The formation of the oxide ion O^{2-}_{(g)} r | vedclass

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(C) The formation of the oxide ion $O^{2-}_{(g)}$ occurs in two steps.In the first step,an electron is added to a neutral oxygen atom,which is an exot

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$O^{-}$ ion will tend to resist the addition of another electron

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(C) The formation of the oxide ion $O^{2-}_{(g)}$ occurs in two steps.In the first step,an electron is added to a neutral oxygen atom,which is an exothermic process $(\Delta H = -142 \, kJ \, mol^{-1})$ due to the attraction between the nucleus and the incoming electron.In the second step,an electron is added to the negatively charged $O^{-}$ ion.Since both the $O^{-}$ ion and the incoming electron are negatively charged,there is a strong electrostatic repulsion between them.To overcome this repulsion and force the electron into the ion,energy must be supplied,making the process endothermic $(\Delta H = 844 \, kJ \, mol^{-1})$.Therefore,the $O^{-}$ ion resists the addition of another electron.

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:

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Which of the following has the maximum electron gain enthalpy?

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$(B)$ $1s^2 2s^2 2p^4$
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Which of the following is the correct order of increasing tendency to gain electron :

The $CORRECT$ order of $\Delta H_{eg}$ of the following elements is

Consider the following conversions:
$(i) \ O_{(g)} + e^- \to O^{-}_{(g)} ; \Delta H_1$
$(ii) \ F_{(g)} + e^- \to F^{-}_{(g)} ; \Delta H_2$
$(iii) \ Cl_{(g)} + e^- \to Cl^{-}_{(g)} ; \Delta H_3$
$(iv) \ Na_{(g)} \to Na^{+}_{(g)} + e^- ; \Delta H_4$
Which of the following statements is incorrect?

The second electron gain enthalpies (in $kJ \ mol^{-1}$) of oxygen and sulphur respectively are

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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: