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Question

(a) (i) Write the disproportionation reaction of H3PO3 .
(ii) Draw the structure of XeF4 .
(b) Account for the following :
(i) Although Fluorine has less negative electron gain enthalpy yet F2 is strong oxidizing agent.
(ii) Acidic character decreases from N2O3 to Bi2O3 in group 15.

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Solution

(a) (i) 4H3PO33H3PO4+PH3

(ii) The structure of XeF4 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.

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