Draw a set of 3D axes and label them x, y, z. On the axes,accurately sketch a 3dxy orbital. Describe any nodes that thisorbital has. Include how many nodes if any a...

Sketch a three-dimensional representation for each of the following orbital sets. Label the $x$ -axis, $y$ -axis, and $z$ -axis. (a) $1 s, 2 s, 2 p_{x}$ (b) $3 p_{x}, 3 p_{Y}, 3 p_{z}$

To draw three dimensional representation of each of the following orbital's. We need to know the shape and their orientation with respect to a traditional Cartesian coordinates where axes are X. Y. And Z. one S. is spherical in shape and it's centered right at the center of the axes. Two PX. It's a piece so it's dumb balance shaped with two lobes, it's P. X. So that means it is centered on the X. Axis three P. Y. It's also P. So it's dumb bell and shaped. But the Y says it's centered on the Y axis. I know in the textbook they have these axes flipped. Z. Is up and why is this way? But normally we have X. And Y. And so I'm just adding the Z. In this dimension. It doesn't matter how you do it so long as you label it correctly. So three py is on the y axis and then four PZ is going to be a p dumbbell shaped orbital on the Z. Axis.

Just focusing on the activity of sketching orbital's. I've got a few examples of listed out We've got the one asked the to P three d x Y three ds that squared. So as you can see here, we've got the one I sorbitol because it's an s shell. We have no nodal planes, so it's a simple, spare, honest recording access. Next, we have the two p which is just drawn down here. So we have one little planes. So we've got one point off zero electron density probability where we have a face change. Next we have the three d x wild, but also this is one of the five deal because we can have. And then lastly, this is the D's that squared where the orbital you can see pretty much sits on our side Orbital on our side axis where it looks like we've got a sort of p orbital with a doughnut ring around it. But this is our desired squared

Nodes are basically finding areas in orbital's where the electrons are not likely to be found and therefore most likely will not form a bond in those areas. The equation to find the number of nodes is equal to enriches your principal quantum number minus l, which is the angular momentum minus one. So if we look at one ass, we know that one is the principal quantum number. So one is end and an l for ass is equal to zero. So let's put that into our equation. Your number of nose is gonna be equal to one minus l, which is zero minus one. So you're a number of nodes. Zero. So for the 2nd 1 here, we know that, and it's too. We know that Al is equal to one because it's a P, and we know that once we put that in equation that the number is going to be equal two to minus one minus one that equals zero. Also for three d you know that three is the energy level for D. L is equal to two some. We put that into the equation. We know that the number is equal to three minus two minus one, which equals zero. The last one is four F. We know that four is gonna be your principal quantum number. That you're al value for an AF is equal to three. So in this equation, we know that it is. The number is equal to four minus three minus one, which also equals zero.

Hi there. In problem 71 we are trying to determine which orbital is being represented by the number of radial notes and angular notes. So we need to know a little bit of information here. First of all, the total nodes are always gonna be equal two and minus one so we can use the total notes to determine n which is the energy level. And then the number of angular nodes is equal to L. And then, of course, tells us the sub level. So let's get started with letter A Here in letter A were given one radial node and one angular note. Good. Therefore total notes one plus one gives us too. Two is gonna be equal two and minus one. Therefore solving for N and must equal three. So third energy level. The angular note. Here we are, given just one of those and that is equal to l and L is a P shaped hill is a P shaped orbital. Therefore, our answer is going to be energy level three and it's gonna be a p moving on to let her be and let her be. We have zero radial notes and we have to angular notes, adding, These together we get to That's our total nodes. And we know that our total nodes is equal to end minus one. So again in part B and it's gonna be three. So the third energy level, our angular nodes in this is to So that means l equals two, and that is a D orbital. So putting these two together, we find out that we have a three D in part C. Yeah, we have two radial nodes and we have three angular notes. Two plus three gives us five. So five is the total nodes and we know that five is equal to and minus one. Therefore, n must be six our angular nodes. We have three of those. So that means that L is equal to three because l equals the number of angular notes. And when we have l equals three, it's an F sub level. So the orbital identified here is a six f. There we go. Thank you so much for watching