Question

(a)Account for the following:
(i)Acidic character increase from $HF$ to $HI$.
(ii)There is a large difference between the melting and boiling points of oxygen and sulphur.
(iii) Nitrogen does not form pentahalide.
(b) Draw the structures of the following:
(i) $Cl{F}_3$
(ii) $Xe{F}_4$

Answer 1

(a)

(i) Size of halide ions order $F^{-}<C{l}^{-}<B{r}^{-}<I^{-}$ with increase size negative charge is dispered throughout ions and ions gets stabilized. Thus acidity order $HF<HCl<HBr<HI$

(ii) Due to present of vacant d-orbital and combining forms of sulphur ($S_8$) which are not occurs in oxygen, the cohesive energy of sulphur more than oxygen leading to large melting point and boiling point

(iii) Due to absent of d-orbital, nitrogen can not extend its octate to form pentahalide.