p-Block Elements — JEE Main practice

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.

1. Finding A

   The first reaction is a double displacement (precipitation) reaction:

   $$\mathrm{PbCl_2 + K_2CrO_4 \rightarrow PbCrO_4 + 2KCl}$$

   Here, $\mathrm{Pb^{2+}}$ combines with $\mathrm{CrO_4^{2-}}$ to form lead chromate, a yellow precipitate.

   Therefore,

   $$\boxed{\mathrm{A = PbCrO_4}}$$

2. Finding B

   Substitute $\mathrm{A}$ into the second reaction:

   $$\mathrm{PbCrO_4 + NaOH \rightleftharpoons B + Na_2CrO_4}$$

   Lead compounds are amphoteric, so in excess $\mathrm{NaOH}$ they dissolve to form the tetrahydroxoplumbate complex.

   The balanced reaction is:

   $$\mathrm{PbCrO_4 + 4NaOH \rightarrow Na_2[Pb(OH)_4] + Na_2CrO_4}$$

   Hence,

   $$\boxed{\mathrm{B = Na_2[Pb(OH)_4]}}$$

3. Finding X

   Ammonium acetate acts as a complexing agent. Four acetate ions coordinate with $\mathrm{Pb^{2+}}$ to form the tetraacetatoplumbate(II) complex.

   The reaction is:

Therefore,

4. Combining all three results:

   • $\mathrm{A = PbCrO_4}$
   • $\mathrm{B = Na_2[Pb(OH)_4]}$
   • $\mathrm{X = (NH_4)_2[Pb(CH_3COO)_4]}$

   Hence, Option C is correct.

1. The binary cation $(EH_x)^+$ must correspond to a stable hydrogen-containing cation.

   Nitrogen forms the well-known ammonium ion:

   $$NH_4^+$$

   Hence, $E$ is nitrogen and $x=4$.

2. The compound $EH_3$ is therefore:

   $$NH_3$$

   Ammonia reacts with alkaline potassium tetraiodomercurate(II), $K_2HgI_4$, to give a brown precipitate of basic mercury(II) amido-iodide (Nessler's test).

   This confirms that $E$ is nitrogen.

3. The first elements of groups 13, 14, 15, and 16 are:

   • Group 13: $\mathrm{B}$
   • Group 14: $\mathrm{C}$
   • Group 15: $\mathrm{N}$
   • Group 16: $\mathrm{O}$

4. Nitrogen has the highest first ionisation enthalpy among these because of its stable half-filled electronic configuration:

   $$1s^2,2s^2,2p^3$$

5. The first ionisation enthalpy of nitrogen is approximately:

   $$1402\ \mathrm{kJ,mol^{-1}}$$

6. Therefore, the correct answer is Option A (1402).