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Ionization Energy and Electron Affinity

The first and...

Question

The first and the second ionisation energies of magnesium are 7.646 eV and 15.035 eV respectively. The amount of energy in kJ needed to convert all the atoms of magnesium into $M g^{2 +}$ ions present in 24 mg of magnesium vapour will be:
(Given $1 e V = 96.5 k J m o l^{- 1}$ ).

A

2.455 k J

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B

2.188 k J

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C

1.094 k J

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D

4.370 k J

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Solution

The correct option is B $2.188 k J$
Total energy needed to convert one $M g$ atom into $M g^{2 +}$ gas ion,
$= I . E_{I} + I . E_{I I} = 22.681 e V = 2188.71 k J m o l^{- 1}$ .

24 mg of $M g$ = $1 \times 10^{- 3}$ mole.
Total energy = $1 \times 10^{- 3} \times 2188.7 = 2.188 k J$

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Ionization Energy and Electron Affinity

Standard XIII Chemistry

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