Answer
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Hint: Electron gain enthalpy of an element is the energy release when a neutral isolated gaseous atom accepts an extra electron to form the gaseous negative ions i.e. anion. We denote it by $\Delta_{eg}H$. Greater the amount of energy release in the above process, higher is the electron gain enthalpy of the element.
Complete step by step solution:
The electron gain enthalpy of an element is a measure of the firmness or strength with which an extra electron is bound to it. It is measured in electron volt per atom or $kJ\,per\,mole$.
The electron gain enthalpy of $O$ and $F$ less than that of $S$ or $Cl$ is due to small size of $O/F$ the inter electronic repulsion in compact 2p-orbital is much more than the repulsion in 3p-orbital of $S/Cl$ hence, the incoming new electron feels less attraction in $O/F$ and more attraction in $S/Cl$. Due to this the addition becomes difficult in $O/F$ and electron gain enthalpy becomes less then $S/Cl$.
So, When an electron is added to O or F, if it goes to a smaller$(n = 2)$ level it will suffer more repulsions than the electrons in S or Cl in larger level $(n = 3)$.
Hence the correct option is (B).
Note: Sulphur is a chemical substance with symbol $S$ and atomic number 16. It is abundant, multivalent and nonmetallic. Under normal conditions, sulfur atoms form cyclic octatomic molecules with chemical formula $S_8$. Element sulfur is a bright yellow, crystalline solid at room temperature.
Chlorine is a chemical element with the symbol $Cl$ and atomic number 17. largest of the It appears between fluorine and bromine in the periodic table and its properties are mostly intermediate between them. Chlorine is a yellow-green gas at room temperature. It is an extremely reactive element and a strong oxidizing agent.
Complete step by step solution:
The electron gain enthalpy of an element is a measure of the firmness or strength with which an extra electron is bound to it. It is measured in electron volt per atom or $kJ\,per\,mole$.
The electron gain enthalpy of $O$ and $F$ less than that of $S$ or $Cl$ is due to small size of $O/F$ the inter electronic repulsion in compact 2p-orbital is much more than the repulsion in 3p-orbital of $S/Cl$ hence, the incoming new electron feels less attraction in $O/F$ and more attraction in $S/Cl$. Due to this the addition becomes difficult in $O/F$ and electron gain enthalpy becomes less then $S/Cl$.
So, When an electron is added to O or F, if it goes to a smaller$(n = 2)$ level it will suffer more repulsions than the electrons in S or Cl in larger level $(n = 3)$.
Hence the correct option is (B).
Note: Sulphur is a chemical substance with symbol $S$ and atomic number 16. It is abundant, multivalent and nonmetallic. Under normal conditions, sulfur atoms form cyclic octatomic molecules with chemical formula $S_8$. Element sulfur is a bright yellow, crystalline solid at room temperature.
Chlorine is a chemical element with the symbol $Cl$ and atomic number 17. largest of the It appears between fluorine and bromine in the periodic table and its properties are mostly intermediate between them. Chlorine is a yellow-green gas at room temperature. It is an extremely reactive element and a strong oxidizing agent.
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