Assume that the rules for quantum numbers are different and that the spin quantum number ms can have any of three values, ms = -1/2, 0, +1/2, while all other rules remain the same. (a) Draw an orbital-filling diagram for the element with Z = 25, showing the individual electrons in the outer-most subshell as up arrows, down arrows, or 0. How many partially filled orbitals does the element have?

Video transcript

Hello everyone today. We have been given the following problem. Suppose the rules for quantum numbers were changed in the first three quantum numbers are still the same. But the spin quantum number can have three values of negative one half, zero and positive one half provide the orbital filling diagram for the element with an atomic number 41 determine the number of partially filled orbital's. This element has Use up arrows, zero and negative arrows to represent electrons. So the first thing you wanna do is they want to go over some principles and definitions. Speaking of principles, our first quantum number is known as a principle quantum number or N. And this is essentially just the energy levels that are in our orbital's and it can be anywhere from any positive integer to infinity. You can have infinite numbers of energy levels. Our second definition is going to be known as an angular momentum quantum number or L. For short in L just represents the sub shells and this is going to be our S P D and F. Sub shells. And for reference our L value must be in -1. So it's L. But it's also in -1. And so for examples our our sub shells so we can have our L equal zero. We can have an equal one, two and three. And so if illegal zero we are in the S sub shell if it equals one we are in the p sub shell. If it equals two we are in the D sub shell and of L equals three. We are in the f sub shell. Thirdly, we have our what's known as a magnetic quantum number or M sub S M sub L. And so this is just going to be a range of values from negative L. So whatever the value of L is too positive L. And so for our examples here, if we are dealing with the S sub shell, we can have one orbital. If we're dealing with the p sub shell, we can have three orbital's. If we are dealing with the D sub shell, we can have five orbital's. And lastly with our F sub shell, we can have a maximum of seven orbital's and each orbital holds two electrons last but not least. We have our spin quantum number. It is M sub S. And this according to our question is going to have values ranging from negative one half, zero and positive one half. Generally the values only are positive plus positive one half and negative one half. But this question calls into examining to see what would happen if we had zero. And so given the question stem, we have an atomic number 41 And we have to recall that when we have a neutral species with an atomic number of 41, that's going to equal the number of protons and the number of electrons. So we have electrons that we can use. And so we're going to fill up our orbital diagrams and we're going to complete an orbital diagram and fill it up as we go. So we're going to construct an orbital diagram. So the very first orbital that we start off with as stated before is R S orbital and more specifically our one S orbital. And we're going to completely fill that up by drawing 1/2 arrow up and 1/2 arrow downwards. We then move on to the second row or second period of the periodic table where we have our two s. We're going to completely fill that up. We noted that our p orbital's that's what's come next. That is what comes next on the right hand side of our periodic table can hold three orbital's and each line is an orbital. So we have our two P orbital and we're going to completely fill those up. We're going to pay attention to huns rule that states that we have to fill each orbital up with one electron before pairing them up. The next row or period is our three us. And we filled that up. We move right along on the right hand side of the periodic table where we have our three p orbital's we pay attention to hunt's role, repair those up and then we move on to our fourth row or fourth period. We pair that electron up and then after the four s orbital we get into our transition metals where we have our d orbital's and our first d orbital is our three D orbital. And so after filling these electrons up, we know now we only have six electrons left. So we're gonna do huns rule which is filling one orbital up at a time with one up arrow followed by we have eight electrons to disperse. Excuse me. So we go along by Huntsville, we fill them up and then we see that we stop when we get to our first orbital here. And so we can see here that we have 1234 partially filled orbital's. So we have four partially filled orbital's whenever we see a lone electron that represents a partially filled orbital, since an orbital is only completely filled if it has two electrons total. So we had we have our orbital diagram as follows and then we have our answer of having four partially filled orbital's. And with that we've answered the question overall, I hope that this helped. And until next time.