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Lattice Energy

In the prepar...

Question

In the preparation of compounds of Xe, Bartlett had taken $O_{2}^{+} P t F_{6}^{-}$ as a base compound. This is because:

both

O_{2}

and Xe have same size

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both

O_{2}

and Xe have same electron gain enthalpy

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both

O_{2}

and Xe have almost same ionisation enthalpy

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both Xe and

O_{2}

are gases

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Solution

The correct option is C both $O_{2}$ and Xe have almost same ionisation enthalpy

bartlett observed that the first ionization potential of Xenon is almost equals to that of the Oxygen molecule and made Born-Haber calculation to predict that there would be a stable compound: $X e^{+} P t F_{6}^{-}$ .
Hence option C is correct.

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Similar questions

O_{2}^{+} P t F_{6}^{-}

O_{2}^{+} P t F_{6}^{-}

X e^{+} P t F_{6}^{-}

O_{2}

P t F_{6}

O_{2} + P t F_{6} \to O_{2}^{+} P t F_{6}^{-}

X e + P t F_{6} \to X e^{+} P t F_{6}^{-}

_{2}

N a < N O < X e = O_{2} . O_{2}

P t F_{6}

O_{2} [P t F_{6}] .

P t F_{6}

O_{2}

O^{+}

X e

X e^{+}

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