Reproduced below is a copy of the periodic chart. Noble

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1 Goldilocks The eriodic hart Reproduced below is a copy of the periodic chart. A A V V V V V A VA VA VA oble VA ses.0. a..0. s. r () e.0. 0 a 0.0 r. a. Ra.0 c. Y. *La. Ac.0 Ti. 0 Zr. f. 0 Rf () V 0. b. Ta 0. 0 Db () r.00 Mo. W. 0 g () Mn. Tc. Re. 0 h () e. Ru 0.0 s 0. 0 s () o. Rh 0. r. 0 M t () i. d 0. t.0 0 () u. Ag 0. Au. () 0 Zn. d. 0 g 00. () i.0. 0 n. b 0. () b e..0 o (0) () r.0.0 (0) e.00 0 e 0. Ar. r.0 Xe. Rn () () *Lanthanide series Actinide series e 0. 0 Th. r 0. a.0 0 d. U.0 m () p.0 m 0. u () Eu. Am () Gd. m () Tb. k () Dy.0 f () o. Es () Er. 00 m () Tm. 0 Md () 0 Yb.0 0 o () Lu. 0 Lr () A look at this chart shows that there is a space (a square) for each element, each space containing three entries: an integer the "atomic number" for that element; a one or two letter "symbol" the initial letter capitalized; and a decimal number the "atomic mass" for that element. You may also have noticed sections--we will call them "blocks" within the chart, which we will name as shown here: e e e 0 e a s 0 a c Ti V r Mn e o i u 0 Zn i p e r Ar r s r a Y *La 0 Zr f b Ta Mo W d Tc Re Ru s Rh r d t Ag Au d 0 g 0 n b b o Xe Rn r Ra Ac 0 Rf 0 Db 0 g 0 h 0 s 0 M t 0 *Lanthanide series Actinide series e 0 Th r a 0 d U m p m u Eu Am f Gd m Tb k Dy f o Es Er 00 m Tm 0 Md 0 Yb 0 o Lu 0 Lr

2 Goldilocks: The eriodic hart p. hemists view the atom as composed of a tiny nucleus containing protons and neutrons, with electrons filling the rest of the space around the nucleus. (This is oversimplified, as any physicist would be happy to explain, but it is effective for our needs to simplify things this way) The charge is the same in magnitude on protons and electrons but opposite in sign with protons defined as + and electrons as. eutrons are neutral. hemists use the symbols, p +, e, and n, for these subatomic particles. l atoms of a given element have something in common the number of protons in the nucleus. The number of electrons in a neutral atom, then, must equal the number of protons in the nucleus. The number of neutrons can vary, giving rise to "isotopes"; a given element might have only one or as many as 0 or more stable isotopes, and abundances for these isotopes can vary widely. dividual isotopes are identified by their symbol (representing the proton count) and their "mass number", the total number of protons and neutrons in the nucleus. The atomic mass is an average, weighted for abundance, of the masses of the stable isotopes of that element. f an element has no stable isotopes, then the mass number for its most stable (least unstable) isotope is given, in parentheses. Thus atoms of copper (element, symbol u or " u") all have protons in their nuclei, and if neutral, electrons surrounding each nucleus. An atom of copper with more than electrons would be negative, an extra for each extra electron, and with fewer than electrons, it would be positive, an extra + for each missing electron. When you looked at the periodic chart that was shown at the first of this discussion, you may have noticed two dark lines, the first a stair-step line across the p-block, from upper left to lower right, and the other a vertical line within the p-block separating the far right column from atoms in the other five columns of that block. There are two "Goldilocks diagrams" included in the periodic chart, one associated with each line. A. The oble ses Elements in the last column of the p-block are called the "oble ses", because of the inert, non-reactive nature of these elements. We speak of these elements as "inert" or "non-reactive" because atoms of these elements do not usually or easily gain or lose or share electrons when allowed the chance to react with other elements. This leads us to a mostly reliable and always simple predictor for the chemical behavior of elements not in this column: atoms of elements located just before a noble gas will tend to gain enough electrons so as to have the same number of electrons as that noble gas atom would have when neutral. atoms of elements located just after a noble gas will tend to lose enough electrons so as to have the same number of electrons as that noble gas would have when neutral. an you see the Goldilocks pattern? oms in the noble gas column are "just right". oms of elements in columns just before have "too few" electrons, while those in columns just after have "too many". Well, maybe you don't see this yet, because the idea of "before" and "after" is hard to see from the periodic chart as we normally display it. o let's try to show the periodic chart in another way, an admittedly silly but useful way. Let's put the noble gas column in the middle of the page, and show, all on a line, the elements with lower atomic numbers and the elements with higher atomic numbers, on the left and right, respectively. urther, and with only a little justification right now, let's for the moment leave out elements in the d- and f-blocks, and just show the s- and p-block elements in our silly chart. [ doing this, we are

