Question

Oxygen reacts with platinum (VI) fluoride, $Pt{F}_6$ as follows:
$O_2+Pt{F}_6\to O_2^{+}Pt{F}_6^{-}$
It was suggested that xenon should react similarly and, in this way,the first noble gas compound was produced.
$Xe+Pt{F}_6\to X{e}^{+}Pt{F}_6^{-}$
The most likely reason for the suggestion being made, is:

• A. $O$ and $Xe$ have similar atomic radii
• B. $O$ and $Xe$ have similar electron affinities
• C. $O$ and $Xe$ have similar electronic configurations
• D. $O$ and $Xe$ have similar first ionisation energies

Answer 1

Answer: (D)

$O$ and $Xe$ have similar first ionisation energies

$O_2+Pt{F}_6\to O_2^{+}Pt{F}_6^{-}$

In the same way, $Xe+Pt{F}_6\to {Xe}^{+}Pt{F}_6^{-}$

Now, 1st ionisation energy of $O_2$ is $\sim 1167kJ/mol$

$O_2-e⟶O_2^{+}$

and 1st ionisation energy of $Xe$ is $1170kJ/mol$. So, both $O_2$ and $Xe$ have similar ionisation energies. $O_2$ molecule has paired electrons, that is why its 1st ionisation is very, similar to that of nobel gas $Xe$.