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Ionisation Energy

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Question

In the formation of a chloride ion from an isolated gaseous chlorine atom, 3.8 eV energy is released, which would be equal to:

electron affinity of

C l^{-}

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ionisation potential of

C l

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electronegativity of

C l

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ionisation potential of

C l^{-}

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Solution

The correct option is C ionisation potential of $C l^{-}$
In the formation of a chloride ion from an isolated gaseous chlorine atom, 3.8 eV energy is released which would be equal to Ionisation potential of $C l^{-}$ .
The electron affinity or electron gain enthalpy of the chlorine atom is equal in magnitude and opposite in sign to the ionization potential of chloride ion.

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Similar questions

3.7

1

C l^{-}

1

= 23.06

m o l^{- 1}

+ 37 e V

k J

3.55 g

C l

(1 e V = 96.49 k J / m o l e)

(C l \to C l^{+} + e^{-})

C l + e^{-} \to C l^{-}

6.023 \times 10^{23}

(I P f o r C l = 1250 k J

m o l e^{- 1} a n d E A = 350 k J m o l e^{- 1}

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