Q. 275.0( 1 Vote )

(a) Account for t

Answer :

(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+


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