 ###### Jonathan Fong

U.C.Berkeley
M.Ed.,San Francisco State Univ.

Jonathan has been teaching since 2000 and currently teaches chemistry at a top-ranked high school in San Francisco.

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# Energy in Hydrogen Emission Spectrum Problem - Concept

Jonathan Fong ###### Jonathan Fong

U.C.Berkeley
M.Ed.,San Francisco State Univ.

Jonathan has been teaching since 2000 and currently teaches chemistry at a top-ranked high school in San Francisco.

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Here we have a challenging problem about identifying elements on the periodic table. So here we have four different cases. Letter A says we want to identify a member of the same family as oxygen, whose most stable ion contains 36 electrons. What I always tell my students is, figure out what kind of information you have that can help you figure out how the answer to the problem.

Here we actually have two pieces of information. So the same family of oxygen. If you take a look on the periodic table, oxygen is in group 6A, the same family that would be group 6A. That narrows it down to Oxygen, Sulphur, Selenium, Tellurium, or Lead. You have five choices. 20% chance you could guess, but let’s make it a little bit easier.

The most stable ion contains 36 electrons. Look at which period which would have closest to 36. So 34 you have Selenium, a neutral atom. And then Sulphur would have 16, Oxygen would have 8, Tellurium would have 42 and then Polonium would have 84. When you take a look here, the closest one would be Selenium but let’s check take a look. Group 6A all of them form ions with a -2 charge for its ion.

If you have a -2 charge, that means you’re going to add two electrons to each of those numbers that we have here. So instead of 34 electrons, for an atom of selenium, it would be 34 plus 2 which is 36. Which is what we wanted to find here. And so for each of these, remember electrons are negative, so if you have a -2 charge, for the ions, that means you’re adding 2 electrons. So Oxygen Oxide would have 10 and then Sulphur would have 18. Tellurium, the ion would have 54 and then Polonium for the ion would have 86 just like the normal gases that you would have there.

So the answer for letter A would be selenium, because it’s in group 6A which satisfies the first part, and the ion would have 34 plus 2, which is 36 electrons. That would satisfy the second part. So letters b, c and d that we have here, basically we do the same thing.

Alkali metals. Let’s take a look. They are in group 1A and then they all form a +1 charge for the ion. So if the ion contains 54 electrons, that means that the real element would have 55 electrons. 55 electrons for the atom and then since it has a +1 charge, so we’re going to take away one electron because of +1 charge. So we'd have ion 54 and atom 55. So the atom would have 55. You take a look at 55 on the periodic table and that would be Cesium, because Cesium has 55 for the atom. And then its stable ion has a +1 charge, so you take away 1 from the ion charge and then you’ll get 54 electrons, total for the ion.

For the noble gas, noble gases they are in group 8. So this one’s easy. 54 protons, so you take a look at the one with the atomic number 54. And 54 is Xenon. That one was simple.

Halogen, group 7A. 53 protons, so you take a look at the atomic number 53, and you actually didn’t need the information halogen here. And it is Iodine. On these last two examples, instead of looking at the group number we could have gone straight for the number of protons. Remember the protons always represent the element that you have there.

Remember elements have unique number of protons, so the protons always tell you which element it is. Hopefully, this helps you out in identifying elements on the periodic table. Remember, if you have the protons, go to the protons first, because that will definitely tell you what the element is, without even spending time looking for the other information. But sometimes you need the other clues to help you in isolating the element. Have a good one.