How many unpaired electrons are present in each of the following ...

Question

How many unpaired electrons are present in each of the following ground-state atoms?
(a) O 
(b) Si 
(c) K 
(d) As

Accepted Answer

welcome back everyone. We need to identify the number of unpaid electrons present in each of our below atoms. Beginning with example A. We have our atoms sulfur where we want to write out its condensed electron configuration and recall that that would be finding our noble gas that proceeds sulfur on the periodic table where sulfur we recall is located in group six A. And lies across period three. So we would find the noble gas neon which proceeds sulfur. And we place this in brackets and begin our configuration where to get to sulfur. We pass through the third period of R. S. Block and we fill in for two electrons where continuing on we hit the third energy level of R. P. Block and we count for a total of four units to land on our atom of sulfur where we should recognize that our exponents here represent our electrons filled in. And so because we have even numbers of electrons, we want to recall that even number of exponents in our configuration is going to correspond to paired electrons so an electron filled in an orbital where we have an upward spin electron and an downward spin electron. And so here for example A we would say we have zero unpaid electrons present. Moving on to example B. We have our atom cobalt. So riding out cobalt condensed electron configuration, we find the noble gas that precedes cobalt on the periodic table which we see corresponds to our atoms of argon and from its configuration we move into the third energy level of our D block where we will count for a total of seven units to land on our atom of cobalt. And we also are going to pass through our fourth energy level of R. S block where we would fill in for all two electrons for that sub level. And we want to recognize that our D block at the fourth energy level will begin sorry at the fourth energy level of R. S block. R D block begins at the third energy level. So that's just something to remember. And we want to write our configuration so that we have our outermost electrons in the higher energy level written last. And so here we can see we have an even number of electrons in our outermost energy level. But then in our three D sub level we have an odd number of electrons as seven here and so here because we have seven electrons, we can imagine our orbital where for our d sub d sublevel recall that our D orbital has a total of five orbital's. So 12345 lines represent our five orbital's of our d sublevel. And within each orbital we would fill in our seven electrons accordingly where we have according to Hunt's role. Each orbital filled with one spin direction of of an electron. So we have 12345. And then we would have now pairing up our electrons six and then seven which would leave us with a total of 123 unpaid electrons. And so we can confirm that for example, B we have three unpaid shared electrons. So now moving on to example, see for our adam calcium writing out it's condensed configuration, we find the noble gas that precedes calcium on the periodic table which also happens to be argon for calcium. And going into the S block at our fourth energy level of our periodic table. We would count for a total of two units to land on our adam calcium in group two A of our periodic table, which is why we fill in that exponents to And yet again we have an even number in our exponents representing our electrons being paired, meaning that we have zero Owen paired electrons for example, C for our adam calcium. So now moving on to example, D for our adam chlorine, writing out it's condensed configuration. Beginning with the noble gas that precedes chlorine on the periodic table. We would find neon as that noble gas. And so moving into our configuration, we would pass through the third energy level of R. S. Block and fill in for all two electrons for that orbital. Now moving up in energy or rather to our next energy level orbital which is our three p orbital where we would count for a total of five units to land on our atom chlorine. So meaning we would fill in five electrons in our p orbital. And because as we stated earlier, we want to follow huns rule, we're going to imagine our p orbital which we should recall consist of a total of three orbital's. So we have our three p orbital's and according to Hunt's rule we're going to fill in our five electrons so that we have each orbital filled in with an electron of any direction spin. So we would have our upward spin, let's say we start with. So for 123 electrons so far and now pairing them up we have four and then five. And this would leave us with a total of just one unpaid electron for example for chlorine. So we would confirm that we have one um paired electron present. And so for our final answers we have for example a zero on paired electrons present corresponding to sulfur. For example, B. We have for cobalt three unpaid electrons present. For example, C. For calcium we have zero unpaid electrons present. And then for example, D for our adam chlorine, we determined that we have one unpaid electron present. So it's highlighted in yellow represents our unpaid electrons present in each of our corresponding atoms highlighted in yellow. If you have any questions, please leave them down below and I'll see everyone in the next practice video

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