In what way are the electron configurations of H, Li, Na, K, R...

Question

In what way are the electron configurations of H, Li, Na, K, Rb, and Cs similar?

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Welcome back everyone. We're told that for the element barium provide the four quantum numbers, N. L, M. L and M. S. That describe its outermost two electrons. So recall that our outermost electrons are our valence electrons. And so we're going to want to consider the shorthand or condensed electron configuration for barium by finding our noble gas group recall that's in group eight A. On the periodic table. And we would see that for the condensed electron configuration of barium, we find our noble gas which proceeds barium being our noble gas xenon. We place that in brackets and passing xenon. We go into period six of our periodic table where barium is located and we land into our S block where we count for two units to land on our atom barium. So we would have for the sixth period six of our energy level, our S block being R. S. Orbital. And we would account for two units which stands for the two electrons that allow us to land on our atom barium. And this would be our condensed configuration for barium. We're going to use this configuration to get our quantum numbers. So beginning with our value for N. We want to recall that N is our principal quantum number, which we should recall can range from one to infinity and describes the energy level of our orbital. So if we know that based on our configuration, we're in the s orbital at the sixth period. So the sixth energy level, we would say that our principal quantum number is an equal to six for barium. Next we want to find our quantum number for L. Which we should recall is our angular momentum quantum number and recall that this number ranges from zero to n minus one. So what we would have is our end value, which we stated is the sixth energy level for our s orbital minus one, which gives us five. So we would say that for barium we can have angular momentum quantum number values that range Up to the value five. But because we recognize that we are in the s orbital for barium, we would recall that when L is equal to a value of zero. This corresponds to the S sub shell continuing on. We can recall that when our L value equals one. This corresponds to our P sub shell when L is equal to two. This tells us that we are in R D sub shell and when L is equal to three, this tells us we're in our f sub shell. And so because we can see that we are in our s sub shell or orbital, we would say that L is equal to zero. So now moving forward we want our M L. And M. S values. So ml we want to recognize is our magnetic quantum number and this can range from negative L two positive L. And so because we stated that our barium is located in the S sub shell, we would recall that our S sub shell only consists of one orbital, which is why we can only fill in a maximum of two electrons. And if we know that our ml values range from negative L. Two positive L. And because we're in the S sub shell R. L. Value is equal to zero, then we would therefore deduce that our Ml value is also equal to just zero. So now we want to get our M. S. Value which we should recall represents our spin quantum number. And this refers to the spin of our electrons where our upward spin would correspond to a value of positive one half and a downward spin corresponds to a value of negative one half. And as we stated our configuration was xenon six S. To where we can see that for these two electrons we filled in in our single s orbital we have one electron with an upward spin And following huns rule, we're going to recall that we can complete filling this orbital by filling it in with a downward spin electron last or the opposite has been electron last. And so let's make a note to huns role here as a reminder. So because we have an upward and downward spin we would say that R. M. S. Or spin quantum number values is equal to positive one half. And we also have an M. S value equal to negative one half. And so for our final answers, we would confirm that our four quantum numbers for electrons of barium we'll say for the valence electrons of barium are going to be any equal to six L. Equal to zero M. L. equal to zero M. S. Equal to positive one half. So this covers one of our electrons. And then for our second electron we have our quantum numbers where N is also equal to six L. Is also equal to zero Ml is equal to zero, and M. S. Is now equal to negative one half. So the opposite this is our first electron. And so what's highlighted in yellow are our two final answers as our quantum numbers for the two electrons in the valence orbital of barium. I hope everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.

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In what way are the electron configurations of H, Li, Na, K, Rb, and Cs similar?

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The Electron Configuration: Quantum Numbers 연습 세트