$$HI$$  being a stronger reducing agent than $${H_2}S{O_4},$$  reduces $${H_2}S{O_4}$$  to $$S{O_2}$$  and is itself oxidised to $${I_2}$$  ( violet fumes ).
$$\eqalign{ & MI + {H_2}S{O_4} \to MHS{O_4} + HI \cr & {H_2}S{O_4} \to {H_2}O + S{O_2} + \left[ O \right] \cr & 2HI + \left[ O \right] \to {H_2}O + \mathop {{I_2}}\limits_{{\text{(violet fumes)}}} \cr} $$

In the laboratory dinitrogen is prepared by heating an aqueous solution containing an equivalent amount of ammonium chloride and sodium nitrite.
$$N{H_4}Cl\left( {aq} \right) + NaN{O_2}\left( {aq} \right)$$      \[\xrightarrow{\text{Heat}}{{N}_{2}}\left( g \right)+2{{H}_{2}}O\left( l \right)+NaCl\]

The products of the concerned reaction react each other forming back the reactants.
$$Xe{F_6} + 3{H_2}O \to Xe{O_3} + 6HF.$$

$$4HCl + \mathop {{O_2}}\limits_{air} \to \mathop {2C{l_2} + 2{H_2}O}\limits_{cloud\,of\,white\,fumes} $$

The cross linked polymers will be formed by $$RsiC{l_3}$$

Due to inert pair effect the lower $$O.S.$$  is more common at the bottom of group

Nitrogen does not form $$NC{l_5}$$  ( nitrogen pentachloride ) because nitrogen does not have vacant $$d$$-orbital, so it can form only $$NC{l_3}.$$

All boron trihalides except boron trifluoride are hydrolysed to boric acid.

$$N{O_2}$$  is paramagnetic and readily dimerises to $${N_2}{O_4}$$  which is diamagnetic.

sbond dissociation enthalpy of halogens follows the sequency as :
$$C{l_2} > Br > {F_2} > {I_2}$$
Enthalpy of dissociation decreass as the bond distance increases from $${F_2}$$  to $${I_2}$$  due to a corresponding increase in size of the atom as one move down the group from $$F$$  to $$I.$$  However, the $$F-F$$  bond dissociation enthalpy is smaller than that of $$Cl - Cl$$   ( even than that of $$Br-Br$$  ) because $$F$$ -atom is very small and hence electron-electron repulsion between the lone pairs of electrons are very large.