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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

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The correct Answer is:
A, B
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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

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    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,

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    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

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