3 Goldilocks: The eriodic hart p. concentrating our attention on what chemists call the "main group elements", that is, elements in the s- and p-blocks.] oble ses A A e A VA VA VA VA e 0 e A a A e A VA VA VA VA i a i Ar 0 a e r 0 a e r r r 0 n b r 0 n b Xe s a b o s a b o Rn r Ra r Ra Thus we think of the noble gases as having a special and desirable number of electrons. And so, if other elements are to be, well, can we say "happy"?, they'd have to gain or lose electrons to get to that number: or or r or, each located one space before the noble gas column, would tend to gain one electron, forming or or r or, while or a or or, each located one space past the noble gas column (at the first of the next row), would tend to lose one electron, forming + or a + or + or + and from this sort of analysis, we can begin to make predictions: if a were to react with (which we note in passing exists in nature as the diatomic molecules, ), we would expect to see a become a + while becomes, producing a product with the formula a. We could go further, and guess the behavior for elements in the columns just beyond these two: elements in the column that begins with oxygen ( ) would be tend to gain two electrons, forming, or, etc, while elements in the column that begins with beryllium ( e) would tend to lose two electrons, forming e +, +, etc, and so forth to other columns beyond. redictions that would now be possible would include: that the product of the reaction of with (or, a source of atoms) would produce + and ions, which would combine as, while the reaction of with might be expected to produce, containing + and ions.

4 Goldilocks: The eriodic hart p. Extending this would allow us to predict that: the reaction of with (which natures provides in the form of ) would be predicted to yield, containing + and ions, while a reacting with (which nature provides as ) would be predicted to give a, containing a + and ions. ow, these predictions do not always work perfectly. or example, a and (as ) react to give a mixture of two compounds, one as we expected, a, but the other with an unexpected formula, a. We cannot, therefore, be absolutely certain that predictions of this sort will always be proven right in the realm of experimental outcome, but we can be confident that these predictions will tell us how best to place our bets on what will happen, and we would usually win those bets. The key to all of this is the ability of atoms to match up with other atoms, some to gain electrons that the others lose, creating species that are "isoelectronic" with ( = have the same number of electrons, doing, roughly, the same things as) atoms of the nearest noble gas. ========== efore we go any farther, let's express these same predictions using the language of atom symbols, as found in "dot diagrams". To do this, we need to take note of three things: () neutral atoms in the s- and p-blocks have electrons only in s- and p-orbitals; () there is one orbital in each s-subshell and three orbitals in each p-subshell; () therefore, other than and e, all s- and p-block elements have outer shell s- and p- orbitals in each atom. Thus, if we display outer shell electrons in the outer shell of an atom as dots in the space around the symbol of that atom, and if we use the spaces left, right, above, and below the symbol as representing the s- and p-orbitals of the atom, we have the capacity to display orbitals and up to electrons around each atom. or s- and p-block atoms, electrons fill these orbitals in this order (see und's rule, elsewhere): one, then two, in the s-orbital, then one in each p-orbital, then a second in each p-orbital, until all the outer shell electrons are used. Thus: while: a + yields a compound a and + yields a compound + and + yields a compound and + + yields a compound + + and a + yields a compound a and +

