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Standard XI
Chemistry
Intermolecular Forces
Question

Reason
The addition of second electron to a monovalent anion is difficult because both have the same charge and experience more repulsion.
Assertion
The formation of $$F^-(g)$$ from $$F(g)$$ is exothermic, whereas that of $$O^{2-}(g)$$ from $$O(g)$$ is endothermic.
The formation of $$F^-(g)$$ from $$F(g)$$ is exothermic, whereas that of $$O^{2-}(g)$$ from $$O(g)$$ is endothermic.

A
Both Assertion and Reason are incorrect
B
Assertion is correct but Reason is incorrect
C
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
D
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Solution
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Correct option is A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

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Similar Questions
Q1
Reason
The addition of second electron to a monovalent anion is difficult because both have the same charge and experience more repulsion.
Assertion
The formation of $$F^-(g)$$ from $$F(g)$$ is exothermic, whereas that of $$O^{2-}(g)$$ from $$O(g)$$ is endothermic.
The formation of $$F^-(g)$$ from $$F(g)$$ is exothermic, whereas that of $$O^{2-}(g)$$ from $$O(g)$$ is endothermic.
View Solution
Q2
The formation of the oxide ion O2−(g) requires first an exothermic and then an endothermic step as shown below. O(g)+e−=O−(g);ΔH0=−142kJmo1−1
O(g)−+e−=O2−(g);ΔH0=844kJmo1−1 This is because:
View Solution
Q3
The formation of the oxide ion O2−(g) requires first an exothermic and then an endothermic step as shown below :
O(g)e→O−(g);ΔH=−142kJ/mol
O(g)e→O−(g);ΔH=844kJ/mol
This is because :

View Solution
Q4
The formation of the oxide ion $$O^{2-}_{(g)}$$ requires first an exothermic and then an endothermic step as shown below.
$$O_{(g)} + e^- = O^-_{(g)}, \Delta H^o = -142\ kJ\ mol^{-1}$$
$$O^-_{(g)} + e^- = O^{2-}_{(g)}, \Delta H^o = +844 \ kJ\ mol^{-1}$$
This is because
View Solution
Q5
The formation of the oxide ion O2−(g) requires first an exothermic and then, an endothermic step as shown below,
O(g)+e⟶O−(g); ΔH=−142kJ/mol.
O−(g)+e⟶O2−(g); ΔH=844kJ/mol.
This is because:
View Solution