structure is N-N-O.
   • The central N atom forms two sigma bonds and has no lone pairs (structure N≡N⁺-O⁻).
   • Sigma bonds = 2, Lone pairs = 0.
   • SN = 2 + 0 = 2.
   • Hybridization: sp.
   • Shape: Linear.
   • Conclusion: It is linear and its hybridization (sp) does not involve d-orbitals. (This molecule is counted).

4. NO${}_2^+$ (Nitronium ion)

   • Central atom: Nitrogen (N).
   • The structure is $[O=N=O]^+$. The central N atom forms two sigma bonds with the two O atoms and has no lone pairs.
   • Sigma bonds = 2, Lone pairs = 0.
   • SN = 2 + 0 = 2.
   • Hybridization: sp.
   • Shape: Linear.
   • Conclusion: It is linear and its hybridization (sp) does not involve d-orbitals. (This ion is counted).

5. O${}_3$ (Ozone)

   • Central atom: Oxygen (O).
   • The central O atom is bonded to two other O atoms and has one lone pair.
   • Sigma bonds = 2, Lone pairs = 1.
   • SN = 2 + 1 = 3.
   • Hybridization: $sp^2$.
   • Shape: Bent (V-shape).
   • Conclusion: It is not linear.

6. SCl${}_2$ (Sulfur dichloride)

   • Central atom: Sulfur (S).
   • Sulfur has 6 valence electrons. It forms two single bonds with two Cl atoms and has two lone pairs.
   • Sigma bonds = 2, Lone pairs = 2.
   • SN = 2 + 2 = 4.
   • Hybridization: $sp^3$.
   • Shape: Bent (V-shape).
   • Conclusion: It is not linear.

7. ICl${}_2^-$

   • Central atom: Iodine (I).
   • Iodine has 7 valence electrons. With a -1 charge, the central atom effectively has 8 electrons in its valence shell for VSEPR purposes. It forms two bonds with Cl atoms.
   • Bond pairs = 2. Lone pairs = (8 - 2*1)/2 = 3.
   • Sigma bonds = 2, Lone pairs = 3.
   • SN = 2 + 3 = 5.
   • Hybridization: $sp^3d$.
   • Shape: With 2 bond pairs and 3 lone pairs, the electron geometry is trigonal bipyramidal. The lone pairs occupy the equatorial positions to minimize repulsion, resulting in a Linear shape.
   • Conclusion: It is linear, but its hybridization ($sp^3d$) involves d-orbitals. So, it is not counted.

8. I${}_3^-$ (Triiodide ion)

   • Central atom: Iodine (I).
   • Similar to ICl${}_2^-$. The central I is bonded to two other I atoms.
   • Sigma bonds = 2, Lone pairs = 3.
   • SN = 2 + 3 = 5.
   • Hybridization: $sp^3d$.
   • Shape: Linear.
   • Conclusion: It is linear, but its hybridization ($sp^3d$) involves d-orbitals. So, it is not counted.

9. XeF${}_2$ (Xenon difluoride)

   • Central atom: Xenon (Xe).
   • Xenon has 8 valence electrons. It forms two single bonds with F atoms.
   • Sigma bonds = 2. Lone pairs = (8 - 2*1)/2 = 3.
   • SN = 2 + 3 = 5.
   • Hybridization: $sp^3d$.
   • Shape: Linear.
   • Conclusion: It is linear, but its hybridization ($sp^3d$) involves d-orbitals. So, it is not counted.

Final Count: The molecules/ions that are linear and have a central atom hybridization without d-orbital contribution are:

• BeCl${}_2$
• N${}_3^-$
• N${}_2$O
• NO${}_2^+$

The total number is 4.