# NEET Chemistry The p-Block Elements (XII) Questions Solved

BOARD

(a) Draw the molecular structures of the following compounds :

(i) ${N}_{2}{O}_{5}$                             (ii) $XeO{F}_{4}$

(b) Explain the following observations :

(i) Sulphur has a greater tendency for catenation than oxygen.

(ii) ICI is more reactive than ${I}_{2}.$

(iii) Despite lower value of its electron gain enthalpy with negative sign, fluorine $\left({F}_{2}\right)$ is a stronger oxidising agent than $C{l}_{2}.$

(a) (i) ${N}_{2}{O}_{5}$

Planar

(ii) $XeO{F}_{4}$

Shape : Square pyramidal

(b) (i) It is because S-S bond is stronger than O-O bonds as there is more interelectronic repulsion in O-O due to small size than in S-S.

(ii) ICI is more reactive than ${I}_{2}$ because I-Cl bond is polar and weaker while I-I bond is stronger and non-polar.

(iii) It is due to (a) low enthalpy of dissociation of F-F bond, (b) high hydration enthalpy of F.

Difficulty Level:

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