Question

(a) (i) Write the disproportionation reaction of $H_{3}P{O}_3$ .
(ii) Draw the structure of $Xe{F}_4$ .
(b) Account for the following :
(i) Although Fluorine has less negative electron gain enthalpy yet $F_2$ is strong oxidizing agent.
(ii) Acidic character decreases from $N_2{O}_3$ to $B{i}_2{O}_3$ in group 15.

Answer 1

(a) (i) $4H_{3}P{O}_3⟶3H_{3}P{O}_4+P{H}_3$

(ii) The structure of $Xe{F}_4$ is square planar with two lone pairs. The fluorine atoms occupy the four equatorial positions with the lone pairs occupying the axial positions.

(b)(i)Oxidizing power depends on the electron gain enthalpy and the hydration energy. Fluorine has less negative electron gain enthalpy and small size. Due to its smaller size, it has very high hydration energy and thus acts as a strong oxidising agent.

(ii) The acidic strength of an oxide depends on the electronegativity of the central atom. Thus from $N$ to $Bi$ electronegativity decreases and thus acidic strngth also decreases.