
The successive ionization energies for element X is given below
$ I{E_1}:\;250kJmo{l^{ - 1}}$
$ I{E_2}\;:\;820kJmo{l^{ - 1}}$
$ I{E_3}:\;1100kJmo{l^{ - 1}}$
$ I{E_4}:\;1400kJmo{l^{ - 1}}$
Find out the number of valence electrons for the element X.
$(A)3$
$(B)4$
$(C)2$
$(D)1$
Answer
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Hint: We know that ionization energy of an element or ion is defined as the energy required to remove an electron from the outermost orbital (valence orbital) of an isolated gaseous atom. In order to determine the number of valence shell electrons in the above element, we will understand the change in electronic configuration of the atom after ionization.
Complete answer:
As we know, the ionization energy of an atom or ion is defined as the energy required to remove an electron from the outermost orbital (which is most loosely bound) of an isolated gaseous atom. After one ionization process, the atom gets converted into ion and if that ion again undergoes ionization to form a dictation, then it is known as the second ionization energy of the atom. If this process continues for a number of times, then it is known as successive ionization energies.
Now, if an atom is loosely bonded to the atom and after losing that electron the species attains noble gas configuration, then it will have a low ionization energy. So, it is easy to remove electrons from the valence shell of the atom.
Once the species attains noble gas configuration, then it will be very difficult to remove an electron from the outermost orbital of the atom and the ionization energy is very high.
Now, for element X the difference between the first and the second ionization enthalpies is large as compared to the difference between other ionization enthalpies. So, we can conclude that after losing one electron, the ion will attain the noble gas configuration.
Hence, there is only one valence electron in element X.
Therefore, the correct option is $(D)1.$
Note:
We should remember that there is one more type of energy which is also known as electron gain enthalpy. It is defined as the amount of energy which is released when an isolated gaseous atom accepts an electron to form an anion. The first electron gain enthalpy is always negative because the energy is released in this process of accepting electrons.
Complete answer:
As we know, the ionization energy of an atom or ion is defined as the energy required to remove an electron from the outermost orbital (which is most loosely bound) of an isolated gaseous atom. After one ionization process, the atom gets converted into ion and if that ion again undergoes ionization to form a dictation, then it is known as the second ionization energy of the atom. If this process continues for a number of times, then it is known as successive ionization energies.
Now, if an atom is loosely bonded to the atom and after losing that electron the species attains noble gas configuration, then it will have a low ionization energy. So, it is easy to remove electrons from the valence shell of the atom.
Once the species attains noble gas configuration, then it will be very difficult to remove an electron from the outermost orbital of the atom and the ionization energy is very high.
Now, for element X the difference between the first and the second ionization enthalpies is large as compared to the difference between other ionization enthalpies. So, we can conclude that after losing one electron, the ion will attain the noble gas configuration.
Hence, there is only one valence electron in element X.
Therefore, the correct option is $(D)1.$
Note:
We should remember that there is one more type of energy which is also known as electron gain enthalpy. It is defined as the amount of energy which is released when an isolated gaseous atom accepts an electron to form an anion. The first electron gain enthalpy is always negative because the energy is released in this process of accepting electrons.
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