Use arrows to show electron pairing in the valence p subshell of ...

Question

Use arrows to show electron pairing in the valence p subshell of a. Sulfurb. Brominec. Silicon

Accepted Answer

Welcome back, everyone. We need to identify the number of paired and unpaired electrons in the valence D subs shell of cadmium. And then two Cobalt choice A states that for Cadmium, we have two paired electrons and eight unpaired electrons. And it says that for Cobalt, we have six paired electrons and one unpaired electron. And choice B it states that for a cadmium, we have 10 paired electrons and zero unpaired electrons. And then for Cobalt, we have four paired electrons and three unpaired electrons. In choice C it states that for Cadmium, we have eight paired electrons and two unpaired electrons. And for Cobalt, we have two pair of electrons, five unpaired electrons. And then choice D it states that for Cadmium, we have six paired electrons and four unpaired electrons. And then for Cobalt, we have four paired electrons and three unpaired electrons. So we're going to need to consider our orbital diagrams which will consist of the valence electrons for each of our atoms. Beginning with Cadmium, recall that Cadmium has an atomic number of our periodic table or Z value equal to 48. And to draw out or rather write out its electron configuration, we want to find the nearest preceding noble gas in which honor period that we can refer to our noble gas krypton. And so we would place krypton in brackets and move forward in our configuration to land on Cadmium where we would hit the fifth period, specifically in our S block and fill for a total of the two maximum electrons that our S block can hold and moving forward in our configuration, we would hit our D block, which at the fifth period of our periodic table begins at the fourth energy level where we would have four D and we would count for a total of 10 units to land on our atom of cadmium in group two B of our D block. So we would fill in all 10 electrons that our D block can hold. Now because cadmium is located in our D block, we're going to focus our orbital diagram on this block or subs shell rather where we would express a total of five boxes representing each of our five orbitals that make up our D subs shell. And we will fill in our maximum of 10 electrons where honoring our poly exclusion principle, we would fill in one electron each in any direction, spin arrow in each orbital where we would have five singly filled in electrons in each orbital and then pairing them up, we would have the opposite spin electron in each orbital giving us our total of 10 orbitals and 410 as the ending of our valence electron configuration. And we can see from our orbital diagram that we have zero unpaired electrons for the configuration of cadmium. So now we can also count that we have a total of 10 paired electrons in the configuration of cadmium. Since we have two electrons filled in each of our orbitals for our four D subs shell. So moving on to our atom Cobalt recall that on our product table, Cobalt corresponds to an atomic number or Z value equal to 27. So we want to write out its valence or noble gas configuration in which we would find the noble gas that preceeds Cobalt on the periodic table, which in period three would be argon. So we would place argon in brackets and to land on Cobalt, we would hit our S block at the fourth energy level giving us four S and fill in its maximum of two electrons in our exponent. And then moving forward in our configuration, we find Cobalt in our D block where our transition metals are located. And recall that at the fourth energy level or fourth period, our D block begins at the third energy level. So we would have 3d and count for a total of seven units to land on our atom Cobalt corresponding to seven electrons that we would fill in our D block. So drawing our orbital diagram as before, for the D block, we have a total of five orbitals which I'll represent as boxes. And we, we would have 3d where we would fill in a total of seven valence electrons. Again, honoring the poly exclusion principle, we're going to fill in each electron with any direction spin singly first. So 12345 in each orbital and then counting our last three electrons with the opposite spin, we would pair the first orbital up with the 6th electron. And then we have just one more electron to add for a total of seven where we would place another downward spin electron in the second orbital. And here we can count a total of three unpaired electrons. And then in our first two orbitals, we would count a total of four paired electrons for our configuration of Cobalt. Now looking at our answer choices, we would be able to rule out choice a choice C as well as choice D. And this leaves us with choice B which again states that for cadmium, we have 10 paired electrons which we would agree with and zero unpaired electrons. And then for Cobalt, we have four paired electrons which we agree with and three unpaired electrons. So choice B is an accurate choice as our final answer for consideration of the paired and unpaired electrons in our valance D subs shelf for cadmium and cobalt. I hope this made sense. And let us know if you have any questions

Use arrows to show electron pairing in the valence p subshell of a. Sulfurb. Brominec. Silicon

Key Concepts

Valence Electrons

Electron Configuration

Lewis Structures

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