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Compare the stability of Ni^(4+)and Pt^(4+)from their ionisation enthalpy values. {:(IE,Ni,Pt),(I,737,864),(II,1753,1791),(III,3395,2800),(IV,5297,4150):} |
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Answer» Solution :(i) `Ni^(4+)I.E.=737+1753+3395+5297=11182"kJ mol"^(-1)` (ii) `PT^(4+)I.E. =864+1791+2800+4150=9605 "kJ mol"^(-1)` `Pt^(4+)`compounds are stable than `Ni^(4+)` compounds because the ENERGY needed to remove 4 electrons in Pt is LESS than that of Ni. |
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