
The ionisation potential of ${X^ + }$ ion is equal to
(A) the electron affinity of X atom
(B) the electronegativity of X atom
(C) the ionisation energy of X atom
(D) the electron affinity of ${X^{2 + }}$ ion
Answer
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Hint:Ionisation potential is defined as the minimum amount of energy required to remove an electron from the outermost shell of an isolated gaseous atom. On the other hand electron affinity is the energy released when an electron is added to a valence system. Both the processes involve the release and gain of electrons which are equal in magnitude but opposite in signs.
Complete answer:
We know that ionisation potential is equal to the minimum energy required to remove the electron from an isolated gaseous atom. So at first we need to find the ionisation potential of ${X^ + }$ ion.
So the reaction will be as follows
${X^ + } \to {X^{2 + }} + {e^ - }$ here the ionisation potential is I.P.
Here the electron has been removed from ${X^ + }$ ion. So the enthalpy is equal to the ionisation potential of ${X^ + }$ ion. Similarly if we reverse the reaction then an electron is added to the system.
Then the reaction is as follows
${X^{2 + }}(g) + {e^ - } \to {X^ + }$
Here electron has been added to the system. So the electron gains enthalpy here is the electron affinity. Here the enthalpy is equal to the electron affinity of ${X^{2 + }}$ ion.
From the above two equations we can see that one reaction is the reverse of the other .They are numerically equal in magnitude .
Therefore ionisation potential of ${X^ + }$ = electron affinity of ${X^{2 + }}$ ion.
Hence the correct answer is option D
Note:
When an electron is added to an atom the process is exothermic due to the attraction to the nucleus . On the other hand when an electron is removed from an atom the process is endothermic due to repulsion from the nucleus.
Complete answer:
We know that ionisation potential is equal to the minimum energy required to remove the electron from an isolated gaseous atom. So at first we need to find the ionisation potential of ${X^ + }$ ion.
So the reaction will be as follows
${X^ + } \to {X^{2 + }} + {e^ - }$ here the ionisation potential is I.P.
Here the electron has been removed from ${X^ + }$ ion. So the enthalpy is equal to the ionisation potential of ${X^ + }$ ion. Similarly if we reverse the reaction then an electron is added to the system.
Then the reaction is as follows
${X^{2 + }}(g) + {e^ - } \to {X^ + }$
Here electron has been added to the system. So the electron gains enthalpy here is the electron affinity. Here the enthalpy is equal to the electron affinity of ${X^{2 + }}$ ion.
From the above two equations we can see that one reaction is the reverse of the other .They are numerically equal in magnitude .
Therefore ionisation potential of ${X^ + }$ = electron affinity of ${X^{2 + }}$ ion.
Hence the correct answer is option D
Note:
When an electron is added to an atom the process is exothermic due to the attraction to the nucleus . On the other hand when an electron is removed from an atom the process is endothermic due to repulsion from the nucleus.
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