Write an appropriate Lewis structure for each compound. Make certain to distinguish between ionic and molecular compounds.
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 Step 2 of 5 The molecule ClF is formed by Chlorine (Electronegativity=3.1 ) and Fluorine 5 (Electronegativity=4.0) Electronegativity= Electronegativity of F - Electronegativity of Cl=4.0-3.1= 0.9 Since 0.4