Home
Class 12
CHEMISTRY
The formation of the oxide ion O^(2-)(g)...

The formation of the oxide ion `O^(2-)(g)` requires first an exothermic and then an endothermic step as shown below:
`O(g) + e^(-) to O^(-)(g) , DeltaH^(@) = -142kJ mol^(-1)`
`O^(-)(g) + e^(-) to O^(2-)(g), DeltaH^(@) = +844kJ mol^(-)1`
 Which is not the cause of the above fact? 

A

Oxygen is more electronegative.

B

Oxygen has high electron affinity.

C

`O^(-)` ion has comparatively larger size than oxygen atom.

D

`O^(-)` ion will tend to resist the addition of electron.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the formation of the oxide ion \( O^{2-}(g) \) and the associated enthalpy changes, we need to analyze the steps involved in the process and identify which statement is not a cause for the observed enthalpy changes. ### Step-by-Step Solution: 1. **Understanding the Reaction Steps**: - The first step involves the formation of the \( O^{-}(g) \) ion from neutral oxygen gas: \[ O(g) + e^{-} \rightarrow O^{-}(g), \quad \Delta H = -142 \, \text{kJ mol}^{-1} \] This step is exothermic, meaning it releases energy. The negative enthalpy change indicates that energy is released when an electron is added to neutral oxygen. 2. **Analyzing the Second Step**: - The second step involves the formation of the \( O^{2-}(g) \) ion from the \( O^{-}(g) \) ion: \[ O^{-}(g) + e^{-} \rightarrow O^{2-}(g), \quad \Delta H = +844 \, \text{kJ mol}^{-1} \] This step is endothermic, meaning it requires energy. The positive enthalpy change indicates that energy is absorbed when a second electron is added to the negatively charged ion. 3. **Identifying the Causes**: - The first electron addition (to form \( O^{-} \)) is favorable due to oxygen's high electronegativity and electron affinity, which explains the exothermic nature of this step. - The second electron addition (to form \( O^{2-} \)) is unfavorable due to the increased electron-electron repulsion. The \( O^{-} \) ion already has an extra electron, and adding another electron leads to repulsion between the negatively charged particles, which requires energy to overcome. 4. **Evaluating the Options**: - The question asks which statement is NOT a cause of the observed enthalpy changes. - The options likely include statements about oxygen's electronegativity and electron affinity. While these factors explain the first step's exothermic nature, they do not account for the endothermic nature of the second step. 5. **Conclusion**: - The correct answer is that the high electronegativity of oxygen and its high electron affinity do not explain why the second electron gain is endothermic. The repulsion between electrons in the \( O^{-} \) ion is the primary reason for the positive enthalpy change in the second step. ### Final Answer: The cause that is NOT responsible for the observed enthalpy changes is related to oxygen's high electronegativity and electron affinity, as these factors do not explain the endothermic nature of the second electron gain.
Doubtnut Promotions Banner Mobile Dark
|

Similar Questions

Explore conceptually related problems

The formation of the oxide ion, O^(2-) (g), from oxygen atom requires first an exothermic and then an endothermic step as shown below: O(g) + e^(-) to O^(-) (g), DeltaH^(@) = -141 "kJ mol"^(-1) O^(-)(g) + e^(-) to O^(2-) (g), DeltaH^(@) = +780"kJ mol"^(-1) Thus process of formation of O^(2-) in gas phase is unfavourable even though O^(2-) is isoelectronic with Neon. It is due to the fact that,

The formation of the oxide ion O_((g))^(2-) requires first an exothermic and then an endothermic step as shown below. O_((g)) +e^(-) = O_((g))^(-) DeltaH^(@) =- 142 kJ mol^(-1) O_((g))^(-) + e^(-) = O_((g))^(2-) DeltaH^(@) = 844 kJ mol^(-1) This is because of :

