Home
Class 11
CHEMISTRY
The formation of the oxide ion O^(-)(g) ...

The formation of the oxide ion `O^(-)(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)`
this is because :

A

oxygen is more electronegative

B

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

C

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

D

oxygen has high electron affinity.

Text Solution

Verified by Experts

The correct Answer is:
C

This is because `O^(-)` will tend to resist the addition of another electron.
Promotional Banner

Similar Questions

Explore conceptually related problems

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

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

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

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)

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 :

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)

Calculate enthalpy of formation of methane (CH_4) from the following data : (i) C(s) + O_(2)(g) to CO_(2) (g) , Delta_rH^(@) = -393.5 KJ mol^(-1) (ii) H_2(g) + 1/2 O_(2)(g) to H_(2)O(l) , Deta_r H^(@) = -285.5 kJ mol^(-1) (iii) CH_(4)(g) + 2O_(2)(g) to CO_(2)(g) + 2H_(2)O(l), Delta_(r)H^(@) = -890.3 kJ mol^(-1) .