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Question: What inspired N. Bartlett for carrying out a reaction between Xe and \[Pt{{F}_{6}}\] ?...

What inspired N. Bartlett for carrying out a reaction between Xe and PtF6Pt{{F}_{6}} ?

Explanation

Solution

Neil Bartlett, in the year 1962 was successful in the forming of an ionic compound. Recall the reaction between molecular oxygen and platinum hexafluoride (PtF6Pt{{F}_{6}}). The resultant product was formed as platinum hexafluoride is a strong oxidizing agent. The observations were keenly noted by N. Bartlett.

Complete answer:
N. Bartlett didn’t directly carry out the reaction between Xenon and Platinum hexafluoride.
He was experimenting with the reaction between Oxygen andPtF6Pt{{F}_{6}}. So, when molecular oxygen (O2{{O}_{2}}) and PtF6Pt{{F}_{6}} reacted it resulted in the formation of an ionic compound called dioxygenyl hexafluoroplatinate (V). in this reaction, PtF6Pt{{F}_{6}} oxidizes the O2{{O}_{2}} molecule to O22+{{O}_{2}}^{2+}. The subsequent product was red.
Let’s understand the chemical equation between oxygen and PtF6:
PtF6(g) + O2(g)  O2+[PtF6]Pt{{F}_{6}}_{\left( g \right)}\text{ + }{{\text{O}}_{2}}_{\left( g \right)}\text{ }\to \text{ O}_{2}^{+}{{[Pt{{F}_{6}}]}^{-}}
However, later he realized that the first ionization enthalpies of oxygen and xenon were almost the same. The First Ionization enthalpy of Oxygen is 1175 kJ/mol and that of Xenon is 1170 kJ/mol. He thought of substituting Xenon with oxygen in the above reaction. He assumed PtF6Pt{{F}_{6}} would tend to oxidize Xenon just like oxygen. Another similarity was that the molecular diameter of oxygen and atomic radius of Xenon happens to be 4 Ao. So, they not only have similar Ionization enthalpies but also the same atomic sizes. Therefore, Bartlett was inspired to form a compound with Xenon and PtF6Pt{{F}_{6}}.
When PtF6Pt{{F}_{6}} was made to react with Xenon, a rapid reaction occurred and it formed Xenon hexafluoroplatinate (V).
Let’s see the reaction between PtF6Pt{{F}_{6}} and Xenon:
PtF6 + Xe(g)  Xe+[PtF6]Pt{{F}_{6}}\text{ + X}{{\text{e}}_{\left( g \right)}}\text{ }\to \text{ X}{{\text{e}}^{+}}{{[Pt{{F}_{6}}]}^{-}}
Here in the above reaction, PtF6Pt{{F}_{6}} oxidized Xe to Xe+. Xe+[PtF6]\text{X}{{\text{e}}^{+}}{{[Pt{{F}_{6}}]}^{-}}is a crystalline red compound. It also is the first formed compound of a noble gas.

Note:
You might find this reaction impossible, as noble gases are inert to any reactions. But as the properties of Xenon matched with oxygen the reaction was successful. However, keep in mind that the temperature for the reaction has to be 278 K.