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Byju's Answer

Standard XII

Chemistry

Nature of Oxides

The formation...

Question

The formation of the oxide ion $O^{2 -} (g)$ requires first an exothermic and then an endothermic step as shown below.

$O_{(g)} + e^{-} = O^{-} (g), Δ H^{0} = - 142 k J m o l^{- 1}$
$O^{-} (g) + e^{-} = O^{2 -} (g), Δ H^{0} = + 844 k J m o l^{- 1}$

This is because :

O^{-}

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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O^{-}

ion has comparatively larger size than oxygen atom

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Solution

The correct option is A $O^{-}$ ion will tend to resist the addition of another electron
$O^{-}$ ion tends to resist the addition of another electron due to large repulsion of e $^{-} -$ e $^{-}$ in small sized $O^{-}$ ion. Thus, energy is required to overcome this force of repulsion.

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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^{-} \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 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 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:

Q. The formation of the oxide ion

(O^{2 -} (g))

required first an exothermic and then an endothermic step as shown below.

O + e^{-} \to O^{-}_{(g)} + 142 k J m o l^{- 1}

O^{-} + e^{-} \to O^{- 2}_{(g)} - 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); △ 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:

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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),ΔH0=−142 kJmol−1O−(g)+e−=O2−(g),ΔH0=+844 kJmol−1This is because :

The correct option is A O− ion will tend to resist the addition of another electronO− ion tends to resist the addition of another electron due to large repulsion of e−− e− in small sized O− ion. Thus, energy is required to overcome this force of repulsion.

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 :

The correct option is A $O^{-}$ ion will tend to resist the addition of another electron
$O^{-}$ ion tends to resist the addition of another electron due to large repulsion of e $^{-} -$ e $^{-}$ in small sized $O^{-}$ ion. Thus, energy is required to overcome this force of repulsion.