This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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(a) NaCl Step 1: Lewis dot symbols of atoms. Sodium (Group 1): 1 valence electron. Na · Chlorine (Group 17): 7 valence electrons. · Cl · Step 2: Electron transfer for octet stability. Na loses 1 electron to Cl (Na achieves He configuration, Cl achieves octet). Na · + · Cl · \ \ [Na]^+ + [\ :\!Cl\!:\ ]^- Step 3: Ionic bond formation. Oppositely charged ions [Na]^+ and [\ :\!Cl\!:\ ]^- are held by electrostatic attraction. Final Lewis dot diagram: [Na]^+ \ [\ :\!Cl\!:\ ]^- (b) MgO Step 1: Lewis dot symbols of atoms. Magnesium (Group 2): 2 valence electrons. · Mg · Oxygen (Group 16): 6 valence electrons. · O · Step 2: Electron transfer for octet stability. Mg loses 2 electrons to O (Mg achieves Ne configuration, O achieves octet). · Mg · + · O · \ \ [Mg]^2+ + [\ :\!O\!:\ ]^2- Step 3: Ionic bond formation. Oppositely charged ions [Mg]^2+ and [\ :\!O\!:\ ]^2- are held by electrostatic attraction. Final Lewis dot diagram: [Mg]^2+ \ [\ :\!O\!:\ ]^2- (c) CaF₂ Step 1: Lewis dot symbols of atoms. Calcium (Group 2): 2 valence electrons. · Ca · Fluorine (Group 17): 7 valence electrons (×2). · F · Step 2: Electron transfer for octet stability. Ca loses 2 electrons (1 to each F). Ca achieves Ar configuration, each F achieves octet. · Ca · + 2(· F ·) \ \ [Ca]^2+ + 2[\ :\!F\!:\ ]^- Step 3: Ionic bond formation. Oppositely charged ions [Ca]^2+ and 2[\ :\!F\!:\ ]^- are held by electrostatic attraction. Final Lewis dot diagram: [Ca]^2+ \ 2[\ :\!F\!:\ ]^-