Iron is commonly found as Fe, Fe2+, and Fe3+.
(b) What are the n ...

Question

Iron is commonly found as Fe, Fe2+, and Fe3+.
(b) What are the n and l quantum numbers of the electron removed on going from Fe2+ to Fe3+?

Accepted Answer

Hello everyone today. We are being given the falling problem. Mercury exists as nature as mercury. Mercury one plus and Mercury two plus, Mercury loses an electron when it goes from Mercury one plus two. Mercury two plus. Give the principle number and the angular momentum. Quantum numbers of the electron lost from this transition. So the first thing we want to do is we want to recall that electrons are removed from the outermost orbital first. So when an element loses an electron, it's always moved from the outermost orbital first. And so for our first one here we have just regular mercury. Its atomic number, so its atomic number is 80. and so therefore it has 80 electrons and is therefore considered neutral. It's regular condensed electron configuration per the periodic table is going to be xenon. Since we go we see where mercury is. And then we go backwards until we hit the first noble gas. We then go to R six S orbital which is in the sixth period six S to to be specific because we are filling in all of those orbital's all of those the entire orbital for the S orbital. We then dip into the four f orbital four F 14. And then we go back to our five D 10. And this is our condensed electron configuration. And so we have to note then our highest orbital here That isn't going to be 6 S2, essentially. Our six is the highest number here. So it's considered the highest orbital. So we're gonna just keep that number or answer safe for now. Next thing you wanna do is you wanna write the electron configuration for our Mercury one plus and The configuration for our Mercury two plus for our Mercury plus. The condensed formula is once again xenon because that's the closest noble gas. And then we have six S one, four F 14 and then five D 10. So notice that we are missing an electron in the six S orbital, as we said before, electrons are removed from the outermost orbital first and the outermost is synonymous with highest orbital. And so therefore, an electron was lost in The six S orbital. If we look at Mercury two plus, once again we have our Xenon, We simply just have four F 14, five D 10. And so we also note that we lost two electrons in R six S orbital because R six S orbital can hold a maximum of two electrons and the neutral form already had two electrons. So we simply removed one and then we removed too. So ultimately our N. or our principle quantum number is going to be six. And now we have to find our angular momentum and our angular momentum Is essentially going to be in -1. So you're already designated in as being six. So it ranges from 0 to N -6, sort of L is equal to zero. We have an S sub cell if L is equal to one, we have a p sub shell If L is equal to two, we have a. D. Sub shell, L can be equal to three, and we'll have an S. Sub shell and we'll stop there notice this was a six S. So we have an S. Orbital, meaning that our L. Is going to equal zero. So our answers here are going to be our principle number, or in is gonna be equal to six, and our angular momentum is going to be equal to zero. And ultimately we have solved the problem. I hope this helped, and until next time.

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