Question

Would you expect the second electron to gain enthalpy of oxygen as positive, more negative or less negative than the first? Justify your answer.

Answer 1

Hint – In this question use the concept that when an electron is added to oxygen it leads in first electron gain enthalpy but after successful addition of one electron when another electron is added then the electrostatic repulsion is very large as the already present valence electrons repel the new coming electron. This will help commenting upon the second electron gain enthalpy of oxygen with respect to first electron gain enthalpy.

Complete answer:

As we know oxygen belongs to group 16 of p block elements.
It is also the first member of group 16.
It is non-metal and most abundant in earth crust.
So when an electron is added to an Oxygen (O) atom to form ${O^ - }$ ion, energy is released.
$ \Rightarrow O + {e^ - } \to {O^ - } + {\text{energy}}$
Thus the first electron gain enthalpy is negative.
And the value of this enthalpy is $\Delta {H_1} = - 141{\text{ kJmo}}{{\text{l}}^{ - 1}}$
Now when another electron is added to ${O^ - }$ it will form an ${O^{2 - }}$ ions and energy is absorbed to overcome the strong electrostatic repulsion between the negatively charged ${O^ - }$ ion and the second being added.
Thus the second electron gain enthalpy of oxygen is positive (as the energy is absorbed).
 $ \Rightarrow {O^ - } + {e^ - } + {\text{energy}} \to {O^{2 - }}$
And the value of this enthalpy is $\Delta {H_1} = + 780{\text{ kJmo}}{{\text{l}}^{ - 1}}$
So this is the required answer.

Note – The basic definition of electron gain enthalpy helps understanding problems of this kind. Electron gain enthalpy of an element is the energy released when a neutral isolated gaseous atom accepts an extra electron to form the gaseous negative ion. Electron gain enthalpy is denoted as ${\vartriangle _{eg}}H$. Greater is the magnitude of energy released higher is the electron gain enthalpy of the element.