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Hi there.
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This is problem number 105.
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In problem 105, we are looking back at the elements given to us in problem 95, where we did the orbital notation, and we are now writing their electron configurations using the shorthand method.
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So in the shorthand method, we put a noble gas in brackets to represent the core electrons, and then write out the configuration for all of the electrons past those core electrons.
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So starting with, off we have flooring, which is atomic number nine.
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So it has nine electrons and it is in group, i'm sorry, it is in period number two.
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So the element that goes into the brackets is the noble gas from the period before that.
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So that would be h .e.
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So that represents the first two electrons since h .e is atomic number two.
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So now we need to just work across that second period until we get to flooring.
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In other words, 2s2, 2p5.
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And this would be the shorthand electron configuration for flooring.
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Let's go on to part b.
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Part b, we have vanadium, which is number 23, and it is in group four.
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So i look for the noble gas in group three, and that is argon.
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So argon goes in brackets.
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That takes care of the first 18 electrons.
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It looks like we have five more electrons to put in here.
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So moving across the fourth period, we're going to fill in the s first.
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So that's 4s2.
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And then after 4s comes 3d.
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And 3d can hold up to 10.
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However, we only need three more electrons to give us the 23 electrons...