Among [I₃]⁺, [SiO₄]⁴⁻, SO₂Cl₂, XeF₂, SF₄, ClF₃, Ni(CO)₄, XeO₂F₂, [PtCl₄]²⁻, XeF₄, and SOCl₂, the total number of species having sp³-hybridised central atom is:

To determine the total number of species having an sp³-hybridised central atom, we can calculate the steric number (SN) for the central atom of each given chemical species.
The formula for steric number is given by:

$\text{Steric Number (SN)}=\frac{1}{2}\times [V+M-C+A]$

where:
• V = number of valence electrons of the central atom
• M = number of monovalent atoms attached to the central atom
• C = charge on the cation
• A = charge on the anion

The relationship between Steric Number and Hybridisation is:
• SN = 4 ⇒ sp³ hybridisation
• SN = 5 ⇒ sp³d hybridisation
• SN = 6 ⇒ sp³d² hybridisation
• SN = 4 (for transition metal complexes with strong field ligands/dsp²) ⇒ dsp² hybridisation

Let us determine the hybridisation for each species step-by-step:

1. [I₃]⁺:
Central atom: Iodine (V = 7)
Monovalent atoms attached (M) = 2
Cationic charge (C) = 1

$\text{SN}=\frac{1}{2}\times [7+2-1]=4$

Since SN = 4, the hybridisation of the central iodine atom is sp³.

2. [SiO₄]⁴⁻:
Central atom: Silicon (V = 4)
Monovalent atoms attached (M) = 0 (Oxygen is divalent)
Anionic charge (A) = 4

$\text{SN}=\frac{1}{2}\times [4+0+4]=4$

Since SN = 4, the hybridisation of Silicon is sp³.

3. SO₂Cl₂:
Central atom: Sulfur (V = 6)
Monovalent atoms attached (Cl, M) = 2
Oxygen is divalent (doesn't count in M)

$\text{SN}=\frac{1}{2}\times [6+2]=4$

Since SN = 4, the hybridisation of Sulfur is sp³.

4. XeF₂:
Central atom: Xenon (V = 8)
Monovalent atoms attached (F, M) = 2

$\text{SN}=\frac{1}{2}\times [8+2]=5$

Since SN = 5, the hybridisation of Xenon is sp³d.

5. SF₄:
Central atom: Sulfur (V = 6)
Monovalent atoms attached (F, M) = 4

$\text{SN}=\frac{1}{2}\times [6+4]=5$

Since SN = 5, the hybridisation of Sulfur is sp³d.

6. ClF₃:
Central atom: Chlorine (V = 7)
Monovalent atoms attached (F, M) = 3

$\text{SN}=\frac{1}{2}\times [7+3]=5$

Since SN = 5, the hybridisation of Chlorine is sp³d.

7. Ni(CO)₄:
Central atom: Nickel (d⁸s² system, oxidation state = 0)
CO is a strong field ligand, so it causes pairing of electrons resulting in d¹⁰ configuration.
The 4 pairs of electrons from 4 CO ligands enter one 4s and three 4p orbitals.
Hence, hybridisation of Nickel is sp³.

8. XeO₂F₂:
Central atom: Xenon (V = 8)
Monovalent atoms attached (F, M) = 2

$\text{SN}=\frac{1}{2}\times [8+2]=5$

Since SN = 5, the hybridisation of Xenon is sp³d.

9. [PtCl₄]²⁻:
Central atom: Platinum (5d⁸ system, Pt²⁺ ion).
Since 5d transition series complexes always form low-spin square planar complexes regardless of ligand field strength, it forms a square planar complex with dsp² hybridisation.

10. XeF₄:
Central atom: Xenon (V = 8)
Monovalent atoms attached (F, M) = 4

$\text{SN}=\frac{1}{2}\times [8+4]=6$

Since SN = 6, the hybridisation of Xenon is sp³d².

11. SOCl₂:
Central atom: Sulfur (V = 6)
Monovalent atoms attached (Cl, M) = 2

$\text{SN}=\frac{1}{2}\times [6+2]=4$

Since SN = 4, the hybridisation of Sulfur is sp³.

Summary:
The species having an sp³-hybridised central atom are:
1. [I₃]⁺
2. [SiO₄]⁴⁻
3. SO₂Cl₂
4. Ni(CO)₄
5. SOCl₂

Therefore, the total number of species with sp³-hybridised central atom is 5.

The correct answer is 5.