Knowledge Check

  • The formation of the oxide ion O^(2-) (g) requires first an exothermic and then an endothermic step as shown below : O (g) + e^(-) to O^(-) ( g) : Delta H^(@) =- 142 k J mol^(-1) O^(-) (g) + e^(-) to O^(2-) (g) , DeltaH^(@) = + 844 kJ mol^(-1) This is because

    A
    Oxygen is more electronegative
    B
    Oxygen has high electron affinity
    C
    `O^(-) ` will tend to resist the addition of another electron
    D
    `O^(-)` ion has comparatively larger size oxygen atom.
  • The formation of the oxide ion O_(g)^(2-) requires first an exothermic and then an endothermic step as shown below: O_(g)+e^(-) = O_(g)^(-), DeltaH^(@)=-142 kJmol^(-1) O_(g)^(-)+e^(-) = O_(g)^(2-), DeltaH^(@)= 844 kJmol^(-1) This is because

    A
    `O^(-)` ion will tend to resist the addition of another electron
    B
    oxygen has high electron affinity
    C
    oxygen is more electronegative
    D
    `O^(-)` ion has comparatively larger size than oxygen atom.
  • The formation of the oxide ion O_((g))^(2-) requires first an exothermic and then an endothermic step as shown below O_((g))+e^(-)= O_((g))^(-) DeltaH^(@)= -142 " kJ mol"^(-1) O^(-)(g)+e^(-) = O_((g))^(2-) DeltaH^(@)=844 " kJmol"^(-1) This is because

    A
    `O^(-)` ion will tend to resist the addition of another electron
    B
    Oxygen has high electron affinity
    C
    Oxygen is more electronegative
    D
    `O^(-)` ion has comparatively larger size than oxygen atom
  • Similar Questions

    Explore conceptually related problems

    The formation of the oxide ion O_(g)^(2-) requires first an exothermic and then an endothermic step as shown below: O_(g)+e^(-) rarr O_(g)^(-) , DeltaH=-142 kJ mol^(-1) O(g)+e rarr O_(g)^(2-) , DeltaH=844kJ mol^(-1) This is because:

    The formation of the oxide ion O_((g))^(2-) requires first an exothermic and then an endothermic step as shown below: O_((g)) +e^(-) rarr O_((g))^(-), DeltaH =- 142 kj mol^(-1) O_((g))^(-) + e^(-) rarr O_((g))^(2-), DeltaH = 844 kj mol^(-1) This is because:

    The formation of the oxide ion, O^(2-) (g) from oxygen atom requires first an exothermic and than an endothermic step as shown below O(g) + e^(-)rarr O^- (g), DeltaH^(-) = - 141 kJ mol^(-1) O^(-)(g) + e^(-)rarr O^(2-) (g), DeltaH^(-) = + 780 kJ mol^(-1) Thus, process of formation of O^(2-) in gas phase is unfavourable even though O^(2-) is isoelectronic with neon. It is due to the fact that,

    The formation of the oxide ion, O^(2-) (g), from oxygen atom requires first an exothermic and then an endothermic step as shown below : O ( g) + e^(-) to O^(-) , Delta H^(Theta) = - 141 kj mol^(-1) O^(-) (g) + e^(-) to O^(2-) ( g) , Delta H^(Theta) = + 780 kJ mol^(-1) Thus process of formation of O^(2-) in gas phase is unfavourable even though O^(2-) isoelectronic with neon. If is due to the fact the ,

    The formation of the oxide ion O^(2-) (g), from oxygen atom requires first an exothermic and then an endothermic step as shown below : O(g)+e^(-) to O(g): Delta_(f)H^(Ѳ)= - 141 "kJ mol"^(-1) O^(-)(g)+e^(-) to O^(2-) (g), Delta_(f)H^(Ѳ)= + 780 "kJ mol"^(-1) Thus process of formation of O^(2-) in gas phase in unfavourable even though O^(2-) is isoelectronic with neon. It is due to the fact that