
What is the order of Ionization Energy of \[F,Cl,{F^ - },C{l^ - }\]?
Answer
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Hint:Ionization energy (I.E.) refers to the minimum amount of energy that is required to remove or detach the most loosely bound electron from an isolated neutral gaseous molecule or atom.
Complete answer:
In the periodic table, the ionization energy generally decreases from top to bottom in a group while it increases across a period. Actually, from top to bottom (i.e. down a group) in a periodic table, electronegativity decreases because atomic number usually increases down a group which leads to an increase in the distance between the outer shell electrons and the nucleus resulting in a greater atomic radius. And smaller the atomic radius, higher is the ionization energy.
Therefore, the size of anion is greater compared to the size of a neutral atom. Thus, the attraction of nucleus for valence electrons is larger in neutral atoms. Consequently, the ionization energy of a neutral atom is larger in comparison to anion. While in case of halogens, addition of an electron destroys the noble gas configuration.
Among the given elements or anions, ionization energy of fluoride ion (\[{F^ - }\]) is the lowest owing to the electron-electron repulsion.
Among the given elements ionization potential of \[F > Cl\;\] since size of \[F\] is less than \[Cl\].
Among the given anions, ionization potential of \[C{l^ - } > {F^ - }\] because there is a vacant d-orbital in case of chlorine, which makes it capable of holding the extra electrons. On the other hand, Fluorine does not possess an extra vacant orbital, so it has the capacity to lose the extra electron easily.
Hence, the correct order of ionization energies (I.E.) is \[{F^ - } < C{l^ - } < Cl < F\].
Note:
In the atomic structure, a positive nucleus is located in the centre while negatively charged electrons are around it, so electrostatic force by the nucleus acting on the electron will affect the ionization energy. Thus closer the electron to the nucleus, higher is the attraction force which means higher is the energy required to remove the electron.
Complete answer:
In the periodic table, the ionization energy generally decreases from top to bottom in a group while it increases across a period. Actually, from top to bottom (i.e. down a group) in a periodic table, electronegativity decreases because atomic number usually increases down a group which leads to an increase in the distance between the outer shell electrons and the nucleus resulting in a greater atomic radius. And smaller the atomic radius, higher is the ionization energy.
Therefore, the size of anion is greater compared to the size of a neutral atom. Thus, the attraction of nucleus for valence electrons is larger in neutral atoms. Consequently, the ionization energy of a neutral atom is larger in comparison to anion. While in case of halogens, addition of an electron destroys the noble gas configuration.
Among the given elements or anions, ionization energy of fluoride ion (\[{F^ - }\]) is the lowest owing to the electron-electron repulsion.
Among the given elements ionization potential of \[F > Cl\;\] since size of \[F\] is less than \[Cl\].
Among the given anions, ionization potential of \[C{l^ - } > {F^ - }\] because there is a vacant d-orbital in case of chlorine, which makes it capable of holding the extra electrons. On the other hand, Fluorine does not possess an extra vacant orbital, so it has the capacity to lose the extra electron easily.
Hence, the correct order of ionization energies (I.E.) is \[{F^ - } < C{l^ - } < Cl < F\].
Note:
In the atomic structure, a positive nucleus is located in the centre while negatively charged electrons are around it, so electrostatic force by the nucleus acting on the electron will affect the ionization energy. Thus closer the electron to the nucleus, higher is the attraction force which means higher is the energy required to remove the electron.
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