What inspired $N. \ Bartlett$ to carry out the reaction between $Xe$ and $PtF_6$?
(N/A) $Neil \ Bartlett$ initially carried out a reaction between oxygen $(O_2)$ and $PtF_6$. This resulted in the formation of a red compound,$O_2^+ [PtF_6]^-$.
Later,he realized that the first ionization energy of oxygen $(1175 \ kJ \ mol^{-1})$ and $Xe$ $(1170 \ kJ \ mol^{-1})$ is almost the same.
Thus,he hypothesized that if $O_2$ could react with $PtF_6$,$Xe$ should also react with $PtF_6$. He was successful,and a red-coloured compound,$Xe^+ [PtF_6]^-$,was formed.
Later,he realized that the first ionization energy of oxygen $(1175 \ kJ \ mol^{-1})$ and $Xe$ $(1170 \ kJ \ mol^{-1})$ is almost the same.
Thus,he hypothesized that if $O_2$ could react with $PtF_6$,$Xe$ should also react with $PtF_6$. He was successful,and a red-coloured compound,$Xe^+ [PtF_6]^-$,was formed.
Explore More
Similar Questions
Ionisation potential values of noble gases decrease down the group with increase in atomic size. Xenon forms binary fluorides by the direct reaction of elements. Identify the correct statement$(s)$ from below.
Match List-$I$ with List-$II$ and select the correct answer using the codes given below the lists:
List-$I$ List-$II$ $(A) \ XeF_4$ $(1) \ \text{Distorted octahedral}$ $(B) \ XeF_6$ $(2) \ \text{Tetrahedral}$ $(C) \ XeO_3$ $(3) \ \text{Square planar}$ $(D) \ XeO_4$ $(4) \ \text{Pyramidal}$
| List-$I$ | List-$II$ |
|---|---|
| $(A) \ XeF_4$ | $(1) \ \text{Distorted octahedral}$ |
| $(B) \ XeF_6$ | $(2) \ \text{Tetrahedral}$ |
| $(C) \ XeO_3$ | $(3) \ \text{Square planar}$ |
| $(D) \ XeO_4$ | $(4) \ \text{Pyramidal}$ |
Medium
View SolutionThe inert gases are
Easy
View SolutionWhich one of the following does not exist?
$(i)$ $XeOF_{4}$
$(ii)$ $NeF_{2}$
$(iii)$ $XeF_{2}$
$(iv)$ $XeF_{6}$
$(i)$ $XeOF_{4}$
$(ii)$ $NeF_{2}$
$(iii)$ $XeF_{2}$
$(iv)$ $XeF_{6}$
Easy
View SolutionWhich of the following is not correct for the given reaction sequence:
$XeF_6$ $\xrightarrow{+ H_2O} A$ $\xrightarrow{+ H_2O} B$ $\xrightarrow{+ H_2O} C$
$XeF_6$ $\xrightarrow{+ H_2O} A$ $\xrightarrow{+ H_2O} B$ $\xrightarrow{+ H_2O} C$
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo