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Hi there.
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This will be question number 96.
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And for 96, we need to write the orbital notations for the ground state electron configurations for each of the elements.
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So let's go ahead and get started.
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Our first one is for phosphorus.
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So the first thing we need to do is look on the periodic table to determine the atomic number, which is 15.
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We need that because that tells us the number of protons, but more importantly for this question, since that's the number of protons, that is also the number of electrons.
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So we need to do the orbital notation for 15 electrons.
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So let's go ahead and get started.
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We're going to fill the electrons in order from lowest energy upward as alf bile predicted.
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So we'll start with the 1s.
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There's only one 1s orbital.
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So i'm going to put my two electrons of opposite spin into that orbital.
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I'm drawing half arrows just to speed this up a little bit.
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It as we go through here.
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All right.
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Next is 2s.
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2s is again an s, so there's only one s orbital, so the next two electrons go there.
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Next we come to 2p.
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Anytime we have a p, there are three p orbitals of equal energy, and the electrons enter one into each of those before they double up.
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So our next six electrons go like that.
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So so far we have placed 10 electrons, we have five more.
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So continuing on, next would be 3s, 3s holds two, and then 3p.
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Again, it's a p orbital, so i need to draw the three p orbitals, and i only have three electrons remaining because we have 12 arrows here so far, so i only have three remaining.
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So remember, one goes into each of the empty orbitals before they double up.
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So our final configuration would look like this.
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So this would be our final answer for phosphorus.
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We're going to repeat this process for the next three elements, but we're going to have to continue on a little further for each of those because they have higher atomic numbers.
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So let's go on to letter b.
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Letter b is nickel, and nickel has 28 electrons because its atomic number is 28.
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So i'm going a little more quickly this time.
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The two electrons in the 1s, next two electrons in the 2s.
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Then we come to the 2p, then the 3s, followed by the 3p.
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All right, and then after the 3p, we don't go to the 3d next in order of energy.
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The 4s has lower energy than the 3d.
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So we fill up the 4s next.
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Then we come to 3d.
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And since it's a d, there are five orbitals of equal energy.
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And we are getting close at this point.
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We only need eight more electrons...