Heating elemental cesium and platinum together for two days at 97...

Question

Heating elemental cesium and platinum together for two days at 973 K gives a dark red ionic compound that is 57.67% Cs and 42.33% Pt. 
(c) What are the charge and electron configuration of the platinum ion?

Accepted Answer

Hello everyone today. We have been given the following problem. An ionic compound composed of 71.36% rubidium and 28.64% of rhodium is produced when elemental rubidium and ramon iem or rhodium is heated together for 17 days at 500 degrees kelvin provide the charge and electron configuration of the rhodium ion. So first we need to pay attention to our percentages that we have and we have to assume That we are working with a 100 g sample. And so this simplifies things a lot. And so if we do that, we can just turn our percentages into masses. So for example, for our rubidium instead of 71.36%, we have 71.36 g of rubidium. And for rhodium we have 28.64 g. We're gonna keep those in mind. Then we have to find our moles of both rubidium and our rhodium. And so how do we find the most of that? Well, we start with our masses 71.36 g of rubidium. And we have to use the molar mass according to the periodic table. So if we go to the periodic table, we would note that one mole of rubidium is equal to 85.4678 g of rubidium. Our g of rubidium canceled out. And we're left with 0. moles of rubidium. We're gonna do the same procedure for our rhodium. So we have our 28.64 g of rhodium. And if we use the periodic table, we can note that one mole of rhodium is going to be equal to 102.9055g of rhodium. And so this is going to give us . moles of rhodium when our units of g for rhodium cancel out. And so we're gonna save these numbers for later steps. The next step is to find our mole ratio and essentially we're just going to divide by the smallest number of moles that we have. So for example, if we want to find the mole ratio for rubidium, we're going to take the moles that we found for rubidium. So if you want to find it for rubidium, we're gonna take our Our 0.8349 moles. And we're going to divide by the smallest number of moles we have between the rubidium and rhodium and that's going to be our rhodium. So we're gonna take that number and we're gonna divide that by . moles. And that's gonna give us three For Rodeo. We're just going to simply divide .2783 moles By itself .2783 moles. And that's going to give us one. So this essentially tells us how much of each we have. So our formula, it's going to be rubidium three rhodium and that does not get a coefficient because it is just one of them, so we have our formula there. And so if our ions were if this formula were to associate we have to find out what ions we would get. So ions if we had that dissociation and so we're gonna have a cat eye on a positively charged species and an an eye on for a cat eye on that's going to be our rubidium and it's gonna form a plus one charge because there's only one of them are an I on the other hand, is going to form a negative charge and that's gonna be our rhodium and that's going to form a three minus charge because it has to cancel out with the three rubidium positive charges that are there. And so let's use a general chemistry concept that when we have a neutral atom, the number of protons equals the number of electrons in a neutral atom. And so the question asked us to provide the charge and the electron configuration for the rhodium ion. So we have the charge here being minus or three minus. So now we have to find the condensed formula. So we're gonna go down here to step six and to finally condensed electron configuration, we're going to take the preceding noble gas. So the noble gas that appears just before we reach rhodium and that's going to be krypton. And so after we have krypton we have to count how many electrons we're going to have. And so when we're riding out our formula, we're going to have our noble gas in brackets followed by our four D and R five s orbital's now, which one comes first? It's actually going to be our four D orbital because it is a lower energy level than our five S. Four simply comes before five. And that is why that is a lower energy. So we have our four D 75 s to even though conventionally you will usually see the five S or the S orbital come first. This is the special case. So we have our condensed electron formula configuration here as krypton in our brackets with our noble gas, followed by four D 75 S two. And then we have the charge of rhodium, which would be three minus or negative three charge. And with that we've answered the question overall, I hope that this helped. And until next time.

Heating elemental cesium and platinum together for two days at 973 K gives a dark red ionic compound that is 57.67% Cs and 42.33% Pt. (c) What are the charge and electron configuration of the platinum ion?

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