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Byju's Answer

Standard XII

Chemistry

Anomalous Behaviour of Oxygen

The formation...

Question

The formation of the oxide ion $O^{2 -} (g)$ requires first an exothermic and then an endothermic step as shown below:
$O (g) + e^{-} ⟶ O^{-} (g)$ ; $Δ H = - 142 k J / m o l$
$O^{-} (g) + e^{-} ⟶ O^{2 -} (g)$ ; $Δ H = 844 k J / m o l$
This is because:

O^{-}

has comparatively larger size than oxygen atom

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Oxygen has high electron affinity

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O^{-}

will tend to resist the addition of another electron

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Oxygen is more electronegative

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Solution

The correct option is D $O^{-}$ will tend to resist the addition of another electron
This is because $O^{-}$ will tend to resist the addition of another electron, so because of the charge repulsion second step is endothermic.

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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 ⟶ O^{-} (g)

;

Δ H = - 142 k J / m o l

O^{-} (g) + e ⟶ O^{2 -} (g)

;

Δ H = 844 k J / m o l

.
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

Q. The formula 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 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^{-} = O^{-} (g); Δ H^{0} = - 142 k J m o 1^{- 1}

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

This is because:

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The formation of the oxide ion O2−(g) requires first an exothermic and then an endothermic step as shown below:O(g)+e−⟶O−(g); ΔH=−142kJ/molO−(g)+e−⟶O2−(g); ΔH=844kJ/molThis is because:

The correct option is D O− will tend to resist the addition of another electronThis is because O− will tend to resist the addition of another electron, so because of the charge repulsion second step is endothermic.

The formation of the oxide ion $O^{2 -} (g)$ requires first an exothermic and then an endothermic step as shown below:
$O (g) + e^{-} ⟶ O^{-} (g)$ ; $Δ H = - 142 k J / m o l$
$O^{-} (g) + e^{-} ⟶ O^{2 -} (g)$ ; $Δ H = 844 k J / m o l$
This is because:

The correct option is D $O^{-}$ will tend to resist the addition of another electron
This is because $O^{-}$ will tend to resist the addition of another electron, so because of the charge repulsion second step is endothermic.