An unidentified element is found to have an electron configuratio...

Question

An unidentified element is found to have an electron configuration by shell of 2 8 18 8 2. To what group and period does this element belong? Is the element a metal or a nonmetal? How many protons does an atom of the element have? What is the name of the element? Write its electron-dot symbol.

Accepted Answer

Welcome back everyone Consider the following electron configuration by shell of an unknown element with the following electrons where we have to in the first shell eight in the second shell 18 in the third shell 18 in the fourth show. And seven electrons in the fifth shell starting with part one of our prompt. We need to define which group and period in the periodic table. Our atom is a part of. So beginning with the first shell for part one, we're told that we have two electrons in this first show, so that should correspond to one single orbital in our s sub shell where we have a maximum of two electrons filled in. Now let's move on to our second shell, we're told that according to the prompt, we have eight electrons for our second show corresponding to one s sub shell filled in with two electrons. And then we would have our P sub shell with three orbital's Where we would fill in a total of six electrons because that is the maximum that it can hold. Now moving on to our 3rd Shell We have our according to the prompt, 18 electrons in this shell and that's going to correspond to the following where we have one s sub shell filled in with two electrons. Three P sub shells, sorry, three P orbital's a part of our P sub shell filled in with six electrons. And then we have a d sub shell which we should recall contains five orbital's And holds a maximum of 10 electrons. And notice that I'm representing the electrons by arrows with opposite spins. So we have 10 electrons in R. D sub shell, six electrons max in our p sub shell and two electrons max in our s sub shell. Now let's move on to the fourth shell or energy level in which we're told according to the prompt that we still have another 18 electrons a part of this energy level. So beginning with our S sub shell we have one s sub shell with two electrons. We have our three P orbital's filled in with six electrons and then we have our D orbital's which we have five of a part of our d sub shell filled in with maximum electrons to give us our 18 electrons. And that leaves us with our fifth shell or fifth energy level in which we're told according to the pump that we have seven electrons for our atom filled in this shell. And so we have one s orbital. And then we have our p orbital. And sorry, this was the orbital here. But we have our p orbital which are are three p orbital's filled in with a total of and I'll show you how to properly fill it in. So we would fill each orbital in with one electron first of the same spin and then pair our last two electrons last because we only need five electrons filled in. So we would have our fourth electron paired in the first orbital and the fifth electron paired in the second orbital leaving our third p orbital for the fifth shell with only one electron to give us our total of seven electrons. And based on these shells that we've outlined, we can say that therefore our configuration is going to be one S 22, S 22 P 63, S 23 P 63 D 10, four, S 24, P six. Just making more room after four P six. We have four D 10. Then we have our five US two and then five P. And then our exponent we have five because we only filled in five electrons in R P orbital of our seventh shell for this atom. And so this will correspond for our first answer for part one being an element that is at period five and Group seven A. We know that because when we reach our fifth energy level where our P block is, we would count for five units and land on a specific atom that is within period five, which is our rows across the periodic table and then Down Group seven a. So this would be our first two answers for part one of our prompt. Now moving on to part two of our prompt, we need to determine whether the element is a metal or a non metal. So because we know that our element is a group seven a adam, which in our understanding we would see that it's a halogen Recall that group seven is our halogen group. We can understand that based on this position are halogen group would be considered atoms that are all non metals. And that would be our answer for part two. So now moving on to part three determine the number of protons in an atom of this element. Let's recall that for neutral atoms only our number of protons equal our number of electrons. So in this case we've determined that according to our configuration and also according to the electrons for each shell given in the prompt, we can add the sum up to get our number of electrons. We can add the exponents in our configuration to get our number of electrons or by counting the arrows we filled in for each of our shells in the orbital's. But in this case let's make it simple on ourselves. We'll take our sum of electrons equal to two plus eight plus 18 plus 18 plus seven. And this is going to give us our total electrons equal to a value of 53. And because we're not told that we have a charged atom, this means that we have a neutral atom And so therefore we have 53 electrons which also corresponds to 53 protons. So this would be our answer for part three. And now let's move on to part four which is to provide the identity of the element. So as I stated earlier, once we get to our p block of our valence shell, which is our fifth shell, we filled in a total of five electrons counting five units to land on a specific atom which is in Group seven A. So because we determined that our number of protons is 53 protons for our answer in part, three that tells us that we also can assume or understand that the atomic number represented by Z is equal to and beyond our electron configuration that we determined in part one. We can also determine that by looking at the number 53 on our periodic table, we would find the atom iodine with the following symbol and the atomic number 53 above it. And so the identity of our element is going to be iodine and this would be our answer for part four. So now moving on to part five, we need to draw the electron dot symbol of this element, recall that our electron dot symbol consists of our valence electrons for our atom. And because we know that iodine is in or is a group seven a halogen. This tells us that we therefore have seven valence electrons for this atom. So drawing out its electron dot structure or symbol rather, we have the symbol of our atom which on the periodic table for iodine is just a capital I. And then we have our seven valence electrons filled around and we would fill it in as follows where we have one single electron around each part of iodine counting for four electrons. And then we pair them up so that we have and seven, which leaves one single lone electron on one side of the ID in here. And this would be our answer for part five of the prompt with the electron dot symbol for Dean. So everything highlighted in yellow represents our final answers to complete this example corresponding to choice A and the multiple choice as the correct answer. So I hope that this made sense and let us know if you have any questions.

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An unidentified element is found to have an electron configuration by shell of 2 8 18 8 2. To what group and period does this element belong? Is the element a metal or a nonmetal? How many protons does an atom of the element have? What is the name of the element? Write its electron-dot symbol.

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