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# In which process energy is released?(A) $Be \to B{e^ - }$ (B) $F \to {F^ + }$ (C) $Ne \to N{e^ - }$ (D) $B \to {B^ - }$

Last updated date: 29th Feb 2024
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Hint: Ionization energy is the energy required to remove a valence electron from an atom or ion in its gaseous state causing to form a cation whereas electron affinity is the amount of energy released when an electron is added to a neutral atom or molecule in its gaseous state causing to form a anion.

In option a. $Be \to B{e^ - }$ , the atomic number of beryllium is 4 and it is a group 2 element which generally has much lower electron affinities. The electronic configuration of beryllium is $1{s^2}2{s^2}$ ,it has a filled outer s orbital hence adding an electron to it would make it unstable hence it requires energy to add an electron which makes the reaction endothermic.
In option b. $F \to {F^ + }$ , here ionization energy comes into action since an electron is removed from the halogen atom fluorine. It requires more energy to remove an electron and it has high ionization energy. Ionization energy is always an endothermic process.
In option c. $Ne \to N{e^ - }$ , the atomic number of neon is 10 and it is a noble gas group that is group 18. The electronic configuration of neon is $1{s^2}2{s^2}2{p^6}$ . Since it is a noble gas, it satisfies the octet rule, so adding an electron would disturb its octet rule and make it unstable. Hence it requires a lot of energy to add an electron to the noble gas elements which makes the reaction endothermic.
In option d. $B \to {B^ - }$ , the atomic number of boron is 5 and it belongs to group 13 from where the non-metals group starts. Electron affinity for non-metals is greater compared to metals. The electronic configuration of boron is $1{s^2}2{s^2}2{p^1}$ . It is electron deficient, possessing a vacant p-orbital. Hence electrons added will settle in the empty p-orbital releasing the energy. This is the exothermic process (negative electron affinity).