5 Goldilocks: The eriodic hart p. This works quite well, as long as we choose our atoms so that one is just before and one just after the noble gas column. There are, however, some problems we run into. irst, if we go far enough with this, say to carbon ( ) which is four places away from a noble gas, we reach the prediction that atoms of carbon should lose electrons but also that they should gain electrons. What do we make of this? econd, what if we pick both of our atoms on the same side of the noble gas column, or what if we pick atoms of just one element. an stable chemical substances be made from those choices? The initial clue is in the existence of species like, and the explanation begins with considering what atoms can do if they have only atoms like themselves with which to interact. the case of atoms, each atom tends to gain one electron, but no atom has any more attraction for electrons than does any other atom. The transfer of electrons to make ions, some positive and some negative, is then not possible. What atoms do, in this case, is to share electrons. Each atom gives one electron to a "shared pair" of electrons, resulting in a bond (a "covalent" bond). Using our "dot diagram" pictures, this would be: + yields the molecule which we write as and which we sometimes symbolize as: ote that things can get much more complex, since each chlorine atom could share even more electrons, but in simple terms, we have seen two possibilities emerge for atoms:. take an electron from another atom, giving the ion, or. share a pair of electrons with another atom, the atom providing one of the two electrons in the shared pair. a rule, chlorine atoms tend to take electrons from atoms on the other side of the noble gas column, and share electrons with atoms on the same side. The twist in this emerges when we notice, as we suggested above, that the chlorine atom has more electrons it could share. a rule, chlorine will share other electrons in pairs, and a look at the dot diagram of chlorine suggests that it could share up to seven electrons before running out. What is interesting is that if you go back and look at the atomic number of chlorine, you will see that chlorine has electrons when it is neutral. This line of thinking leads us to conclude that chlorine can use up to seven of those electrons, but no more. We will not provide any justification of this here, but this conclusion is solid. A closer look at the original periodic chart show that there is a Roman numeral at the top of each column (some periodic charts will have different numbers in those locations, but we will use these Roman numbers here). or these main group elements, that is, for atoms in the s- and p-blocks, this number runs from for the lithium () column to in the fluorine () column. Do you see it? That number is the maximum number of electrons that atoms in that column could lose or share. The atom could lose or share less, but not more.

6 Goldilocks: The eriodic hart p. The chart is called the "periodic chart" because periodically you get back to the same number of "usable" electrons. or now, we will call those electrons "outer shell" electrons, noting as we do that some textbooks refer to these as "valence" or "valence shell" electrons. And now, can you guess what the number would be for the noble gases, if we were to put a number at the top of that column. t would be the number, wouldn't it? Thus noble gas atoms (with the exception of e = helium) have outer shell electrons. ther atoms (from among the "main group" elements) may lose (or share) any or all of their outer shell electrons or gain electrons up to the point at which they have outer shell electrons under their control. t is from this latter point that we get the name "octet rule", to indicate that atoms tend to gain or lose electrons so that they have outer shell electrons. A stickler would insist that for and e, this is a "duet rule". (ote that, for reasons to be mentioned later, it is possible for elements beyond e to have more than electrons under their control, but for the moment, this more simplified viewpoint is useful.) There is language we use for this, and it is the language of "oxidation state" (or "oxidation number"--the two terms are used interchangeably). ositive oxidation states come from losing outer shell electrons and negative oxidation states come from gaining, that is, adding electrons to the outer shell. There is a bit more to this, but for now, you need only to note that this means that the oxidation states possible for a main group atom range from the group number (a positive oxidation state) to the group number minus (a negative oxidation state). Let's try some examples: a. consider phosphorus = -- this element is in group VA, so its group number is. This gives it access to oxidation states ranging from + to. ur prediction would be the same for or, etc b. consider gallium = -- the element is in group A, so the group number is and we could expect it to have possible oxidation states ranging from + to. c. consider bromine = r -- in group VA, so the group number is and the range of possible oxidation states is from + to. ow are you doing so far? Are you able to make predictions for other atoms within the main group elements? f so, then we are ready to continue, with an example you've already considered. What do you expect for carbon =? You'd expect a range of possible oxidation numbers from + to, and that is just what the silly chart above led us to suspect. ======= this point, there is some fine-tuning we need to do in order to identify the "probable" oxidation states expected for main group elements. The issue is this: Any oxidation state is possible, even non-integral ones, as long as each such value is within the "range of possibles" as determined above; but some values within that range are more likely than others. t is probably best, right now, just to jump in, by giving the pattern: a. Zero is always probable, but only for the elemental form of that atom. Thus the atoms in,,,, and, all have an oxidation state of zero, as do the atoms in a(s), g(l), and e(s), etc.

