A) \[p{{H}_{1}}>p{{H}_{2}}\approx p{{H}_{3}}>p{{H}_{4}}\]
B) \[p{{H}_{1}}<p{{H}_{2}}<p{{H}_{3}}<p{{H}_{4}}\]
C) \[p{{H}_{1}}<p{{H}_{2}}<p{{H}_{3}}\approx p{{H}_{4}}\]
D) \[p{{H}_{1}}>p{{H}_{2}}>p{{H}_{3}}>p{{H}_{4}}\]
Correct Answer: D
Solution :
The correct order of pH of isomolar solution in sodium oxide \[(p{{H}_{1}})\] sodium sulphide \[(p{{H}_{2}})\], sodium selenide \[(p{{H}_{3}})\] and sodium telluride \[(p{{H}_{4}})\] is \[p{{H}_{1}}>p{{H}_{2}}>p{{H}_{3}}>p{{H}_{4}}\] because in aqueous solution, they are hydrolysed as follows. \[N{{a}_{2}}O+2{{H}_{2}}O\xrightarrow[{}]{{}}2\underset{Base}{\mathop{NaOH}}\,+{{H}_{2}}O\] \[N{{a}_{2}}S+2{{H}_{2}}O\xrightarrow[{}]{{}}\underset{Strong\,base}{\mathop{2NaOH}}\,+\underset{Weak\,acid}{\mathop{{{H}_{2}}S}}\,\] \[N{{a}_{2}}Se+2{{H}_{2}}O\xrightarrow[{}]{{}}\underset{Strong\,base}{\mathop{2NaOH}}\,+\underset{Weak\,acid}{\mathop{{{H}_{2}}Se}}\,\] \[N{{a}_{2}}Te+2{{H}_{2}}O\xrightarrow[{}]{{}}\underset{Strong\,base}{\mathop{2NaOH}}\,+\underset{Weak\,acid}{\mathop{{{H}_{2}}Te}}\,\] Order of acidic strength \[{{H}_{2}}Te>{{H}_{2}}Se>{{H}_{2}}S>{{H}_{2}}O\] Order of neutralisation of NaOH \[{{H}_{2}}Te>{{H}_{2}}Se>{{H}_{2}}S>{{H}_{2}}O\] Hence, their aqueous solutions have the following order of basic character due to neutralisation of NaOH with \[{{H}_{2}}O,\,{{H}_{2}}S,\,{{H}_{2}}Se\] and \[{{H}_{2}}Te\]. \[N{{a}_{2}}O>N{{a}_{2}}S>N{{a}_{2}}Se>N{{a}_{2}}Te\] (\[\because \] pH of basic solution is higher than acidic or least basic solution)
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