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Shown below is the ¹9F nmr of XeF5¹. Given the isotopic abundance of 1⁹F is 100%, with nuclear spin of ½, and 129Xe is 26.4% abundant with nuclear spin of ½, explain the spectrum. 40.0 XeF == 3400 Hz 35.0 d

Shown below is the ¹9F nmr of XeF5¹. Given the isotopic abundance of 1⁹F is 100%, with nuclear spin of ½, and 129Xe is 26.4% abundant with nuclear spin of ½, explain the spectrum. 40.0 XeF == 3400 Hz 35.0 d

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Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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9 Shown below is the ¹⁹F nmr of XeF5¹. Given the isotopic abundance of ¹⁹F is 100%, with nuclear spin of ½2, and 12⁹Xe is 26.4% abundant with nuclear spin of ½, explain the spectrum. 40.0 JXeF = 3400 Hz 35.0 d
Transcribed Image Text:9 Shown below is the ¹⁹F nmr of XeF5¹. Given the isotopic abundance of ¹⁹F is 100%, with nuclear spin of ½2, and 12⁹Xe is 26.4% abundant with nuclear spin of ½, explain the spectrum. 40.0 JXeF = 3400 Hz 35.0 d
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Step 1

Given a 19F NMR spectra of [XeF5]–. Isotopic abundance of 19F is 100% and isotopic abundance of 26.4% with nuclear spin of 1/2 . 

We have to explain the given spectrum.

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