(a) Interpretation: Interpret name of XeO 2 . Concept Introduction: Binary ionic compound is the species contains two ions in the compound to form a chemical species. The name of Binary ion contains the name of cation first then proceeded the name of anion. The name of cation in Binary ion remain same to that of metal while for anion its name ends by suffix ‘ate’ while anion of group 6 and 7 ends with suffix ‘ide’. For cation with lower oxidation state name is ended with ‘ous’ and with high oxidation state is ended with ‘ic’.

Question

Part II. Given two isomers: 2-methylpentane (A) and 2,2-dimethyl butane (B) answer the following: (a) match structures of isomers given their mass spectra below (spectra A and spectra B) (b) Draw the fragments given the following prominent peaks from each spectrum: Spectra A m/2 =43 and 1/2-57 spectra B m/2 = 43 (c) why is 1/2=57 peak in spectrum A more intense compared to the same peak in spectrum B. Relative abundance Relative abundance 100 A 50 29 29 0 10 -0 -0 100 B 50 720 30 41 43 57 71 4-0 40 50 60 70 m/z 43 57 8-0 m/z = 86 M 90 100 71 m/z = 86 M -O 0 10 20 30 40 50 60 70 80 -88 m/z 90 100

Answer 1

Question

Chapter 5, Problem 30CR

Interpretation Introduction

(a)

Interpretation:

Interpret name of XeO₂.

Concept Introduction:

Binary ionic compound is the species contains two ions in the compound to form a chemical species.

The name of Binary ion contains the name of cation first then proceeded the name of anion. The name of cation in Binary ion remain same to that of metal while for anion its name ends by suffix ‘ate’ while anion of group 6 and 7 ends with suffix ‘ide’.

For cation with lower oxidation state name is ended with ‘ous’ and with high oxidation state is ended with ‘ic’.

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Part II. Given two isomers: 2-methylpentane (A) and 2,2-dimethyl butane (B) answer the following: (a) match structures of isomers given their mass spectra below (spectra A and spectra B) (b) Draw the fragments given the following prominent peaks from each spectrum: Spectra A m/2 =43 and 1/2-57 spectra B m/2 = 43 (c) why is 1/2=57 peak in spectrum A more intense compared to the same peak in spectrum B. Relative abundance Relative abundance 100 A 50 29 29 0 10 -0 -0 100 B 50 720 30 41 43 57 71 4-0 40 50 60 70 m/z 43 57 8-0 m/z = 86 M 90 100 71 m/z = 86 M -O 0 10 20 30 40 50 60 70 80 -88 m/z 90 100

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