QJEEM24Ch3b9e: JEE Main 2024 Chemistry - Class 12 JEEM24Ch3b9e Solution: Given below are two statements : Statement (I) : and are acidic …

Both statements need to be evaluated to determine which (if any) are true:

Statement (I) : $\mathrm{SiO}_2$ and $\mathrm{GeO}_2$ are acidic while $\mathrm{SnO}$ and $\mathrm{PbO}$ are amphoteric in nature.

This statement is generally true. The oxides of silicon ( $\mathrm{SiO}_2$ ) and germanium ( $\mathrm{GeO}_2$ ) are indeed acidic. These elements belong to group 14 of the periodic table and as you move down the group, the acidic character of the oxides decreases while the basic character increases. Tin oxide ( $\mathrm{SnO}$ ) and lead oxide ( $\mathrm{PbO}$ ) are lower in the group and they are amphoteric, which means they show both acidic and basic properties depending on the reacting substance.

Statement (II) : Allotropic forms of carbon are due to the property of catenation and $\mathrm{p} \pi-\mathrm{d} \pi$ bond formation.

This statement is false. Catenation is indeed the property that carbon has to form long chains and rings with other carbon atoms, which is responsible for the vast number of organic compounds. However, in the context of pure carbon allotropes—such as diamond, graphite, and fullerene—the different physical structures and properties are mainly due to the $\mathrm{sp}^3$ , $\mathrm{sp}^2$ , and $\mathrm{sp}$ hybridizations of carbon, not to $\mathrm{p} \pi-\mathrm{d} \pi$ bonding. Carbon does not use d-orbitals in forming the allotropes. The $\mathrm{p} \pi-\mathrm{d} \pi$ bonding is typically discussed in the context of transition metal complexes, not allotropic forms of carbon.

Thus, the correct answer is Option D : Statement I is true but Statement II is false.