What is the value of electron gain enthalpy of $N{{a}^{+}}$$+10.2eV$ if $I{{E}_{1}}$ of $Na=5.1eV$?
A. $-5.1eV$
B. $-10.2eV$
C. $+2.55eV$
D. $+10.2eV$
Answer
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Hint: We have to just find out the value of the electron gain enthalpy. Try to state about what is ionization energy and electron gain enthalpy and about the relation between them. Remember, ionization energy is just opposite of electron gain enthalpy.
Complete step by step answer:
As per the given question, given that,
The first ionization energy i.e. $I{{E}_{1}}$ of the sodium atom (i.e. $Na$) is equal to $5.1eV$.
We have to find out the value of electron gain enthalpy of the sodium cation i.e. $N{{a}^{+}}$.
First of all, let us know about what ionization energy and electrons gain enthalpy.
- Ionization energy: Ionization energy can be described as the minimum amount of energy that an electron in a gaseous atom or ion has to absorb to come out from the influence of the nucleus. It is also referred to as ionization potential and it usually undergoes an endothermic process. It is denoted by the symbol $IE$.
- Electron gain enthalpy: Electron gain enthalpy of an element refers to the amount of energy that get released upon the acceptance of an atom from any neutral isolated gaseous atom to form a negative gaseous ion (anion) in the process. It is represented by $H$.
So, we can see that both the terms are opposite to each other i.e. electron gain enthalpy is the negative term of the ionization energy. So, if the value of ionization energy is given then its electron gain enthalpy will be a negative value of the ionization energy having the same magnitude but opposite sign.
Therefore, the electron gain enthalpy of sodium cation will be negative value of the first ionization energy of the sodium atom. So, the value will be $-5.1eV$.
So, the correct answer is “Option A”.
Note: The electron gain enthalpy of an element tells us about the strength of the extra electron that gets bound to the gaseous atom. Remember, the more the amount of energy released in a chemical reaction, the more will be its electron gain enthalpy. Generally, electron gain enthalpy can be both exothermic and endothermic while ionization energy is endothermic only.
Complete step by step answer:
As per the given question, given that,
The first ionization energy i.e. $I{{E}_{1}}$ of the sodium atom (i.e. $Na$) is equal to $5.1eV$.
We have to find out the value of electron gain enthalpy of the sodium cation i.e. $N{{a}^{+}}$.
First of all, let us know about what ionization energy and electrons gain enthalpy.
- Ionization energy: Ionization energy can be described as the minimum amount of energy that an electron in a gaseous atom or ion has to absorb to come out from the influence of the nucleus. It is also referred to as ionization potential and it usually undergoes an endothermic process. It is denoted by the symbol $IE$.
- Electron gain enthalpy: Electron gain enthalpy of an element refers to the amount of energy that get released upon the acceptance of an atom from any neutral isolated gaseous atom to form a negative gaseous ion (anion) in the process. It is represented by $H$.
So, we can see that both the terms are opposite to each other i.e. electron gain enthalpy is the negative term of the ionization energy. So, if the value of ionization energy is given then its electron gain enthalpy will be a negative value of the ionization energy having the same magnitude but opposite sign.
Therefore, the electron gain enthalpy of sodium cation will be negative value of the first ionization energy of the sodium atom. So, the value will be $-5.1eV$.
So, the correct answer is “Option A”.
Note: The electron gain enthalpy of an element tells us about the strength of the extra electron that gets bound to the gaseous atom. Remember, the more the amount of energy released in a chemical reaction, the more will be its electron gain enthalpy. Generally, electron gain enthalpy can be both exothermic and endothermic while ionization energy is endothermic only.
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