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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

oxygen is more electronegative.

B

oxygen has high electron affinity.

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
C
The addition of second electron in an atom or ion is always endothermic because of repulsion between two 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:

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 m

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

The formation of water from H_(2)(g) and O_(2)(g) is an exothermic process because :

H_2(g) + 1/2 O_2(g) to H_2O (l), DeltaH = - 286 kJ 2H_2(g) + O_2(g) to 2H_2O (l), DeltaH = …kJ

State whether each of the following processes is exothermic or endothermic : H_2O (l) to H_2O (g) , DeltaH = + 40 .7 kJ .

O(g) + 2e^(-) to O^(2-)(g) , DeltaH_(eg) = 744.7 kJ/mole . The positive value of DeltaH_(eg) is due to :

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)

Calculate the enthalpy of formation of Delta_(f)H for C_(2)H_(5)OH from tabulated data and its heat of combustion as represented by the following equaitons: i. H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(g), DeltaH^(Theta) =- 241.8 kJ mol^(-1) ii. C(s) +O_(2)(g) rarr CO_(2)(g),DeltaH^(Theta) =- 393.5kJ mol^(-1) iii. C_(2)H_(5)OH (l) +3O_(2)(g) rarr 3H_(2)O(g) + 2CO_(2)(g), DeltaH^(Theta) =- 1234.7kJ mol^(-1)

For the following reaction, the value of K change with N_(2)(g)+O_(2)(g) lt lt 2NO(g), DeltaH=+180 kJ mol^(-1)

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