Short Answer TypeArrange the following in increasing order of their basic strength:
(i) C6H5 – NH2, C6H5 – CH2 – NH2, C6H5 – NH – CH3
Give reasons for the following:
(i) N2 is less reactive at room temperature.
(ii) H2Te is the strongest reducing agent amongst all the hydrides of Group 16 elements.
(iii) Helium is used in diving apparatus as a diluent for oxygen.
Give reasons for the following:
(i) Oxygen is a gas but sulphur is solid.
(ii) O3 acts as a powerful oxidising agent.
(iii) BiH3 is the strongest reducing agent amongst all the hydrides of Group 15 elements.
Explain the following facts giving appropriate reason in each case:
(i) NF3 is an exothermic compound whereas NCl3 is not.
(ii) All the bonds in SF4 are not equivalent.
Long Answer Type(a) Draw the molecular structure of the following compounds.
(i) N2O5
(ii) XeOF4
(b) Explain the following observation:
(i) Sulphur has a greater tendency for catenation than oxygen.
(ii) ICI is more reactive than I2.
(iii) Despite the lower value of its electron gain enthalpy with a negative sign, fluorine (F2) is a stronger oxidizing agent than Cl2.
(a) Complete the following chemical equation
(i) Cu + HNO3 (dilute) --->
(ii) XeF4 + O2F2 -->
(b) Explain the following observation:
(i) Phosphorus has a greater tendency for catenation than nitrogen.
(ii) Oxygen is a gas but sulphur a solid.
(iii) The halogens are coloured. Why?
(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
OR
(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)
a)
(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, O3 becomes 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
Or
(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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