
Which of the following is an energy consuming process?
A. ${\text{O(g) + e}} \to {{\text{O}}^{\text{ - }}}{\text{(g)}}$
B. ${\text{N}}{{\text{a}}^{\text{ + }}}{\text{(g) + e}} \to {\text{Na(g)}}$
C. ${{\text{O}}^{\text{ - }}}{\text{ + e}} \to {{\text{O}}^{{\text{2 - }}}}{\text{(g)}}$
D. ${{\text{O}}^{{\text{2 - }}}}{\text{(g)}} \to {{\text{O}}^{\text{ - }}}{\text{(g) + }}{{\text{e}}^{\text{ - }}}$
Answer
464.4k+ views
Hint: When an electron is added to the valence shell of a neutral atom, a certain amount of energy is required or released. This energy is termed as Electron Gain Enthalpy or Electron Enthalpy Value. Electron Gain Enthalpy is opposite to the Ionization Enthalpy (IE). When addition of electrons requires energy, then Electron Gain Enthalpy gets a positive sign and when it releases energy, Electron Gain Enthalpy gets a negative sign.
Complete step by step answer:
In the option A, ${\text{O(g) + e}} \to {{\text{O}}^{\text{ - }}}{\text{(g)}}$, Oxygen is in gaseous state and it is accepting an electron, thus releasing energy. So, its Electron Enthalpy Value is negative.
In the second option B,${\text{N}}{{\text{a}}^{\text{ + }}}{\text{(g) + }}{{\text{e}}^{\text{ - }}} \to {\text{Na(g)}}$, ${\text{N}}{{\text{a}}^{\text{ + }}}$ is in gaseous state. It is accepting an electron and releasing energy. So, again the Electron Enthalpy Value is negative.
In the third option C, ${{\text{O}}^{\text{ - }}}{\text{ + }}{{\text{e}}^{\text{ - }}} \to {{\text{O}}^{{\text{2 - }}}}{\text{(g)}}$, ${{\text{O}}^{\text{ - }}}$ has excess electrons already and an electron is again added. So, the incoming electron will face repulsion because of which energy needs to be given to add this incoming electron. Hence, the energy absorbed in this process. And the Electron Enthalpy Value is positive here.
If we take the option D (${{\text{O}}^{{\text{2 - }}}}{\text{(g)}} \to {{\text{O}}^{\text{ - }}}{\text{(g) + }}{{\text{e}}^{\text{ - }}}$), it’s a reverse process of option C. In this reaction, energy is released.
Thus, the correct option is C (${{\text{O}}^{\text{ - }}}{\text{ + }}{{\text{e}}^{\text{ - }}} \to {{\text{O}}^{{\text{2 - }}}}{\text{(g)}}$).
Note: When an electron is added to an atom. It gets support from the nucleus in the form of nuclear attraction, but it also faces repulsion from other electrons which is called a screening effect. Hence, Electron Gain Enthalpy depends on the element’s nature as well as its requirements.
Complete step by step answer:
In the option A, ${\text{O(g) + e}} \to {{\text{O}}^{\text{ - }}}{\text{(g)}}$, Oxygen is in gaseous state and it is accepting an electron, thus releasing energy. So, its Electron Enthalpy Value is negative.
In the second option B,${\text{N}}{{\text{a}}^{\text{ + }}}{\text{(g) + }}{{\text{e}}^{\text{ - }}} \to {\text{Na(g)}}$, ${\text{N}}{{\text{a}}^{\text{ + }}}$ is in gaseous state. It is accepting an electron and releasing energy. So, again the Electron Enthalpy Value is negative.
In the third option C, ${{\text{O}}^{\text{ - }}}{\text{ + }}{{\text{e}}^{\text{ - }}} \to {{\text{O}}^{{\text{2 - }}}}{\text{(g)}}$, ${{\text{O}}^{\text{ - }}}$ has excess electrons already and an electron is again added. So, the incoming electron will face repulsion because of which energy needs to be given to add this incoming electron. Hence, the energy absorbed in this process. And the Electron Enthalpy Value is positive here.
If we take the option D (${{\text{O}}^{{\text{2 - }}}}{\text{(g)}} \to {{\text{O}}^{\text{ - }}}{\text{(g) + }}{{\text{e}}^{\text{ - }}}$), it’s a reverse process of option C. In this reaction, energy is released.
Thus, the correct option is C (${{\text{O}}^{\text{ - }}}{\text{ + }}{{\text{e}}^{\text{ - }}} \to {{\text{O}}^{{\text{2 - }}}}{\text{(g)}}$).
Note: When an electron is added to an atom. It gets support from the nucleus in the form of nuclear attraction, but it also faces repulsion from other electrons which is called a screening effect. Hence, Electron Gain Enthalpy depends on the element’s nature as well as its requirements.
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