Q.ii. Give two reactions showing oxidising property of concentrated H$_2$SO$_4$.

SO3 is trigonal planar with three sigma bonds and no lone pair on S -> sp2 hybridisation.

In sulphur trioxide, the central sulphur atom forms three sigma bonds to three oxygen atoms and has no lone pair. The number of electron domains around S is 3, which corresponds to sp2 hybridisation and a trigonal planar shape (O-S-O bond angle 120 degrees). The additional pi bonding is delocalised over the three S-O bonds.

Oleum, or fuming sulphuric acid, has the formula $H_2S_2O_7$.

When sulphur trioxide ($SO_3$) is absorbed in concentrated sulphuric acid ($H_2SO_4$) during the Contact Process, it forms pyrosulphuric acid, commonly called oleum:

$H_2SO_4 + SO_3 \longrightarrow H_2S_2O_7 \ (\text{oleum})$

Oleum is a fuming liquid because on contact with atmospheric moisture it regenerates $H_2SO_4$, releasing visible white fumes:

$H_2S_2O_7 + H_2O \longrightarrow 2H_2SO_4$

Its structure has two sulphur atoms joined through a bridging oxygen ($HO_3S{-}O{-}SO_3H$), which is why it is also called disulphuric or pyrosulphuric acid.

H2SO4 gives two H+ ions, so it is dibasic.

Basicity of an acid = number of replaceable hydrogen atoms (ionisable protons) it can furnish. H2SO4 ionises in two steps:

H2SO4 → H+ + HSO4^-

HSO4^- → H+ + SO4^2-

Two protons are released per molecule, so its basicity is 2 → dibasic acid.

Oleum = H2SO4 + SO3 (fuming sulphuric acid, H2S2O7).

In the Contact process, SO3 is absorbed in concentrated H2SO4 to give oleum (pyrosulphuric acid, H2S2O7), rather than being absorbed directly in water:

H2SO4 + SO3 → H2S2O7 (oleum)

Oleum is later diluted with water to obtain sulphuric acid of the desired concentration. So the constituent dissolved in H2SO4 is SO3.