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Consider the orbitals shown here in outline.(a) What is the maximum number of electrons contained in an orbital of type (x)? Of type (y)? Of type (z)?(b) How many orbitals of type (x) are found in a shell with n = 2? How many of type (y)? How many of type (z)?(c) Write a set of quantum numbers for an electron in an orbital of type (x) in a shell with n = 4. Of an orbital of type(y) in a shell with n = 2. Of an orbital of type (z) in a shell with n = 3.(d) What is the smallest possible n value for an orbital of type (x)? Of type (y)? Of type (z)?(e) What are the possible l and ml values for an orbital of type (x)? Of type (y)? Of type (z)?

Chemistry 101

Chapter 6

Electronic Structure and Periodic Properties of Elements

Electronic Structure

Periodic Table properties

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Question 41 is a review of quantum numbers. The first part of this question asks, What is the maximum number of electrons contained in an orbital of type X? Why and Z? They identify X, y and Z with these pictures access spherical. So this would be an S orbital. Why is dumb Bell shaped? So it's a P orbital, and then this is a D orbital. Knowing this, we should be able to answer this first question and the rest. Also, the maximum number of electrons that can be found in any orbital is too, so the maximum number that could be found in all three of these is to the next question is how maney orbital's of type X are found in a shell with n equals. Two one esque orbital is found in every shell, so there's only one s orbital present how money can be found of type Y. So how Maney P Orbital's are found in the second energy level, there are three p orbital's in the second energy level, there are no D orbital's, so there would be no orbital's of type Z. The next question is right. A set of quantum numbers for an electron in an orbital of type X in a shell where an equals four. Well, if n equals four, then s corresponds to quantum number zero. So it needs to be 40 m l can Onley b zero and then m s could be plus or minus one half. So the only possible combinations would be 400 plus one half or 400 minus one half. The next question asks, How many for type? Why when and equals four? I mean, when an equals two. So if an equals two, then a p corresponds to an l value of one. And then we could have an ML value of negative 10 or one so you could choose. And then we could have an M s value of plus one half or minus one half and you could choose there. And then for the last part, it's orbital Z where n equals three. If n equals three, then we'll start with three. De has an L value, which is the one that comes next that corresponds to to. So then ml could be 210 negative one or negative to you could choose and then m s could be plus one half or minus one half and you could choose for part D. What is the smallest possible and value for an orbital of type X Y and P? Well, an s value is found in all energy levels from one all the way up to, in theory infinity. So we would have and equal one as the smallest possible and value for the S orbital P is only found in energy levels two and above. So the smallest one would be to for the P orbital and the D orbital's air only found in energy levels three and above. So the smallest and value would be three for E. What is the what are the possible l and M l values for each of these? Well, if we look at the s orbital, the L value is always zero for an s orbital. So the M l can only be zero. Also, in the case of P a p orbital, the L value is one. So that means ML could be positive. They'll all the way down to negative ale or negative 1011 and then the l value for a D orbital is two. So m l could be negative to negative 101 or two

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