$O(g) + e^- \to O^-(g); \Delta H = - 142 \,kJ \,mol^{-1}$
$O^-(g) + e \to O^{2-} (g); \Delta H = 844\, kJ \,mol^{-1}$
આ કોના કારણે છે?
\(O _{( g )} e \rightarrow O _{( g )}^{-} ; \Delta H =-142 \,kJ / mol\)
\(O _{( g )}^{-} e \rightarrow O _{( g )}^{2-} ; \Delta H =844\, kJ / mol\)
This is because when an electron is added to negatively charged ion, it experiences more repulsion rather than attraction.
Hence the addition of the second electron usually requires energy. As a result, second electron affinity values are positive i.e. endothermic.
${{O}^{-}}_{\left( g \right)}+{{e}^{-}}\to O_{\left( g \right)}^{2-};\Delta {{H}^{o}}=844\,kJ\,mo{{l}^{-1}}$
આમ થવાનું કારણ ...
$M(s) \to M(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\, ........(1)$
$M(s) \to M^{2+} (g) + 2e^-\,\,\,\,\,\,\,\,.......(2)$
$M(g) \to M^+(g) + e^-\,\,\,\,\,\,\,\,\,\,\,.........(3)$
$M^+ (g) \to M^{2+} (g) + e^-\,\,\,\,\,\,\,\,\,.........(4)$
$M(g) \to M^{2+} (g) +2e^-\,\,\,\,\,\,\,\,\,\,\,..........(5)$
$M$ની બીજી આયનીકરણ ઊર્જાની ગણતરી ક્યા ઊર્જા મૂલ્યોથી કરી શકાય છે?