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Now we'll discuss the solution for problem 76 from chapter 6.
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In this problem, we're asked to write the condensed electron configuration for the following atoms and indicate how many unpaired electrons each has.
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So let's go ahead and start for part a, where we're given magnesium.
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So the noble gas configuration for magnesium would be neon, and then the valence electrons, we have only two, so it's a 3s2 orbital.
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And because both of these electrons are paired in the s -orbital, there are zero unpaired electrons.
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For part b, we're given germanium, ge, and so the nova gas would be argon, and then we have a 4s2, 3d10, and 4p2.
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And since we have two electrons in the p orbital, they'll both be unpaired, so it's two unpaired electrons.
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For part c, we're given bromine.
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So this will be, again, argon, followed by 4s2, 3d10, and 4p5.
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And since we have 5 p electrons, 2 will have two pairs, and then we'll have one unpaired.
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In part d, the element we're given is vanadium.
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So we have argon as our novel gas electron configuration...