Chemical queries. Especially for introductory chemistry teachers

chemical queries. San Diego Stole College. San Dieao. California. Question. I would like t,o know what causes the nucleus of an atom to give off parti...
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chemical queries San Diego Stole College

. . . especially

San Dieao. California

for

introductory chemistry teachers

Question I would like t,oknow what causes the nucleus of an atom to give off particles or become unstable when extra, neutrons are added. 1.131 is radiaaotive while 1-127 is stable. Why do four extra neutrons make 1-131 unstable? Is it the increase in the neutron to proton ratio? If so, haw does one account for the instability of technetium? This element has no stable isotopes, yet the neutron to proton ratio does not seem to me to he so very high. Is there more to stability than only the neutron ta proton ratio? Answer

by Gregory R. Ch,oppin, T h e Florida State University, Tallahassee, Florida Yes, the neutron to proton ratio ( n / p ) is only part of the story. For a more complete answer, we must also consider the balance between attractive and repulsive forces in the nucleus, the shell model of the nucleus, the even and odd numbers of protons and of neutrons in the nucleus, and the existence of stable adjacent isobars. Empirically, it is usually found that stability in light nuclei is related to a neutron to moton (nlv)ratio near unity. Examples are '$N, 'ZO, $ ~ e E:, Ca (with n / p = 1.2) ; all of these 1.0) and ':0, :iNe, itCa (with n / p are stable. On the other hand, ::Na, ;!A1 and i;Ca are not stable; and the n / p ratio is unity, or near unity, for each of these. I t is obvious that the n / p ratio is a useful indicator but no explanation of a complex situation. Moreover, it is also true that as the atomic number increases, stability is associated with n / p values progressively greater thau unity-e.g., the stable isotope ';:Bi has an n / p of 1.5. To gain insight into the origin of radioactivity (i.e., nuclear instability), let us consider the forces present in the nucleus. The macroscopic world provides ready evidence of two types of force--gravitational and coulombic. Gravitational force is too weak to play a n y role in the nucleus; however, since nuclei contain protons at very short distances (-10-l2 to 10-l3 cm), the coulombic force must exert a considerable repulsion. The mere existence of nuclei is convincing evidence that an attractive force of equal or greater magnitude must also be present to offset the coulombic repulsion. This attractive force has been named the nuclear force. Unfortunately, it can only be studied in nuclei since it opcrates only over a very short distance (- 10-la cm). Over the past three decades considerable progress has been made in understanding it (though much remains to be learned). One important property of the nuclear

force is its charge independence; the attraction between two protons, between two neutrons, or between a proton and a neutron is essentially the same. Of course, between the first pair, a coulombic repulsion is also present. With this model of two opposed forces, one repulsive and one attractive, we can begin to understand nuclear stability and instability. Stability, obviously, reflects a situation where the attractive force at least balances the repulsive force. In fact, the half-life of radioactive nuclei reflects the degree of imbalance. If the attractive force is weaker thau the repulsive force in a nucleus, the nuclear instability can be rectified by radioactive decay which increases the former and/or decreases the latter. Either positron emission (0+ decay) or electron capture (E. C.) accomplishes this change. In ffNa,stability is achieved by the reaction For ':;Bi,

electron capture decay produces stability

The net effect of these changes is to reduce the number of protons by one, thereby reducing the coulombic repulsion, and to increase the number of neutrons by one, thereby increasing the nuclear attraction. Of course in nuclei in which the balance is far off, several such changes may be required before stability is achieved. From the n / p values for stable nuclei we learn that in lieht nuclei. balance is achieved when the number of neutrons is approximately equal to the number of protons. As the number of protons increase in heavier nuclei, relatively more neutrons must be present and the value of n / p related to stability increases above unity. However, this explanation of stability is insufficient, it does not account for nuclear instability when n / p values are high. Such instability is evidenced by negatron (0-) decay. For example ~

-

::Na ,,Pb

208

--

::Mg %Bi

+ 8+ 8-

The net effect of 0- decay is to increase by one the number of protons and to decrease similarly the number of neutrons. To understand this we turn to the shell model of the Volume

47, Number I , January 1970

/

73

nucleus. Neutrons and protons occupy independent sets of energy levels in a nucleus. If the number of neutrons exceeds that of the number of protons by very much, there will be a considerable energy difference between the highest energy level occupied by neutrons and that occupied by protons, as shown schematically

This nucleus can lower its energy difference, compare AEz with MI,by conversion of a neutron to a proton, with electron emission.

