Q.a) Nitrogen forms number of oxides in the different oxidation states. Write the names and structural formulae of any four oxides of nitrogen.

Nitrogen forms several oxides (N2O, NO, N2O3, NO2/N2O4, N2O5) of differing oxidation states; H2O boils far above H2S because of extensive hydrogen bonding; ozone is quantitatively estimated by liberating iodine from KI and titrating it against standard sodium thiosulphate.

a) Oxides of nitrogen (any four, with names and structures)

1. Nitrous oxide, N2O (N = +1): linear structure, N≡N→O
2. Nitric oxide, NO (N = +2): a paramagnetic molecule with structure :N=O (odd electron on N)
3. Nitrogen dioxide, NO2 (N = +4): angular (bent) molecule, O=N–O with an unpaired electron on N
4. Dinitrogen pentoxide, N2O5 (N = +5): structure O2N–O–NO2 (two planar NO3 units joined by a bridging oxygen)

(Other valid oxides: N2O3, N = +3, structure O=N–NO2; N2O4, N = +4, dimer of NO2, O2N–NO2.)

b) Why H2O boils far above H2S

Oxygen is much smaller and far more electronegative than sulphur. This lets water molecules form extensive, strong intermolecular hydrogen bonds (O–H⋯O), which link H2O molecules into a large associated network. Breaking this hydrogen-bonded network requires a lot of extra energy, raising the boiling point to 373 K. Sulphur's lower electronegativity and larger size mean H2S cannot form significant hydrogen bonds, so only weak van der Waals forces hold H2S molecules together, and it boils at a much lower temperature (213 K).

c) Quantitative estimation of ozone

Ozone is a strong oxidising agent. It is passed into (or reacted with) a neutral, potassium iodide (KI) solution, where it oxidises iodide ion to iodine and itself is reduced to oxygen:

O3 + 2KI + H2O → 2KOH + I2 + O2

…