A) \[P{{b}^{4+}}+S{{n}^{4+}}\]
B) \[P{{b}^{4+}}+S{{n}^{2+}}\]
C) \[P{{b}^{2+}},S{{n}^{2+}}\]
D) \[P{{b}^{2+}},S{{n}^{4+}}\]
Correct Answer: D
Solution :
\[\overset{+4}{\mathop{Pb}}\,{{O}_{2}}\xrightarrow{{}}\overset{+2}{\mathop{Pb}}\,O\,\Delta {{G}_{298}}<0\,\Delta {{G}_{298}}<0\] For this reaction\[\Delta G\]is negative, hence\[P{{b}^{2+}}\]is more stable than \[P{{b}^{4+}}.\] \[\overset{+4}{\mathop{Sn}}\,{{O}_{2}}\xrightarrow{{}}\overset{+2}{\mathop{Sn}}\,{{O}_{2}}\,\Delta {{G}_{298}}>0\] For this reaction\[\Delta G\]is positive, hence\[S{{n}^{4+}}\]is more stable than\[S{{n}^{2+}}\]since for spontaneous change \[\Delta G\] must be negative.You need to login to perform this action.
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