Q27 of 27 Page 1

(a) Account for the following observations:

(i) SF4 is easily hydrolysed whereas SF6 is not easily hydrolysed


(ii) Chlorine water is a powerful bleaching agent.


(iii) is a stronger oxidizing agent than Sb(V)


(b) What happens when


(i) White phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO2.


(ii) XeF6 undergoes partial hydrolysis.


(Give the chemical equations involved).


OR


(a) What inspired N. Bartlett for carrying out a reaction between Xe and PtF6?


(b) Arrange the following in the order of property indicated against each set:


(i) F2, I2, Br2, Cl2(increasing bond dissociation enthalpy)


(ii) NH3, AsH3, SbH3, BiH3, PH3(decreasing base strength)


(c) Complete the following equations:


(i)


(ii)

(a) (i) SF4 is easily hydrolysed whereas SF6 is not easily hydrolysed because the S atom in SF4 is not sterically protected as it is surrounded by only four F atoms, so attack of water molecules can take place easily. In SF6 the S atom is protected by 6 F atoms. Thus atoms cannot take place easily.


(ii) Chlorine water is a powerful bleaching agent because it produces nascent oxygen(causes oxidation) which is responsible for bleaching action.


Cl2 + H2O → 2HCl + [O]


(iii) Bi(V) is a stronger oxidising agent than Sb(V)due to the inert pair effect in Bi(V) can accept a pair of electrons to form more stable Bi(III) . The + 3 oxidation state is more stable than its + 5 oxidation state.


(b) (i) Phosphorous undergoes disproportionation reaction to form phosphine gas.



(ii) On partial hydrolysis,


XeF6 gives oxyfluride and XeOF4 and HF


XeF6 + H2O XeOF4 + HF


OR


(a) N. Bartlett first prepared a red compound O2+PtF6-. He then realised that first ionisation enthalpy of molecular oxygen was almost identical with Xenon. So he thought that this reaction would also be possible with Xenon and therefore he carried out this reaction.


(b) (i) I2 <F2 <Br2 < Cl2


Reason:


Since the atomic size increases from chlorine to iodine(As we go down a group), bond length also increases from chlorine to iodine. As bond length increases it requires less energy to break the bond, and therefore their bond dissociation energy is less.


Fluorine here is an exception to the rule. Due to its small and compact size,it experiences inter electronic repulsions. These repulsions therefore decrease its bond dissociation energy.


(ii) NH3>PH3>AsH3>SbH3>BiH3


Reason:


Bases are the substances that donate electrons. All the hydrides of group 15 have a lone pair of electrons in their molecule. As we go down a group,the size of the atom increases and the electron density decreases. This in turn obviously says that the substance does not lose electrons easily and hence the basic strength also decreases.


(c) (i) With cold and dilute NaOH chlorine produces a mixture of chloride and hypochlorite.


NOTE: This applies to all alkalies.


2NaOH + Cl2 ⇒ NaCl + NaOCl + H2O


(ii) Fe3 + gets oxidized to Fe2 + in the presence of sulphur dioxide. Moist Sulphur dioxide here is the reducing agent.
2Fe3+ + SO2 + 2H2O ⇒ 2Fe2+ + SO2-4 + 4H+


More from this chapter

All 27 →
23

Explain the following:

(a) Out of Sc3 + , Co2 + and Cr3 + ions, Sc3 + only is colorless in aqueous solutions. (Atomic no. : Co = 27; Sc = 21 and C = 24)


(b) The E°Cu2 + /Cu for copper metal is positive ( + 0. 34), unlike the remaining members of the first transition series


(c) La(OH)3 is more basic than Lu(OH)3.

 24

A metal complex having composition Cr(NH3)4Cl2Br has been isolated in two forms A and B. form A reacts with AgNO3 to give a white precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia.

(i) Write the formulae of isomers A and B.


(ii) State the hybridization of chromium in each of them.


(iii) Calculate the magnetic moment (spin only value) of the isomer.

 25

(a) Identify A – D


(b) Distinguish between the following pair of compounds:


(i) Aniline and Benzylamine.


(ii) Methylamine and Dimethylamine


(c) Complete the following:



OR


(a) Account for the following:


(i) Direct nitration of aniline yields a significant amount of meta derivative.


(ii) Primary aromatic amines cannot be prepared by Gabriel phthalimide synthesis.


(b) Carry out the following conversions:


(i) Ethanoic acid into methenamine.


(ii) Aniline to p - Bromoaniline.


(c) Arrange the following in increasing order of basic strength: Aniline, p - nitroaniline, and p - toluene.

 26

(a) A cell is prepared by dipping a zinc rod in 1M zinc sulfate solution and a silver electrode in 1M silver nitrate solution. The standard electrode potential is given:
E0Zn2+/Zn = -0.76 V,
 E0Ag2+/Ag = +0.80 V,

What is the effect of an increase in the concentration of Zn2 + on the Ecell?


(b) Write the products of electrolysis of aqueous solution of NaCl with platinum electrodes.


(c) Calculate e. m. f. of the following cell at 298 K:


Ni(s) / Ni2 + (0. 01 M) // Cu2 + (0. 1M) / Cu (s)



Write the overall cell reaction.


OR


(a) Apply Kohlrausch law of independent migration of ions, write the expression to determine the limiting molar conductivity of calcium chloride.


(b) Given are the conductivity and molar conductivity of NaCl solutions at 298K at different concentrations:



Compare the variation of conductivity and molar conductivity of NaCl solutions on dilution. Give reason.


(c) 0. 1 M KCl solution offered a resistance of 100 ohms in a conductivity cell at 298 K. If the cell constant of the cell is 1. 29 cm - 1, calculate the molar conductivity of KCl solution.