7 Goldilocks: The eriodic hart p. b. When two or more elements are present (as in a compound like or in an ion like ) one element is expected to have a positive oxidation state and another (the other, if atoms of only two elements are present) is expected to have a negative oxidation state. c. that case, elements closer to fluorine ( ) on the periodic chart are expected to take on a negative oxidation state, while elements further from fluorine (closer to s) should take on positive oxidation states. d. And whatever the oxidation states are for the atoms within the species, those oxidation states must add up to give the charge on the species; to zero if the species is neutral. There is a more formal basis for this, involving something called "electronegativity", but this will do for the moment. ote, however, that we are suggesting yet another Goldilocks diagram emerging from the periodic chart. Do you remember the stair-step line? Looking only at main group elements and, for the moment, ignoring the noble gases, the stair-step line roughly marks the middle ground for what might be called "the tendency to be positive" vs "the tendency to be negative" in oxidation state. oble VA ses A A A VA VA VA e e 0 e a i Ar 0 a e r r r 0 n b Xe s a b o Rn ote that atoms below or to the left of the stair-step line are called "metals", while those to the right and above are called "non-metals", with and the noble gases included among the non-metals. Metals are expected to show only positive oxidation states, while non-metals may show + or oxidation states. this context, hydrogen is a special case. t is shown in two positions, and its tendency to be + or in oxidation state is the average of those positions, and thus more or less where carbon ( ) is. o let's continue the pattern of probable oxidation states: d. in compounds, will have either a + or a oxidation state, depending on what is the other atom present. With metals, should be ; with non-metals, +. e. atoms in columns A or A (other than ) are expected to have oxidation states = + or +, respectively, in compounds

8 Goldilocks: The eriodic hart p. f. for other atoms within this set (columns A to VA) the most likely positive oxidation states are (group #) or (group # ), while the most likely negative oxidation state is (group # ). ote that atoms below or to the left of the stair-step line are probably going to show positive oxidation states only, while those above or to the right could show positive or negative oxidation states. o let's do some exercises to see how all this works. Remember that we are predicting oxidation states and once we can do that well, we can use those predictions in turn to predict compound formulas. ============================ Exercise A. Give the range of possible oxidation states and then the probable oxidation states within that range for each of the following elements a. b. b c. 0a d. r ============================ Answers: a. is in column VA: range = + to probables = + or + R b. b is in column VA: range = + to probables = + or + only (below stairstep line, so only + probables) c. 0a is in column A: range = + to probables = + only (in compounds) d. r is in column VA: range = + to probables = + or + R ow did you do? an you see the patterns? f so, then let's try the next level. ============================ Exercise : redict the formula of a compound formed by reaction of these two elements e. + f. + g. + d. + ============================

9 Goldilocks: The eriodic hart p. Answers: e. + is closer to than is, so goes negative, while goes positive; or, the expected positive oxidation state is +, for, the expected negative oxidation state is ; to get a neutral compound, we need to find x and y, so that ( + ) x ( ) y is neutral; in this case x = y = works, so we expect to see as the formula of the compound. ther formula might be seen, but this is the one we place our bets on. TE--there is a small twist here. The oxygen was listed as, and that might have led you to try to incorporate the "" in your answer. is a source of, and it is the on which we need to focus. Don't be distracted by the way elements "huddle against the cold and dark of night", when only those atoms are present. Rather, focus on the probably oxidation states expected for such atoms and go from there. f. + reactions, we always expect to go to a negative oxidation state, so would go positive; thus is expected, while could be + or +. This would predict either or. TW TE: () you can get a first draft formula easily by using the value of each atom's oxidation number as a subscript on the other atom, thus ( + ) x ( ) y suggests or. () A little tip for cases for which an atom has two probable positive oxidation states from which to choose: "high in the periodic chart, the higher (more positive) one is more likely, while low in the periodic chart, the lower one is more likely. This fine tuning is very fitting for column A elements. or and, and are not observed; instead nature provides and. or (at the bottom of that column), we would strongly expect to see and not. g. + Don't be distracted by the subscripts; this is just and reacting. and are both above the stairstep line (= both are non-metals), so both would prefer to go negative, but they can't both do that. ince is closer to, we predict that goes negative and that goes positive. ur prediction of probables says goes + or +, while goes in either case. This gives either ( +) x( ) y or ( +) x( ) y ; using our tip from before, this gives or. We are higher on the periodic chart, so we might favor, but either is a good answer. d. + This is +, both are non-metals, but is closer to, so we predict + for and for ; the probables are, then, for, + or +, with + more probable, and in either case. This gives us ( + ) x ( ) y or ( + ) x ( ) y ; which leads us to or, after finding the simplest whole number ratios in each case. You may recognize these two formulas as "carbon dioxide" and "carbon monoxide", both expected, but with as the "more expected", consistent with the "higher in the periodic chart" guideline. ote that it is possible to have combinations of atoms, acting as a unit, but with a net non-zero charge, that is, it is possible to have polyatomic ions. such cases, it is useful to remember that the sum of the oxidation numbers of all the atoms present must equal the charge on that species. or neutral molecules or compounds, the oxidation numbers of the atoms present must add to zero. Thus, in a anion, they must add to.