Among [I_3]^{+},[SiO_4]^{4-},SO_2Cl_2,XeF_2,S | vedclass

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(C) To determine the number of species with $sp^3$ hybridised central atom,we calculate the steric number $(SN)$ for each: $SN = \frac{1}{2} (V + M -

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$5$

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(C) To determine the number of species with $sp^3$ hybridised central atom,we calculate the steric number $(SN)$ for each: $SN = \frac{1}{2} (V + M - C + A)$,where $V$ is valence electrons,$M$ is monovalent atoms,$C$ is cationic charge,and $A$ is anionic charge. $1$. $[I_3]^{+}: SN = \frac{1}{2}(7 + 2 - 1) = 4 \rightarrow sp^3$ $2$. $[SiO_4]^{4-}: SN = \frac{1}{2}(4 + 4 + 4) = 4 \rightarrow sp^3$ $3$. $SO_2Cl_2: SN = \frac{1}{2}(6 + 2) = 4 \rightarrow sp^3$ $4$. $XeF_2: SN = \frac{1}{2}(8 + 2) = 5 \rightarrow sp^3d$ $5$. $SF_4: SN = \frac{1}{2}(6 + 4) = 5 \rightarrow sp^3d$ $6$. $ClF_3: SN = \frac{1}{2}(7 + 3) = 5 \rightarrow sp^3d$ $7$. $Ni(CO)_4: Ni$ is $d^{10}s^2$,$CO$ is neutral ligand,$SN = 4 \rightarrow sp^3$ $8$. $XeO_2F_2: SN = \frac{1}{2}(8 + 2) = 5 \rightarrow sp^3d$ $9$. $[PtCl_4]^{2-}: Pt^{2+}$ is $d^8$ system,$SN = 4 \rightarrow dsp^2$ $10$. $XeF_4: SN = \frac{1}{2}(8 + 4) = 6 \rightarrow sp^3d^2$ $11$. $SOCl_2: SN = \frac{1}{2}(6 + 2) = 4 \rightarrow sp^3$ Species with $sp^3$ hybridisation are $[I_3]^{+}$,$[SiO_4]^{4-}$,$SO_2Cl_2$,$Ni(CO)_4$,and $SOCl_2$. Total count = $5$.

Among $[I_3]^{+}$,$[SiO_4]^{4-}$,$SO_2Cl_2$,$XeF_2$,$SF_4$,$ClF_3$,$Ni(CO)_4$,$XeO_2F_2$,$[PtCl_4]^{2-}$,$XeF_4$,and $SOCl_2$,the total number of species having $sp^3$ hybridised central atom is. . . . .

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