
Which one of the following has the highest ionization potential?
\[ A.\;L{i^ + } \\
B.\;M{g^ + } \\
C.\;He \\
D.\;Ne \\
\]
Answer
509.1k+ views
Hint: Ionization potential is defined as the minimum energy that is required to remove a loosely bound electron of an atom or molecule. It is also called ionization energy. When you move across a period, the ionization energy increases and down the group it decreases.
Complete answer:
In the given options, He and Ne are noble gases which have stable electronic configuration whereas Li and Mg belong to IA and IIA groups respectively. To find out the answer, let us see their electronic configurations. The atomic number of Li is 3 and hence the configuration is \[1{s^2}2{s^1}\] in the options \[L{i^ + }\] is given, which means already one electron has been removed to form a cation and the configuration of which is having a noble gas configuration similar to ‘He’ and now can be ruled out because its configuration is \[1{s^2}2{s^2}2{p^6}3{s^1}\] and it requires less energy to remove the valence electron when compared with noble gases having stable configuration. Among He and Ne, the ionization energy required for ‘He’ is more as we know that down the group ionization energy decreases with Helium requiring more energy compared to other noble gases. Hence ‘Ne’ can also be ruled out. Now, we have two options left which are He and \[L{i^ + }\]. Among these \[L{i^ + }\] requires more energy because it has three protons in its nucleus in contrast Helium contains only two protons and hence greater energy is required to remove further electrons from \[L{i^ + }\].
So, the correct option is A.
Note:
The energy required to remove the first outermost electron is first ionization energy and in the same way, we have second and third ionization energies for removing subsequent electrons. In all the cases, the first ionization energy is less than the second which is in turn less than the third because as you keep on removing the outer electrons, there is greater attraction between remaining electrons and the nucleus.
Complete answer:
In the given options, He and Ne are noble gases which have stable electronic configuration whereas Li and Mg belong to IA and IIA groups respectively. To find out the answer, let us see their electronic configurations. The atomic number of Li is 3 and hence the configuration is \[1{s^2}2{s^1}\] in the options \[L{i^ + }\] is given, which means already one electron has been removed to form a cation and the configuration of which is having a noble gas configuration similar to ‘He’ and now can be ruled out because its configuration is \[1{s^2}2{s^2}2{p^6}3{s^1}\] and it requires less energy to remove the valence electron when compared with noble gases having stable configuration. Among He and Ne, the ionization energy required for ‘He’ is more as we know that down the group ionization energy decreases with Helium requiring more energy compared to other noble gases. Hence ‘Ne’ can also be ruled out. Now, we have two options left which are He and \[L{i^ + }\]. Among these \[L{i^ + }\] requires more energy because it has three protons in its nucleus in contrast Helium contains only two protons and hence greater energy is required to remove further electrons from \[L{i^ + }\].
So, the correct option is A.
Note:
The energy required to remove the first outermost electron is first ionization energy and in the same way, we have second and third ionization energies for removing subsequent electrons. In all the cases, the first ionization energy is less than the second which is in turn less than the third because as you keep on removing the outer electrons, there is greater attraction between remaining electrons and the nucleus.
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