JEE Chemistry 2024 Question with Solution

Given: The species are $NH_3$, $SO_2$, $SiO_2$, $BeCl_2$, $CO_2$, $H_2O$, $CH_4$, and $BF_3$.

Find: The number of species in which the central atom uses $sp^3$ hybrid orbitals in bonding.

Check the hybridization of the central atom in each species:

• $NH_3$: Nitrogen forms three sigma bonds and has one lone pair, so it is $sp^3$ hybridized.
• $SO_2$: Sulfur forms two sigma bonds and has one lone pair, so it is $sp^2$ hybridized.
• $SiO_2$: Silicon forms four bonds with oxygen atoms in tetrahedral local bonding, so it is $sp^3$ hybridized.
• $BeCl_2$: Beryllium forms two sigma bonds in a linear arrangement, so it is $sp$ hybridized.
• $CO_2$: Carbon forms two double bonds in a linear structure, so it is $sp$ hybridized.
• $H_2O$: Oxygen forms two sigma bonds and has two lone pairs, so it is $sp^3$ hybridized.
• $CH_4$: Carbon forms four sigma bonds, so it is $sp^3$ hybridized.
• $BF_3$: Boron forms three sigma bonds in a trigonal planar geometry, so it is $sp^2$ hybridized.

Thus, the species with central atom using $sp^3$ hybrid orbitals are:

Hence, the total number is $4$.

Therefore, the correct option is A.

Given: The same set of species is to be checked for central atom hybridization.

Find: How many central atoms are $sp^3$ hybridized.

Use the idea that:

• steric number $4 \Rightarrow sp^3$
• steric number $3 \Rightarrow sp^2$
• steric number $2 \Rightarrow sp$

Now evaluate each central atom:

• In $NH_3$, nitrogen has steric number $4$ because of three bond pairs and one lone pair. So it is $sp^3$.
• In $SO_2$, sulfur has steric number $3$ because of two sigma bonds and one lone pair. So it is $sp^2$.
• In $SiO_2$, silicon has four bonded directions around it locally, so it is $sp^3$.
• In $BeCl_2$, beryllium has steric number $2$, so it is $sp$.
• In $CO_2$, carbon has steric number $2$, so it is $sp$.
• In $H_2O$, oxygen has steric number $4$ because of two bond pairs and two lone pairs. So it is $sp^3$.
• In $CH_4$, carbon has steric number $4$, so it is $sp^3$.
• In $BF_3$, boron has steric number $3$, so it is $sp^2$.

So exactly four species have central atom hybridization $sp^3$.

Therefore, the correct option is A.