The formation of the oxide ion requires first an exothermic and then an endothermic step as shown belowa O g - e - O g 0- - H circ -142 kjmol -1b O g s - e - O g 2- - H circ -844 kjmol -1This is becauseA- Ion will tend to resist the addition of another electronB- O-Ion has comparatively larger size than oxygen atomC- O-Oxygen has high electron affinityD- Oxygen is more electronegative

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Byju's Answer

Standard X

Chemistry

Ionisation Energy

The formation...

Question

The formation of the oxide ion requires first an exothermic and then an endothermic step as shown below

(a) $O_{g} + e^{-} = O_{g}^{-} △ H^{c i r c} = - 142 k j m o l^{- 1}$
(b) $O_{g} + e^{-} = O_{g}^{2 -} △ H^{c i r c} = 844 k j m o l^{- 1}$

This is because

Ion will tend to resist the addition of another electron

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Oxygen has high electron affinity

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Oxygen is more electronegative

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Ion has comparatively larger size than oxygen atom

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Solution

The correct option is A
Ion will tend to resist the addition of another electron

Look at the data first,

$O^{-}_{g} + e^{-} = O^{-}_{g} \triangle H^{circ} = -142kjmol^{-1} $
$O^{-}_{g} + e^{-} = O^{2-} \triangle H^{circ} = 844kjmol^{-1} $

It says that, in the first reaction, energy is released while in second reaction a good amount of energy is provided for the addition of an electron.

Therefore, we can say that it is because the ion will tend to resist the addition of another electron.

Suggest Corrections

Similar questions

Q. 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); △ H^{o} = - 142 k J m o l^{- 1}

O (g) + e^{-} \to O^{2 -} (g); △ H^{o} = - 844 k J m o l^{- 1}

The correct statement is:

Q. 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^{-} \to O_{(g)}^{-};_{Δ f} H^{⊖} = - 141 {kJ mol}^{- 1}

O^{-} (g) + e^{-} \to O_{(g)}^{2 -};_{Δ f} H^{⊖} = + 780 {kJ mol}^{- 1}

Thus the process of formation of

O^{2 -}

in gas phase is unfavourable even thought

O^{2 -}

is isoelectric with neon. It is due to the fact that,

Q. 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), Δ H^{0} = - 142 k J m o l^{- 1}

O^{-} (g) + e^{-} = O^{2 -} (g), Δ H^{0} = + 844 k J m o l^{- 1}

This is because :

Q. 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); Δ H^{0} = - 142 k J m o l^{- 1}

O^{-} (g) + e^{-} \to O^{2 -} (g); Δ H^{0} = 844 k J m o l^{- 1}

The correct statement is?

The formation of the oxide ion, $O_{(g)}^{2 -}$ from oxygen atom requires first an exothermic and then an endothermic step as shown below:
$O_{(g)} + e^{-} \to O_{(g)}^{-}; Δ_{f} H^{\circ} = - 141 k J m o l^{- 1} O_{(g)}^{-} + e^{-} \to O_{(g)}^{2 -}; Δ_{f} H^{\circ} = + 780 k J m o l^{- 1}$
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 requires first an exothermic and then an endothermic step as shown below (a) Og+e−=O−g △Hcirc=−142kjmol−1 (b) Og +e−=O2−g △Hcirc=844kjmol−1 This is because

The formation of the oxide ion requires first an exothermic and then an endothermic step as shown below

(a) $O_{g} + e^{-} = O_{g}^{-} △ H^{c i r c} = - 142 k j m o l^{- 1}$
(b) $O_{g} + e^{-} = O_{g}^{2 -} △ H^{c i r c} = 844 k j m o l^{- 1}$

This is because