| \[Pb{{O}_{2}}\xrightarrow{{}}PbO\,\,\Delta {{G}_{298}}<0\] |
| \[Sn{{O}_{2}}\xrightarrow{{}}SnO\,\,\Delta {{G}_{298}}>0\] |
| Most probable oxidation state of Pb and Sn will be: [AIPMT 2001] |
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 :
| \[Pb{{O}_{2}}\xrightarrow[{}]{{}}PbO\] \[\Delta G_{298}^{{}}<0\] |
| For this reaction \[\Delta G\] is negative, hence \[P{{b}^{2+}}\] is more stable than \[P{{b}^{4+}}\]. |
| \[Sn{{O}_{2}}\xrightarrow[{}]{{}}SnO\] \[\Delta G_{298}^{{}}>0\] |
| For this reaction \[\Delta G\] is positive, hence \[S{{n}^{4+}}\] is more stable than \[S{{n}^{2+}}\]. because for spontaneous change \[\Delta G\] must be |
| negative. |
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