Which of the following has the highest electron affinity?
(A) K
(B) ${{O}^{-}}$
(C) ${{F}^{-}}$
(D) O

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Hint: Electron affinity and electronegativity follow the same trend. And ionization energy is opposite of electron affinity but they also follow the same trend.

Complete answer:
Electron affinity is defined as the change in energy when an electron is added to an atom or molecule in its isolated gaseous state to form a negative ion. Since this is an exothermic process, the change in energy is found to be negative. Also the electron affinity is numerically equal to the negative change in energy, hence obtained value of the electron affinity is a positive value. Electron affinity increases left to right across periods except the Noble gases, and decreases when moving down groups in the periodic table. Therefore, elements with the high electron affinity would be present in the top right corner (p-block) of the periodic table.
Now from our given options, we have an element of s-block, that is Potassium(K). As we go from left to right the electron affinity decreases, also, it belongs to the second period, and down the group electron affinity decreases, therefore, Potassium will have the lowest electron affinity.
Fluorine and Oxygen have high electronegativity and hence they have high electron affinity. In between, Oxygen and fluorine , fluorine have the high electron affinity but, here ${{F}^{-}}$, since fluorine is smaller in size so ${{F}^{-}}$ will experience greater repulsion and therefore , it will require more energy to give electron to ${{F}^{-}}$. Also ${{F}^{-}}$ will acquire noble gas configuration after addition of one more electron, so it will release more energy, therefore it will have the highest electron affinity.

So, the correct answer is “Option C”.

Note: Electron affinity is the negative of electron gain enthalpy. They have a slight difference. Electron gain enthalpy can be defined as the amount of energy absorbed or released whenever a free electron is added to an isolated gaseous atom.