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Textbook Solutions for Chemistry: A Molecular Approach

Chapter 9 Problem 86E

Question

Write an appropriate Lewis structure for each compound. Make certain to distinguish between ionic and molecular compounds.

a. Al2O3                 b. ClF5                 c. MgI2                 d. XeO4

Solution

Solution 86E Step 1 of 5 The molecule Al O2s3ormed by Oxygen (Electronegativity=3.5 ) and Aluminium (Electronegativity=1.6) Electronegativity= Electronegativity of O - Electronegativity of Al=3.5-1.6= 1.9 Since Electronegativity >1.9 we can expect an ionic bond, thus a total transfer of electrons from Aluminium to Oxygen. Aluminium has three electrons of valence. It can lose them to leave out its inner, electron-complete, energy shell. Oxygen has 6 electrons of valence (thus it needs two electrons to complete its valence shell). Thus we can’t have a 1:1 element’s ratio in this compound. To solve the problem, since there is a 3:2 ratio between the electrons Al can donate and the electrons Oxygen can accept, we have to consider a reverse ratio between the elements. Thus there have to be a 3:2 ratio between the atoms of Al and the atoms of O Thus we have a formula Al O .23 Since we have an ionic bond we have to consider - Since Al donate three electrons it becomes a positive, trivalent, bivalent ion. - Since O accepts two electrons it becomes a negative, bivalent ion. Hence we can write the Lewis structure in this form

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full solution

Title Chemistry: A Molecular Approach 3 
Author Nivaldo J. Tro
ISBN 9780321809247

Solved: Write an appropriate Lewis structure for each

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