Question

Give reasons:
(i) $S{O}_2$ is reducing while $Te{O}_2$ is an oxidising agent.
(ii) Nitrogen does not form pentahalide.
(iii) $ICl$ is more reactive than $I_2$.

Answer 1

(i) $S+O_2\to S{O}_2$

Oxidation number of sulphur changes from 0 to +4. Due to presence of d-orbitals (vacant) sulphur can extend its covalency and oxidation states till +6, which is stable in sulphur (eg: $S{F}_6$).

$Te+O_2\to Te{O}_2$

Oxidation number of the changes from 0 to +4. Unlike sulphur, Te cannot show +6 oxidation state as it is highly unstable due to inert pair effect. Therefore, Te can only show a decrease in its oxidation state (it can decrease to -2, +2). Hence, it acts as an oxidising agent.

(ii) Formation of pentahalides requires $s{p}^{3}d$ hybridization of the central atom. This is obtained by the excitement of electron's of the central atom to empty d orbital. Science, nitrogen has no d orbitals, it cannot undergo $s{p}^{3}d$ hybridization and thus cannot form pentahalides.

(iii) Inter halogen bonds are weaker (it is between two different halogens like ICl) because of its partly ionic character. Due to difference in electronegativities while when same halogens form $X_2$( like $I_2$ ) molecules and they form covalent bonds which are stronger than interhalogen compound and weak bond obviously is more reactive than the stronger bond and that's why $ICl$ is more reactive than $I_2$.