Answer the following questions for the elements from cerium throu...

Question

Answer the following questions for the elements from cerium through lutetium:
a. Are they metals or nonmetals?
b. To what general class of elements do they belong?
c. What subshell is being filled by electrons in these elements?

Accepted Answer

Welcome back, everyone. We need to consider the elements lo hand through mercury and identify one if the elements are non metals, metalloid or metals, two, the class of elements that they belong. And three, the subs shell being filled with electrons in the series below, we have our periodic table and then choices A through D where choice A states that the classification of our series is that they are non metals, they are also representative elements. And then for part three, that the two P subs shell is being filled for choice B it states that they are metals, they are also considered transition metals. And then the subs shell being filled is the five D subs shell. In choice C it states that our series are metals and also considered representative elements. And the subs shell being filled is a six S subs shell. And then choice D states that our series are metalloid also considered transition metals. And the subs shell being filled as the four F subs shell. Now we're going to begin by outlining where our series is located on our periodic table. So we recall that is represented by the symbol L capital L lowercase U and is at atomic number 71. Whereas mercury we recall is represented by the symbol capital H lowercase G at atomic number 80. So our series goes through atomic number 71 through atomic number 80 where we land on mercury. And next, we want to recognize that based on our position of luteum, which is located in our lanthanide series of our F block on our periodic table, as well as based on our elements hafnium at atomic number 72 through atomic number 80 where we have mercury, which are located in our D block known as containing our transition metals. We would observe that based on these two positions of our series, we would confirm that they are all considered to be medals. And that is our first answer so far, our series of considered metals. Now we need to determine part two, the class of elements that they belong. Now recall that our transition metals will include our D block elements as well as our lanthanide series and actinide series from our F block, which are also considered transition metals. So our class of elements state that our series Luther mercury belongs is transition metals. We can differentiate our transition metals from our representative elements, which are our second class of elements which consist of groups one A two a and then three a through groups eight A. So these would be our representative elements which have different properties. Then our transition elements. So transition metals are our second answer so far as the class of our series. And now moving on to part three, we need to determine the subs shell being filled with electrons in the series. So we're going to write out the electron configuration for each of our atoms in our series. Beginning with luteum, we're going to use the shorthand electron configuration beginning with the noble gas that proceeds atomic number 71 on our periodic table. We're going to place that in brackets and that would be xenon at atomic number 54. Now, moving forward in our configuration to land on loam, we're going to need to pass through our six S subs shell in which we would fill in a total of two electrons as our exponent moving forward in our configuration. Recall that at the sixth energy level or sixth period of our periodic table, which our period numbers are the rows going across our periodic table. We begin our D block at the fifth energy level. So we would have five D one in which we would only fill in one electron to get through our atomic number 57 for the atom lanthanum. And then moving on through atomic numbers 58 through 71 where we land on luteum, we're going to fill our four or rather our F block, which begins at the fourth energy level. So we would have four F and we would fill in all 14 electrons that it can contain. And this would complete our electron configuration for luteum xenon and brackets six S 24, 14 5 D one. Now moving forward to our next atom in our series, we have our element being half num represented by its chemical symbol capital H lowercase F. So this should be a lower case f beginning with the noble gas that precedes Hani on our periodic table. We again begin with xenon and we're going to begin with passing through our six S subs shell where we fill in two electrons, we also pass through our four F subs shell where we fill in for electrons. And we would hit at the sixth energy level, our fifth energy level of our D subs shell in which to count, we would count for a total of two units to pass through lanthanum on our product table until we get to half N M in our D block as well, which is our second element listed. So that would give us five D two. So next, moving forward to our next element in our series, we have our atom tantalum represented by its chemical symbol capital T lowercase A in which we begin with. Again, the preceding noble gas being xenon, we have just as before six S two and then four F 14. And then getting to our fifth energy level of our D subs shell across the sixth period of our periodic table. We recognize that tantalum is our third element listed in our D subs shell. So we would have five D three where we fill in for three electrons to land on tantalum moving forward in our configuration. Our next element in our series is our element represented as tungsten by its chemical symbol. Capital W So beginning with its electron shorthand electron configuration, finding the noble gas that proceeds, we have xenon. Again, again, we pass through our six S subs shell filling in for two electrons as well as our four F subs shell filling in for 14 electrons. And then we observe that at the fifth energy level of our D subs shell, we would count for a total of one, sorry, 1234 units to land on our fourth element in our D block at the fifth energy level being tungsten. So we would fill in four electrons. Now, instead of writing out the rest of our remaining configurations, we can observe so far that we have an increase in electrons being filled in our five D subs shell, we go from lute's configuration where we have five D one to half NMS configuration of five D two to tantalum's configuration of five D three to tungsten's configuration with five D four. And so therefore, we can say that our five D subs shell is being filled with electrons within our series. So looking at our answer choices based on all of what we've determined so far we would automatically rule out choice A which states that our series is non meals, as well as representative elements. We know that our series are metals as well as the classification being transition metals. We can also rule out choice D because it states that they are metalloid and then we can also rule out choice C because it states that our series are representative elements and that our six S subs shell is being filled. So after everything, we've ruled out, this leaves us with choice B which correctly states our type of elements in our series being metals, our classification of our elements in our series being transition metals and the subs shell being filled with electrons being the five D subs shell, which we do agree with. So B is our final answer for parts one through three of our series luteum through mercury. I hope that this made sense and let us know if you have any questions.

Answer the following questions for the elements from cerium through lutetium: a. Are they metals or nonmetals? b. To what general class of elements do they belong? c. What subshell is being filled by electrons in these elements?

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