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Ionization Enthalpy

The sum of ...

Question

The sum of ${I E}_{1}$ and ${I E}_{2}$ , ${I E}_{3}$ and ${I E}_{4}$ for $N i$ and $P t$ are given below

${I E}_{1} + {I E}_{2}$ ${I E}_{3} + {I E}_{4}$
$N i$ $2.49$ $8.80$
$P t$ $2.66$ $6.70$
If true enter 1, else enter 0.
It can be predicted that ${N i}^{2 +}$ is more stable than ${P t}^{2 +}$ .
1

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Solution

The correct option is A 1
${N i}^{2 +}$ has lower sum of first and second ionization energy

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^{- 1}

N i

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{I E}_{1} {I E}_{2} {I E}_{3} {I E}_{4} {I E}_{5}

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(I P)_{1} + (I P)_{2} - ------------- -

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I E_{5}

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