Which group elements easily accept $1$ and $2$ electrons to achieve noble gas electronic configurations? Why?
(N/A) Elements of group $17$ accept $1$ electron and elements of group $16$ accept $2$ electrons to form anions and achieve the electronic configuration of the next noble gas. This is because they possess high negative electron gain enthalpy.
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In which of the following processes is energy released?
Difficult
View SolutionThe 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:
Which of the following atoms should have the highest $1^{st}$ electron affinity?
The set containing the elements with positive electron gain enthalpies is
Evaluate the following statements regarding electron affinity $(EA)$:
$A$. Carbon is greater than fluorine
$B$. Sulphur is lesser than fluorine
$C$. Iodine is higher than bromine
$D$. Chlorine is greater than sulphur
Select the correct sequence of truth values ($T$ for true,$F$ for false):
$A$. Carbon is greater than fluorine
$B$. Sulphur is lesser than fluorine
$C$. Iodine is higher than bromine
$D$. Chlorine is greater than sulphur
Select the correct sequence of truth values ($T$ for true,$F$ for false):
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