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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^(-)= O^(-) (g), DeltaH^(@) = - 142 kJ moI^(-1)`
`O(g)+e^(-)toO^(2-)(g),DeltaH^(@)=844kJmol^(-1)`
This is because

A

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

B

Oxygen has high electron affinity

C

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

D

Oxygen is more electronegative

Text Solution

Verified by Experts

The correct Answer is:
C
EGE is positive when an electron is added to anion (as repulsion between negative charges).
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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:

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The formation of oxide ion O^(2-)(g) from oxygen atom requires first an exothermic and then 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 through O^(2-) is isoelectronic with neon. It is due to the fact that A) oxygen is more electronegative B) addition of electron in oxygen results in larget size of the ion C) electron repulsion outweights the stability gained by achieving noble gas configuration D) O^(-) ion has comparatively smaller size than oxygen atom

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Calculate the heat of formation of methane in kcalmol^(-1) using the following thermo chemical reactions C(s)+O_(2)toCO_(2)(g) , DeltaH=-94.2 kcal mol^(-1) H_(2)(g)+1/2O_(2)(g)toH_(2)O(l) , DeltaH=-68.3 kcal mol^(-1) CH_(4)(g)+2O_(2)(g)toCO_(2)(g)+2H_(2)O(l) , DeltaH=-210.8 kcal mol^(-1)

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VMC MODULES ENGLISH-CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES -LEVEL -2
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