Fluorine is more reactive than chlorine. So, bond energy of chlorine is greater than fluorine. Sulphur exhibits 6 coordination number. $$PC{l_5}$$  exists but $$NC{l_5}$$  does not exist as $$N$$  cannot expand its octet due to non-availability of $$d$$ - orbitals. The common oxidation states of group 15 elements are -3, +3 and 5.

$$\eqalign{ & {H_2}{O_2} > {O_3} > {O_2} \cr & B.O.\,{\text{is}}\,{\text{1,1}}{\text{.5}}\,{\text{and}}\,{\text{2}}\,{\text{respectively}}. \cr} $$

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Solid $$C{O_2}$$  is used as refrigerant.

Acidic strength of hydrides increases as the size of central atom increases which weakens the $$M—H$$  bond. Since, the size increases from $$S$$  to $$Te$$  thus acidic strength follows the order.
$${H_2}S < {H_2}Se < {H_2}Te$$
$${\text{Acidic}}\,{\text{nature}} \propto \frac{1}{{{\text{Bond}}\,{\text{dissociation}}\,{\text{enthalpy}}}}$$
$$S$$  to $$Te$$  size increases, bond dissociation enthalpy decreases and acidic nature increases.

Hypophosphorous acid $$\left( {{H_3}P{O_2}} \right)$$   is monobasic acid, so its structure is

$$M{e_3}SiCl$$   is not a monomer for a high molecular mass silicon polymer because it generates $$M{e_3}SiOH$$   when subjected to hydrolysis which contains only one reacting site. Hence, the polymerisation reaction stops just after first step.

Peroxomonosulphuric acid, $${H_2}S{O_5}$$  is also known as Caro’s acid. It has peroxy linkage $$—O—O—$$   which is confirmed by $$X $$ - ray studies.

Generally as the size of the atom increases, bond dissociation energy decreases, so in halogens $${I_2}$$  have lowest bond dissociation energy, but the bond dissociation energy of chlorine is higher than that of fluorine because in fluorine there is a greater repulsion between non-bonding electrons $$(2p).$$
Hence, the order of bond dissociation energy is
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{l_2} > {F_2}\, > B{r_2}\,\, > \,{I_2} \cr & {\text{Bond}}\,\,{\text{dissociation}}\,\,\,\,\,243\,\,\,\,\,\,\,159\,\,\,\,\,\,\,\,193\,\,\,\,\,\,\,\,\,151 \cr & {\text{energy (kJ/mol)}} \cr} $$

$$C{O_2}$$  being more dense covers the igniting material more effectively than $${N_2}.$$