Total number of species from the following wi | vedclass

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(C) To determine the number of species with $sp^2$ hybridization, we analyze the steric number $(SN)$ of the central atom, where $SN = 	ext{number of

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) To determine the number of species with $sp^2$ hybridization, we analyze the steric number $(SN)$ of the central atom, where $SN = 	ext{number of sigma bonds} + 	ext{number of lone pairs}$. For $sp^2$ hybridization, $SN = 3$.$1. NH_3$: $SN = 3 + 1 = 4$ $(sp^3)$$2. SO_2$: $SN = 2 + 1 = 3$ $(sp^2)$$3. SiO_2$: $SN = 2$ $(sp)$$4. BeCl_2$: $SN = 2$ $(sp)$$5. C_2H_2$: $SN = 2$ $(sp)$$6. C_2H_4$: $SN = 3$ $(sp^2)$$7. BCl_3$: $SN = 3$ $(sp^2)$$8. HCHO$: $SN = 3$ $(sp^2)$$9. C_6H_6$: $SN = 3$ $(sp^2)$$10. BF_3$: $SN = 3$ $(sp^2)$$11. C_2H_4Cl_2$: $SN = 4$ $(sp^3)$The species with $sp^2$ hybridization are $SO_2, C_2H_4, BCl_3, HCHO, C_6H_6, BF_3$. The total count is $6$.

Total number of species from the following with central atom utilising $sp^2$ hybrid orbitals for bonding is.................... $NH_3, SO_2, SiO_2, BeCl_2, C_2H_2, C_2H_4, BCl_3, HCHO, C_6H_6, BF_3, C_2H_4Cl_2$

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