Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Among ${\text{[I}}_3{\text{]}}^{+}$, ${\text{[SiO}}_4{\text{]}}^{4-}$, ${\text{SO}}_2{\text{Cl}}_2$, ${\text{XeF}}_2$, ${\text{SF}}_4$, ${\text{ClF}}_3$, ${\text{Ni(CO)}}_4$, ${\text{XeO}}_2{\text{F}}_2$, ${\text{[PtCl}}_4{\text{]}}^{2-}$, ${\text{XeF}}_4$, and ${\text{SOCl}}_2$, the total number of species having $s{p}^3$ hybridised central atom is ______.

When faced with a massive list of molecules and ions, finding the hybridization of the central atom can feel like searching for a needle in a haystack. But fear not! We have a master tool at our disposal: the Steric Number (SN).

The Steric Number is the sum of the number of $\sigma$ (sigma) bonds and the number of lone pairs on the central atom. If the Steric Number equals 4, the geometry is based on a tetrahedron, and the hybridization is strictly $s{p}^3$. Let's break down the given species one by one and separate the $s{p}^3$ champions from the rest.

Let's start with the curious case of the $[I_3{]}^{+}$ ion. The central iodine atom normally has 7 valence electrons. The positive charge indicates the loss of one electron, leaving it with 6. It forms two single bonds with the other two iodine atoms, utilizing 2 electrons. The remaining 4 electrons pair up to form 2 lone pairs. With 2 $\sigma$ bonds and 2 lone pairs, the Steric Number is $2+2=4$. Thus, $[I_3{]}^{+}$ is $s{p}^3$ hybridized and has a bent shape.

Next, we look at the silicate ion, $[Si{O}_4{]}^{4-}$. Silicon, a group 14 element, has 4 valence electrons. It forms four single bonds with four oxygen atoms. There are no lone pairs left. The Steric Number is $4+0=4$, making it a perfect $s{p}^3$ tetrahedron.

Sulfur is a versatile element, and we see it twice here. In $S{O}_{2}C{l}_2$, sulfur (6 valence electrons) forms two double bonds with oxygen and two single bonds with chlorine. Remember, a double bond only contributes one $\sigma$ bond! So, we have 4 $\sigma$ bonds and 0 lone pairs, giving a Steric Number of 4 ($s{p}^3$).

Similarly, in $SOC{l}_2$, sulfur forms one double bond with oxygen and two single bonds with chlorine. This uses 4 of its 6 valence electrons, leaving 1 lone pair. With 3 $\sigma$ bonds and 1 lone pair, the Steric Number is again 4, confirming it as $s{p}^3$ hybridized with a trigonal pyramidal shape.

Not every molecule fits the $s{p}^3$ mold. Molecules like $Xe{F}_2$, $S{F}_4$, and $Cl{F}_3$ all feature expanded octets. For instance, $Xe{F}_2$ has 2 bonds and 3 lone pairs ($SN=5$), making it $s{p}^{3}d$. $S{F}_4$ has 4 bonds and 1 lone pair ($SN=5$), and $Cl{F}_3$ has 3 bonds and 2 lone pairs ($SN=5$). We can safely eliminate these from our count.

Now, we enter the realm of transition metals, where the rules of VSEPR take a backseat to Valence Bond Theory and Crystal Field Theory.

Consider $Ni(CO{)}_4$. Nickel is in the zero oxidation state with a $[Ar]3d^{8}4s^2$ configuration. Carbon monoxide ($CO$) is a notorious strong field ligand. It forces the two $4s$ electrons to pair up in the $3d$ subshell, resulting in a completely filled $3d^{10}$ configuration. This leaves the $4s$ and three $4p$ orbitals completely empty. These four empty orbitals hybridize to form an $s{p}^3$ tetrahedral complex. This is a classic JEE trap!

On the other hand, we have $[PtC{l}_4{]}^{2-}$. You might think that because $C{l}^{-}$ is a weak field ligand, it won't cause pairing. However, Platinum is a heavy 5d transition metal. For 4d and 5d metals, the crystal field splitting energy is so massive that all ligands act as strong field ligands. The $d^8$ configuration of $P{t}^{2+}$ will always result in a square planar $ds{p}^2$ geometry. So, we exclude it.

After carefully analyzing the entire list, we have identified exactly five species that possess an $s{p}^3$ hybridized central atom: $[I_3{]}^{+}$, $[Si{O}_4{]}^{4-}$, $S{O}_{2}C{l}_2$, $Ni(CO{)}_4$, and $SOC{l}_2$.

The final integer answer is 5.