(b) What is the oxidation number and electron
configuration of ca...

Question

(b) What is the oxidation number and electron
configuration of calcium in each product?

Accepted Answer

Welcome back everyone in this example, we need to consider the reaction of magnesium and chlorine and the reaction of magnesium and water. We need to identify the electron configuration and oxidation number of magnesium in the product of each reaction. So let's begin with our reaction of magnesium and chlorine. Recall that magnesium is considered a metal. So it gets the solid phase and our adam chlorine is a diatonic molecule which exists as a gas cl two. This is going to form a single solid product of magnesium chloride, as MG cl two. And this is a solid precipitate product. We want to recognize that this is a redox reaction where we need to figure out how electrons were transferred by analyzing the oxidation states of our magnesium and chlorine atom. And so we want to recall that we have an ionic compound here being our magnesium chloride product which is polly atomic and has a neutral charge of zero. So we'll write neutral poly atomic ionic compounds and in order to determine the oxidation numbers of the atoms that make up this compound, we would take the sum of our oxidation numbers of the atoms making up this ionic compound. And so we would say that the oxidation, sorry the oxidation state of magnesium plus we have two atoms of chlorine. So we would have plus two times the oxidation state of chlorine is equal to the charge of our poly atomic ionic compound, which is neutral here. So it's equal to zero. So recall that magnesium is a group two A elements on the periodic table and elements in group two A will typically have a plus two oxidation state or oxidation number. But because we have magnesium in its elemental form, which is as a solid metal, we would say that its oxidation state is going to be zero for now. So we need to calculate it according to our oxidation state for chlorine. So we want to go ahead and say that we don't know the oxidation state for magnesium. Sorry, this is solid magnesium. So we need to solve for it by saying that the oxidation state for magnesium plus our two moles of our oxidation state for chlorine, which we should recognize is in Group seven A. On the periodic table and therefore will have a oxidation state of minus one because it's in Group seven A is equal to zero. And so solving for our oxidation state of magnesium, we would say that our oxidation state of magnesium plus negative two as our oxidation state of chlorine now is equal to zero. And so to simplify this, we would just add two to both sides, Meaning that our oxidation state of Magnesium is equal to positive two. So, this would be our first answer since we determined our oxidation state of magnesium in our first reaction being with chloride. So now we want to go on to our second reaction, which is our magnesium reacting with according to the prompt water. So magnesium solid magnesium reacting with liquid water yet again, this is going to be another redox reaction where we form our product magnesium hydroxide as a solid precipitate. And then we have a church to gas. Sorry, that's H two gas here. And our magnesium hydroxide here is going to be our poly atomic ionic compound, which has again a neutral charge in this case. So we need to solve for our oxidation state of magnesium yet again in our magnesium hydroxide. So right now, because we recognize that our oxygen Is not in a peroxide or super oxide type of compound. We would say that oxygen has a oxidation state equal to -2, which makes sense since it's in group two A or sorry, six A of our periodic table. And Adams in that group typically have an oxidation state of minus two. So again, that is because it's not a peroxide or super oxide according to our oxidation rules. Next, we want to recognize that hydrogen will always have an oxidation number Or oxidation state equal to plus one. So, with these two facts in mind, we can solve for our oxidation state of magnesium Which is going to be added to the two moles of our oxidation state of our oxygen atom because again, we have this subscript of two in our poly atomic ion attributed to our oxygen and hydrogen and this is added to the two molds of our oxidation state of hydrogen, which because this is a neutral poly atomic compound, it has a charge of zero. So we're setting this equation equal to zero. And so to simplify this, we would have that our oxidation state of magnesium is equal to two times negative two plus two times positive one. As our plugged in oxidation states of oxygen and hydrogen that we noted above set equal to zero. And so to simplify this, we would say that we have our oxidation state of magnesium equal to negative four plus two. And sorry, not an equal sign here. This should be a plus sign equal to zero. And when we simplify this, we have negative four plus two. Which leaves us with our oxidation state of magnesium. When we negative four plus two, which simplifies to negative two equal to zero. Where we can add positive two to both sides to give us our oxidation state of magnesium equal to positive two. And so this would be our second answer for oxidation state of magnesium and the reaction with water. Now we want to figure out our electron configuration. So we need to figure out our value for the atomic number which we recall is represented by the symbol Z. On our periodic or from our periodic tables for magnesium. And we would see that for magnesium we have an atomic number equal to 12, meaning that we have 12 electrons in neutral magnesium. Sorry, that's capital M. There 12 electrons in neutral magnesium for our configuration to fill in our orbital's. And so we can say that we have an electron configuration where we begin at hydrogen on our periodic table to pass through the S. Sub shell as one S. With an expletive to for our first two electrons. We then have lithium and beryllium in the second period of our periodic table where we move into the two S and fill in for two electrons for two S. Two sub shell. We move forward to the p block of our periodic table where we go through our two P six sub shell filling in for six electrons. So sorry, the six should be read here. So right now we can count a total of 6789, 10 of our 12 electrons filled in. Where we then move into the third period of our periodic table where we see sodium and magnesium and we count for a total of two units to land on our atom magnesium. So we would say at the third period we're at the third energy level of R. S. Sub shell and we fill in to our last two electrons for our total of electrons. And this would be our electron configuration of magnesium for neutral magnesium. But in this case according to our oxidation states of magnesium in our above two reactions being both plus two. That means that we have the magnesium two plus kati on that we're dealing with. And so the electron configuration for the magnesium two plus kati on we recall means that we lose two electrons based on that positive charge. And so we would have a configuration where we want to remove those two electrons from the outermost shell or the outermost sub shell. So we would Lose two electrons from the outermost sub shell, which is going to be our valence shell here or valence orbital, which is R three S orbital. And we want to remove these two electrons. So that's going to give us one S 22, S 22 P 63 P zero. Now, since we remove those electrons which we can simplify to one S 22, S 22 P six and that would be our final configuration of our magnesium two plus caddy on. So everything highlighted in yellow represents our three final answers which represent our oxidation states of magnesium as well as our magnesium caddy on configuration based on what we determined. So I hope that everything I explained was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video

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