The p-Block Elements

The p-Block Elements


(a) Account for the following:
(i)Ozone is thermodynamically unstable.
(ii)Solid PCl5 is ionic in nature.
(iii)Fluorine forms only one oxoacid HOF.

(b) Draw the structure of
(i) BrF5
(ii) XeF4

(i)Compare the oxidizing action of F2 and Cl2 by considering parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
(ii)Write the conditions to maximize the yield of H2SO4 by contact process.
(iii)Arrange the following in the increasing order of property mentioned:

(a)H3PO3, H3PO4, H3PO2 (Reducing character)
(b)NH3, PH3, AsH3, SbH3, BiH3 (Base strength)


(i) Ozone  decomposes  into  O2 with  the  evolution  of  heat,  i.e.  ΔH is  negative (exothermic).

O3  -->  O2 + O

ΔH = negative

Since the decomposition of O3 increases the number and freedom of particles, entropy also increases.

Therefore, DS = positive Now, ΔG = ΔH – TΔS

Both −∆H and

-T∆S(since   ∆S is   positive)   result   into   large   negative ∆G.   Hence, Obecomes thermodynamically unstable and decomposes into oxygen easily.

(ii) PCl5 is ionic in solid state because it exists as [PCl4]+ [PCl6]− in which the cation has tetrahedral geometry and the anion has octahedral geometry.

(b) (i) BrF5

(ii) XeF4

(i)   Although electron gain enthalpy of fluorine is less than that of chlorine because of the small size of fluorine,  but  the  oxidising  power  depends  on  other  factors like bond dissociation energy and hydration energy. The smaller the size of  the atom, the greater the hydration enthalpy. Fluorine being small in size has higher hydration enthalpy as compared to chlorine.

Also, fluorine faces greater inter-electronic repulsion among its lone pairs of electrons because of its small size, while there is very less repulsion in chlorine. Hence, the bond dissociation enthalpy of fluorine is lower than that of chlorine.

Thus, the high hydration enthalpy and low bond dissociation enthalpy of fluorine result in its higher oxidising power as compared to that of chlorine.

(ii)   Manufacturing of sulphuric acid via the contact process involves three steps:

(1)  Burning of ores to form SO2

(2)  Conversion of SO2 to SO3 using V2O5 as a catalyst

(3)  Absorption of SO3 in H2SO4 to give oleum

The second step, i.e. conversion of SO2 to SO3  is the key step. Since this reaction is exothermic in nature and two moles of gaseous reactant give one mole of gaseous

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