the most stable of technetium isotopes, as expected, but not as stable as the neighboring isobars which have even atomic numbers. Question Recently, I electrolyzed molten potassium nitrite, using stainless steel electrodes. A colorless gas was evolved at the cathode and a brown solid formed at the anode, which fell to the bottom of the vessel. After separation from the residual ppotsssium nitrite by dissolving it in water, the brown solid was examined. It dissolved in hydrochloric acid to give a yellow solution which gave a positive test for FeS+ions. Is the following explanation (which requires migration of negative ions to the negative electrode) tenable? 6e2Fe Fe

We still have not covered all the complexities. Why are :INa, !A : 1 or i X a unstable? Obviously, we can not argue that the n/p value is SO high that AE is too large since ::Na, etc. is stable or so low that the repulsive force predominates since B:Ne, etc., is stable. The simple fact is that nuclei gain considerable stability when even numbers of neutrons and/or protons are present. When an even number is present, they exist in pairs and the resultant ''pairing euergy" liberated provides an extra degree of stability to the nucleus. Of the stable nuclei, less than 10 have odd (neutron)odd (proton) character, slightly more than 100 have odd-even or even-odd character while more than 160 have even-even character. I n the reactions ::Na

%4l

-

+ENe

ZMg

+ @+

+ @+

nuclei with odd numbers of neutrons and of protons are transforming to nuclei with even numbers. The effect of even character is also evident in the number of stable isotopes of elements. Elements of even atomic number, in general, have 3 or more stable isotopes whereas odd atomic number elements only have 1 or 2 stable isotopes. Finally, let us consider the problem of technetium which has no stable isotopes. First, we note that it is an odd atomic number element (Z = 43) so it would have only 1 or 2 stable isotopes at best. Niobium only has one, ::Nb (n/p = 1.27) as also does rhodium, 'ZRh (n/p = 1.29). Using these examples, we might expect that the even-odd nucleii iiTc (n/p = 1.25) and i;Tc (n/p = 1.30) would be stable while the oddodd nucleus :tTc would have a lower stability. However, an additional postulate of nuclear stability states that adjacent isobars' will not both be stable. Essentially, this says that two isobars are very unlikely to have the same mass (and, hence, energy) and the one of heavier mass (higher energy) is unstable with respect to decay to the lighter. Therefore, :Zc, ETC and :;Tc can only be stable if they do not have stable isobars. But ::Mo, z:Mo, :tRu, ::Ru are all stable; presumably, these are more stable than their Tc isobars due to the odd-even relation. I t is significant that both ::Tc and jiTc have half-lives of over a million years and j:Tc of over 200,000 years whereas all the other known technetium isotopes have half-lives of a few days or less. Obviously, ::Tc, :To and ::Tc are 74

/

Journal of Chemical Education

-

+ +

+

6e2Fe + 2Fe" (at the anode) N2 4OZ(at the cethode) 2 N W + NZ 2Fes+ 402(net reaction) 2KNOn NX Fez01 KIO (overall reaction)

+ ZNOZ-

+ + +

+

+

Also wuld you cite a few references on the suhjert of the electrolysis of molten salb using reactive electrodes? Answer

by J . P. Young, Oak ~ i d i Natiaal e Laboratory, Oak Ridge, Tennessee Although the net and overall reactions might describe the ultimate chemical process (which would iuclude the product separation scheme), the suggested mechanism does not necessarily describe the electrode reactions as such. With any system, and molten salts are no exception, a definitive description of electrode processes can become a very interesting and involved study. It might help in this case, for example, to compartmentalize the KNO, melt in sintered glass tubes in order to study the reactions and reaction products a t each electrode separately. The implications of the parenthetical statement in the question require correction. I n molten KN02, one can be sure, there is no need to be concerned about migrations of negatively charged ions to the negative electrode. I n molten KN02, there is an abundance of anions throughout the melt, including the region near each electrode. Therefore, negative ions might very well be involved in a reduction process at the cathode. I n aqueous solutions, for example it has been well established that negatively charged species can be reduced at a cathode even when their concentration is as low as 10-8 M. Of course, merely because the NO2- ions are there. near the cathode, does not mean that they are therefore the reactive species. The cathodic process could involve some solvated NOr- species, positively charged, or neutrd. Or NO+ might be involved, in which case the colorless gas might be NO, rather than Nr. I n addition to the anodic reaction suggested in the question, two other possibilities should be mentioned. The oxidation of the metallic iron to the Fez+state, or the oxidation of the nitrite ion NO%--NO%

+ e-

If the latter occurs, the NOI then might attack the iron of the anode. This could be checked independently by bubbling NOr, from a cylinder or other source, over a

'

Isobars: nuclei with the same total number of neutrons and protons; loosely, isotopic species of different elements that happen to have ipproximatel~equal atomic weights.

&ce of iron or stainless steel immersed in molten I