Would you expect the second electron gain enthalpy of $O$ to be positive,more negative,or less negative than the first? Justify your answer.
(A) The first electron gain enthalpy of oxygen is negative because energy is released when an electron is added to a neutral oxygen atom to form an $O^-$ ion.
The second electron gain enthalpy of oxygen is positive. This is because after the addition of the first electron,the oxygen atom becomes a negatively charged $O^-$ ion.
The second electron is added to this negatively charged ion,and the incoming electron experiences strong electrostatic repulsion from the existing negative charge.
Therefore,energy must be supplied to overcome these coulombic repulsions to force the second electron into the anion,making the process endothermic (positive electron gain enthalpy).
The second electron gain enthalpy of oxygen is positive. This is because after the addition of the first electron,the oxygen atom becomes a negatively charged $O^-$ ion.
The second electron is added to this negatively charged ion,and the incoming electron experiences strong electrostatic repulsion from the existing negative charge.
Therefore,energy must be supplied to overcome these coulombic repulsions to force the second electron into the anion,making the process endothermic (positive electron gain enthalpy).
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Similar Questions
Explain the following :
$(1)$ $Pb^{4+}$ acts as an oxidizing agent but $Sn^{2+}$ acts as a reducing agent.
$(2)$ Electron gain enthalpy of chlorine is more negative as compared to fluorine.
$(1)$ $Pb^{4+}$ acts as an oxidizing agent but $Sn^{2+}$ acts as a reducing agent.
$(2)$ Electron gain enthalpy of chlorine is more negative as compared to fluorine.
Difficult
View SolutionWhich one of the following arrangements represents the correct order of electron gain enthalpy (with negative sign) of the given atomic species?
MediumAIPMT 2005
View SolutionWhich of the following represents the correct order of increasing electron gain enthalpy (with negative sign) for the elements?
$a.$ Nitrogen
$b.$ Phosphorus
$c.$ Chlorine
$d.$ Fluorine
$a.$ Nitrogen
$b.$ Phosphorus
$c.$ Chlorine
$d.$ Fluorine
The formation of the oxide ion $O^{2-}_{(g)}$ requires first an exothermic and then an endothermic step as shown below:
$O_{(g)} + e^{-} \to O^{-}_{(g)}; \Delta H = -142 \, kJ \, mol^{-1}$
$O^{-}_{(g)} + e^{-} \to O^{2-}_{(g)}; \Delta H = 844 \, kJ \, mol^{-1}$
This is because:
$O_{(g)} + e^{-} \to O^{-}_{(g)}; \Delta H = -142 \, kJ \, mol^{-1}$
$O^{-}_{(g)} + e^{-} \to O^{2-}_{(g)}; \Delta H = 844 \, kJ \, mol^{-1}$
This is because:
The value of $2^{nd}$ electron gain enthalpy for $O$ and $S$ will be (in $kJ/mol$)
Medium
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