Among [I$$_3$$]$$^+$$, [SiO$$_4$$]$$^{4-}$$, SO$$_2$$Cl$$_2$$, XeF$$_2$$, SF$$_4$$, ClF$$_3$$, Ni(CO)$$_4$$, XeO$$_2$$F$$_2$$, [PtCl$$_4$$]$$^{2-}$$, XeF$$_4$$, and SOCl$$_2$$, the total number of species having $$sp^3$$ hybridised central atom is ____.

First write the rule for deciding hybridisation.
For a given central atom, the hybridisation is decided by its steric number (SN).
$$\text{SN}= \text{(number of } \sigma \text{-bonds)} + \text{(number of lone pairs)}$$

• SN = 4  →  $$sp^3$$ hybridisation (tetrahedral electron geometry).
• SN = 5  →  $$sp^3d$$ hybridisation.
• SN = 6  →  $$sp^3d^2$$ hybridisation, etc.

Analyse each species one by one.

Case 1: $$[I_3]^+$$
Central iodine forms two $$I-I$$ $$\sigma$$-bonds and possesses two lone pairs (confirmed by a total valence-electron count of 20).
SN = $$2 + 2 = 4$$ ⇒ $$sp^3$$.

Case 2: $$[SiO_4]^{4-}$$
Silicon is bonded to four $$O^{2-}$$ ions through four $$\sigma$$-bonds and has no lone pair.
SN = $$4 + 0 = 4$$ ⇒ $$sp^3$$.

Case 3: $$SO_2Cl_2$$ (thionyl chloride)
Central sulphur has two $$S-Cl$$ single bonds and two $$S=O$$ double bonds. Each double bond contributes one $$\sigma$$-bond.
Number of $$\sigma$$-bonds = 4, lone pairs = 0.
SN = $$4 + 0 = 4$$ ⇒ $$sp^3$$.

Case 4: $$XeF_2$$
Xe has two $$\sigma$$-bonds and three lone pairs.
SN = $$2 + 3 = 5$$ ⇒ $$sp^3d$$ (not $$sp^3$$).

Case 5: $$SF_4$$
Four $$\sigma$$-bonds + one lone pair → SN = 5 ⇒ $$sp^3d$$ (not $$sp^3$$).

Case 6: $$ClF_3$$
Three $$\sigma$$-bonds + two lone pairs → SN = 5 ⇒ $$sp^3d$$ (not $$sp^3$$).

Case 7: $$Ni(CO)_4$$
Nickel forms four metal-carbon $$\sigma$$-bonds with no lone pair involved in the hybrid orbitals.
SN = $$4 + 0 = 4$$ ⇒ $$sp^3$$ (tetrahedral complex).

Case 8: $$XeO_2F_2$$
Xe has four $$\sigma$$-bonds (two Xe-O, two Xe-F) and one lone pair.
SN = $$4 + 1 = 5$$ ⇒ $$sp^3d$$ (not $$sp^3$$).

Case 9: $$[PtCl_4]^{2-}$$
Square-planar $$[PtCl_4]^{2-}$$ uses $$dsp^2$$ hybridisation (not $$sp^3$$).

Case 10: $$XeF_4$$
Four $$\sigma$$-bonds + two lone pairs → SN = 6 ⇒ $$sp^3d^2$$ (not $$sp^3$$).

Case 11: $$SOCl_2$$ (thionyl chloride)
Sulphur has two $$S-Cl$$ $$\sigma$$-bonds, one $$S=O$$ double bond (one $$\sigma$$), and one lone pair.
SN = $$3 + 1 = 4$$ ⇒ $$sp^3$$.

Species with $$sp^3$$ hybridised central atom:
$$[I_3]^+,\; [SiO_4]^{4-},\; SO_2Cl_2,\; Ni(CO)_4,\; SOCl_2$$

Total number = 5.

Hence the required answer is 5.