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.
Two nitrogen atoms are joined by triple bonds. The nitrogen atom is very small, therefore the bond length is also quite small (109.8pm) and as the result, the bond dissociation energy is quite high (946Kj/mol) Therefore, N2 is less reactive at room temperature.
H2Te is the strongest reducing agent among the hydrides of group 16. The size of Te is very large due to which the bonding between hydrogen and Te is not strong. On the other hand, the electronegativity of Te is very less. So it will easily loose hydrogen. As the size of the elements increases in the order O < S < Se < Te, thus bond strength decreases from H2O to H2Te and therefore, the bond dissociation enthalpy decreases. Hence, due to the increase in the tendency to release proton, the element's reducing tendency also increases.
(iii) Helium is used in diving apparatus as a diluent for oxygen because it is chemically inert and does not participate in the chemical reaction. Helium has low solubility in water than many other gases, such as nitrogen. Due to low solubility means it does not enter the bloodstream, even under pressure commonly experienced by deep sea divers.
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)
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