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Hi there.
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In this question, we have ionic compounds made from a metal and a non -metal, as ionic compounds are.
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And what we know about the formulas for ionic compounds is that the total charges must add to zero.
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Because ionic compounds are neutral.
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So the total positive and negative charge must add to zero when we write the formula.
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Okay, so as we're looking at the periodic table, some of the groups of elements always form ions with certain charges.
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For example, everything in group 1a has 1 valence electron and wants to lose that 1 valence electron, so they will form cat ions or positive ions with a 1 positive charge.
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All of the elements in group 2a will form ions with a 2 positive charge.
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The metals, or especially aluminum, in group 3a, will form ions with a 3 plus charge.
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Group 5a, now we're getting into the next.
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Non -metals.
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They are going to gain electrons because they already have five, six, and seven valence electrons and they want to get to eight.
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So the elements in group 5a will gain three, giving them a three negative charge.
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The group where we see oxygen, group six a, those will form ions with a two negative charge.
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And the halogens form ions with a one negative charge.
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All right, using this information and the fact that the total charges must add up to zero, we can write the correct formulas for each of these.
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So in letter a, we have barium and flooring.
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So that is going to make barium fluoride.
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Remember when we write the name, the suffix or the ending of the anion changes to ide.
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So barium fluoride means we have a barium ion together with a flooring.
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Barium is in group 2a of the periodic table, so it will form an ion with a 2 positive charge, while fluorine forms an ion with a one negative charge.
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So that these will add up to zero, we need to have two florins because two times one negative is two negative.
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When we add that to two positive, we get zero.
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So the correct formula for barium fluoride is ba f2...