Write the full orbital diagram for each element. a. N b. F c. Mg d. Al

Video transcript

Hello everyone today. We have been given four different elements belonging to the 2nd and 3rd periods of the periodic table, were then being tasked to draw a complete orbital diagram for each of them below. So we're gonna start off with neon here and neon has has a periodic number or an atomic number of 10. And the atomic number is directly related to the number of protons and electrons. And so in a neutral atom the number of protons is going to equal the number of electrons. And so since the atomic number is 10, we have 10 protons. We will also have 10 electrons to work with. And so now we want to draw an outline of our diagram. So we're going to have a one s orbital a two S orbital and a two P orbital that has three different orbital's within it. So we have our one S. Or two S and R two P. According to Hunt's rule when filling out these orbital diagrams, we must first fill each orbital once before pairing up that electron. So we can fit two electrons in each orbital. So we can start with one S. This represents the first electron in the one S block and this donna area represents the second electron in the one in the one S block. Now we move on to the two S block and that's going to be period two of the periodic table. Once again, we only have two electrons in that orbital in that period. So we're going to go ahead and move on to our two P for R two P. We have six electrons and neon so happens to be the sixth electron in the p orbital. And so therefore according to Huntsville, we go 123 upward arrow and we're going to pair them up 45 and six. And so now we have our complete orbital diagram for neon Moving on to B or Beryllium. We see that beryllium has an atomic number of four. And so according to our rules that we discussed before, the number of protons is going to equal the number of electrons in a neutral species. And so we have four electrons to work with. And so we're actually only gonna use the one s orbital and the two s orbital. Remember that each orbital can hold two electrons. So if we have four electrons, we only need these two orbital's using the same logic as before. With huns rule, we're going to pair one electron in the one s orbital, which represents the first electron. We're gonna have a downward arrow here, which represents the second electron for to us in our second period, beryllium is the second electron. And so therefore we need to add two arrows to show that. So we do one up arrow and one down arrow. And that is the complete orbital diagram for beryllium moving out to sea or sodium sodium has an atomic number of 11 As before, the number of protons equals the number of electrons in a neutral species. So we have 11 electrons to work with. Now we're going to need a little bit more than a two P orbital to complete this diagram. So we're gonna use our 21 s orbital for two electrons or two S electrons for two more our 32 P orbital's. So we have our one s to us and to P to accommodate six electrons. And we actually need an additional three s electron which gets into the third period or where sodium is with Hunt's role. As usual, we're going to pair up one electron up one down same with two S for two P. sodium is in the third period. So we need to get through and fill every single orbital here in the two P orbital according to Huntsville. We'll do one arrow up across and then we'll pair that up with a downward arrow. Three S represents the third period in the S block. sodium is the first electronic, so therefore we will only have one electron there. And lastly, D We have oxygen and oxygen has an atomic number of eight. So an atomic number of eight protons equals the electrons and a neutral species. And so we have eight electrons to work with. We're going to use R one S orbital for our two electrons. We're gonna use a two S orbital for two more electrons. And then we're going to use just a two P orbital for the remaining as before with huns rule, we're going to put one electron up and then pair it up with a two s same logic and oxygen is the fourth electron in the two P orbital in the second period in the P block. And so therefore we only need four electrons once again, according to Huntsville, we're going to draw one arrow up in each of these orbital's and then pair them up. So we have an additional electron there. And so now we have the four complete orbital diagrams for our problems. I hope this helped. And until next time.