Given below are two statements:

**Statement I:** The number of pairs among $[\mathrm{SiO_2, CO_2}]$, $[\mathrm{SnO, SnO_2}]$, $[\mathrm{PbO, PbO_2}]$ and $[\mathrm{GeO, GeO_2}]$, which contain oxides that are both amphoteric is $2$.

**Statement II:** $\mathrm{BF_3}$ is an electron deficient molecule, can act as a Lewis acid, forms adduct with $\mathrm{NH_3}$ and has a trigonal planar geometry.

In the light of the above statements, choose the correct answer from the options given below:

Question

Given below are two statements:

**Statement I:** The number of pairs among $[\mathrm{SiO_2, CO_2}]$, $[\mathrm{SnO, SnO_2}]$, $[\mathrm{PbO, PbO_2}]$ and $[\mathrm{GeO, GeO_2}]$, which contain oxides that are both amphoteric is $2$.

**Statement II:** $\mathrm{BF_3}$ is an electron deficient molecule, can act as a Lewis acid, forms adduct with $\mathrm{NH_3}$ and has a trigonal planar geometry.

In the light of the above statements, choose the correct answer from the options given below:

Accepted Answer

C. 1. **Check Statement I**

Classify each pair of oxides:

* $\mathrm{[SiO_2,\ CO_2]}$ → both are **acidic** oxides.
   * $\mathrm{[SnO,\ SnO_2]}$ → both are **amphoteric** oxides.
   * $\mathrm{[PbO,\ PbO_2]}$ → both are **amphoteric** oxides.
   * $\mathrm{[GeO,\ GeO_2]}$ → both are **acidic** oxides.

Therefore, only the second and third pairs contain oxides that are both amphoteric.

Hence, the number of such pairs is **2**, so **Statement I is true**.

2. **Check Statement II**

* In $\mathrm{BF_3}$, boron has only six electrons around it, making the molecule **electron deficient**.
   * Because of this incomplete octet, $\mathrm{BF_3}$ readily accepts a lone pair and behaves as a **Lewis acid**.
   * It reacts with $\mathrm{NH_3}$ by accepting the lone pair on nitrogen to form a Lewis acid–base adduct.
   * Boron is $\mathrm{sp^2}$ hybridized in $\mathrm{BF_3}$, giving the molecule a **trigonal planar** geometry.

Therefore, **Statement II is also true**.

3. Since **both Statement I and Statement II are true**, the correct answer is **Option C**.