Write the condensed electron configurations for the following atoms and indicate how many unpaired electrons each has: (a) Mg.

Video transcript

hey everyone in this example, we need to choose the correct shorthand electron configuration and number of unpaid electrons for an atom Silicon. So we should recall that the shorthand electron configuration will involve our noble gas or it can be known as our noble gas configuration For our given Adam. Silicon. We should recognize that silicon on our periodic tables is located in group four a. And is located across period three of our periodic tables. So we want to find the noble gas that comes before period three on the periodic table and that would correspond to our noble gas neon. So we would place that in brackets to begin our configuration. And next we would go right back into period three where we would recognize that we land at the S sub level which is now at the third energy level because we're at the third period of our periodic table. So we would have three s to begin our configuration and we should recall that the s sub level has a maximum of one orbital which can hold a total of two electrons. So we would fully fill this in and have to in our exponents here for three S two. So now that we filled in our S sub level, the next sub level across period three would be our p sublevel And we would recognize that because we're in period three, this would begin at the third energy level. So we would have three p we should recall that our p sub level can have a maximum of three orbital's which can hold a total of six electrons. However, once we get to the p sublevel in period three of our periodic tables to land on our atom of silicon, we would count a total of two units, meaning we would have an exponent of two here and that corresponds to electrons being filled in our p orbital. So following our poly exclusion principle, We would recall that we fill in the electrons one x one in each orbital. So we would have one electron here and the second electron here and this would be our complete configuration for our atoms Silicon. So as you can see, we were not able to pair any electrons because we only needed to fill into for our configuration of silicon. So therefore we have to unpaid electrons in our P block. And so our final answer to complete this example is that this is our configuration for silicon and we were left with a total of two unpaid electrons in our P block. So this is our final answer. I hope that everything I explained was clear. If you have any questions, please leave them down below. Otherwise, I'll see everyone in the next practice video