Preparation of sulphuric acid:
By Contact Process: Burning of sulphur or sulphide ores in presence of oxygen to produce SO₂. Catalytic oxidation of SO₂ with O₂to give SO₃ in the presence of V₂O₅.
\(2 S O_{2}(g)+O_{2}(g) \stackrel{v, 0,}{\longrightarrow} 2 S O_{3}(g)\) 
Then SO₃ made to react with sulphuric acid of suitable normality to obtain a thick oily liquid called oleum.
\(\mathrm{SO}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{SO}_{4}(l) \rightarrow \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}(l)\) 
Then oleum is diluted to obtain sulphuric acid of desired concentration
\(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{4}(l)\) 
The sulphuric acid obtained by contact process is 96-98% pure.

(i) High pressure (2 bar),
(ii) 720 K temperature,
(iii) V₂O₅ as catalyst.

 (i) \(C+2H_{ 2 }SO_{ 4 }(conc).CO_{ 2 }+2SO_{ 2 }+2H_{ 2 }O\)
(ii) \(2XeF_{ 2 }+2H_{ 2 }O\rightarrow 2Xe+4HF+O_{ 2 }\)

(i) BrF₃

(ii) XeF₄
​​​​​

Dilute and conc.  Nitric acid gives different products of oxidation when reacted with copper metal.
3Cu(s) + 8HNO₃(dil.) \(\to\) 3Cu(No₃)₂(aq) + 2NO(g) + 4H₂O(l)
Cu(s) + 4HNO₃(conc.) \(\to\) Cu(NO₃)₂(aq) + 2NO₂ + 2H₂O(l)

\({ P }_{ 4 }{ O }_{ 10 }\) is a strong dehydrating agent. It reacts with \({ HNO }_{ 3 }\) to form metaphosphoric acid  \(\left( { HPO }_{ 3 } \right) \) and dinitrogen pentoxide, \({ N }_{ 2 }{ O }_{ 5 }\) 
\(4HN{ O }_{ 3 }+{ P }_{ 4 }{ O }_{ 10 }\longrightarrow 4HP{ O }_{ 3 }+2{ N }_{ 2 }{ O }_{ 5 }\)     \(\) 

(i) HNO₂ (Nitrous acid)
(ii) HNO₃ (Nitric acid)
(iii) H₂N₂O₂ (Hydronitrous acid)
\(\underset{+3}{3HNO_2 } \xrightarrow{Disproportionation} \underset {+5}{HNO_3} + H_2O + \underset {+2}{2NO}\) 

\(\) 

Nitric oxide [NO] is monomeric and paramagnetic in nature in gaseous state due to presence of unpaired electron but it exists as dimer in solid state which is diamagnetic due to absence of unpaired electron.

(i) \(Xe +\underset {excess} {2F_{2}} \xrightarrow [7 bar] {873 k} XeF_{4}\) 
(ii) \(6Li + N_{2} \rightarrow 2Li_{3}N\)

\(\)

(i) Ca(OCI)₂ + 4HCI \(\to\) CaCI₂ + 2H₂O + 2CI₂.
(ii) 2Ca₃(PO₄)₂ + 6Si O₂ + 10C \(\to\) P₄ + 6CaSi O₃ + 10CO.

(i) It is due to increases in size of group 15 elements, the bond angle decreases as bond length increases and bond pair-bond pair repulsion decreases.
(ii) It is because it is inert.

(i) It is because HF molecules are associated with intermolecular H-bonding whereas HCI is not.
(ii) It is due to low bond dissociation energy due to less effective overlapping.

(i) I₂ + 10HNO₃ (conc.) \(\to \) 2HIO₃ + 10NO₂ +H₂O 
(ii) 3HgCl₂ 2PH₃ \(\to \) Hg₃P₂ + 6 HCl