p-Block Elements

Sulphur does not exist as $\mathrm{S}_2$ molecule primarily because of its inability to constitute $p \pi-p \pi$ bonds.

In the case of oxygen, the small size of $O$ atoms and the high electronegativity allow the $2p$ orbitals to overlap effectively, forming a strong $p \pi-p \pi$ double bond in $\mathrm{O}_2$. As a result, molecular oxygen exists as a diatomic molecule. However, sulphur atoms are larger and less electronegative compared to oxygen. This significant size difference makes the overlap between $3p$ orbitals (the corresponding orbitals in sulphur) less effective for forming strong $p \pi-p \pi$ double bonds.

Instead, sulphur stabilizes itself more effectively by catenating to form larger molecular structures like $\mathrm{S}_8$, where each sulphur atom forms single bonds with adjacent sulphur atoms, creating a puckered ring structure. This results in the elemental sulphur commonly found in the form of octa-atomic rings.

So, the correct answer is:

Option A: it is not able to constitute $p \pi-p \pi$ bonds

The term "amphoteric" refers to a substance that can react both as an acid and as a base. In the context of oxides, an amphoteric oxide can react with both acids and bases to form salts and water. Let's analyze the options presented:

Option A: $\mathrm{SnO}_2$ (Tin(IV) oxide)
This oxide is known to be amphoteric. It can react with acids to form stannic salts and with strong bases to form stannates.

Option B: $\mathrm{SiO}_2$ (Silicon dioxide)
This oxide is acidic in nature. It primarily reacts with bases (such as sodium hydroxide) but not with acids.

Option C: $\mathrm{CO}_2$ (Carbon dioxide)

Option D: $\mathrm{CaO}$ (Calcium oxide)
This oxide is basic. It reacts with acids to form calcium salts and water but not with bases.

Among the given options, the correct answer is Option A: $\mathrm{SnO}_2$, which is amphoteric in character.

Inert pair effect is more pronounced in heavier members like Pb.

Hence, Pb (IV) compounds act as strong oxidising agents and are reduced to more stable Pb (II) compounds.

The structure of diborane is

In this structure, there are two 3C$-$2e^$-$ bonds. 3 atoms, B $-$ H $-$ B share two electrons and form an angular geometry, leading bent bond, also known as banana bond.

The structure of pyrophosphoric acid is

There are 4 P$-$OH bonds, 1 P$-$O$-$P bond and two P==O bonds.

Helium is used as a cryogenic agent due to its very low melting point.

BaF₂ is not an anti-fluorite structure. It is of AB₂ type i.e. fluorite (CaF₂) structure. While Rb₂S, K₂O and Li₂O are anti-fluorite structures.

The reaction can be completed as follows

Na₂B₄O₇ + conc. H₂SO₄ + 5H₂O $\to$ Na₂SO₄ + $\mathop {4{H_3}B{O_3}}\limits_{(A)} $

H₃BO₃ + 3C₂H₅OH $\to$ $\mathop {B(O{C_2}{H_5}) + 3{H_2}O \to 3}\limits_{Volatile\,Burn\,with\,green\,edged\,flame} $

The inorganic halide (A) is SO₂Cl₂.

$\mathop {S{O_2}C{l_2}}\limits_{(A)} + 2{H_2}O\buildrel {} \over \longrightarrow \mathop {{H_2}S{O_4}}\limits_{(B)} + \mathop {2HCl}\limits_{(C)} $

$\mathop {S{O_2}C{l_2}}\limits_{(A)} + 4NaOH\buildrel {} \over \longrightarrow \mathop {N{a_2}S{O_4}}\limits_{(D)} + \mathop {2NaCl}\limits_{(E)} + 2{H_2}O$

$\mathop {N{a_2}S{O_4}}\limits_{(D)} + BaC{l_2}\buildrel {} \over \longrightarrow \mathop {BaS{O_4}}\limits_{White\,ppt.} + 2NaCl$

$\mathop {NaCl}\limits_{(E)} + AgN{O_3}\buildrel {} \over \longrightarrow \mathop {AgCl}\limits_{White\,ppt.} + NaN{O_3}$