[An atomic number puzzle]

earth elements after 63), until ... earths,. Sum of atomic digits —. 3 = 4/ electron number. Now this admittedly strange ... (but not necessarily of...
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LETTERS To the Editor: Given an atomic number alone and no other data, the number of valence electrons or number of electrons in the outer orbits can he found by adding the digits of the atomic number, according to a few simple rules. These rules provide a simple and easy method for remembering the valence of all the elements in the Periodic Tahle. They also have usefulness in deciding a t once the group or subgroup character and number of the particular given element. These "rnles" are: 1. Addition rule Add the digits comprising a given atomic number; keep adding the digits of the new sum, if it is more than 10 (excepting rare earth elements after 63), until the final sum is a single figure.

Correspondence rule If the final figure obtained is unlike the given atomic number with respect to being odd or even, suhtract 1. 2.

Four-differential rule With all atomic numhers after 70 add 4 and proceed as above.

3.

4. Rule for atoms before 10

With all numhers before 10 (except atomic numbers 1 and 2) suhtract 2 from the given atomic number. 5. Rule for CO-Ni groups The corresponding sums 9 and 10 denote the total number of electrons in the outer orbits of the Co-Ni groups, respectively, hence what can he called the group number of these atoms. Exemfilm: Determine the maximum valence (electrons in outer orbits) of atomic numbers 43.13.85.46. etc. 43 sum of 4 13 sum of 1

85 sum of 4

46 sum of 4

+3 = 7 +3 =4 + 8 + 5 = 17 1 +7 = 8

+ 6 = 10

7 and 43 are both odd, hence maximum valence is 7. 13 is odd and 4 is even. They are unlike. Therefore subtract 1 and obtain valence of 3 . Since this number comes after 70, first add 4, then add further according to rule, get 8 and 85 which are unlike. Therefore subtract 1 and get valence of 7 Sum 10 according to rule denotes the Ni group, since 10 and 46 are both even.

6. Rule for finding the group or subgroup character of a

given atomic number With a few exceptions among the helium and neon periods i t is found that if the sum of the digits is unlike the atomic number with respect to being odd or even the atom in question is a subgroup atom belonging on the right side of the Periodic Tahle, whereas if the sum of the digits is similar to the atomic number with respect to being odd or even the atom in question is a main-group atom found on the left side of the Periodic Table. With the rare earths the sum of the di&s of the

atomic numbers (carried to two figures after 63) equals the sum of the electrons in the outermost orbits, namely, the 4f 5d 6s orbits. Now the 5d 6s orbits are the valence orbits and amount to 3 for all rare earth atoms. Hence the 4f orbit for each rare earth atom is ohtained by subtracting 3 in each case; that is, for rare earths,

+

+ +

Sum of atomic digits

-3

=

41electron number

Now this admittedly strange result produces the question, why should summation of digits of a given atomic number indicate the number of electrons in the outer orbits of that particular atom? A clue to the above may be discovered in the fact that the summation process is the same as division by 9. Example: Take any given number such as 173=1+7+3=11; 1+1 =2

- - - Quotient 9

+R =2

Thus the sum of the digits of any given numher added to a single term is equal to the remainder ohtained by dividing the given number by 9. Hence adding atomic numbers to a single term is equivalent to dividing each atomic number by 9. The important thing is that this correspondence between remainders and sums of digits occurs only with 9 as a divisor or factor in the divisor. If any other divisor is used there is no such correspondence; Further, the Co-Ni groups and rare earths after 63 conform with the rule if the quotient is put a t one less than the arithmetical value. Thus, Rh has 9 outer electrons. Example: % = 4 + R = 9 9

We have here what may be called "an atomic number puzzle." I t would he interesting to have your readers' solution. NATHANPORITZ TEE PANAMA CANAL 'WaSn1NG.rON. D. C.

To the Editor: While you are correcting an interpretation of the first order reaction law might i t not be well to correct an even more flagrant misstatement of fundamental principle that appears on p. 176 of the April issue? One reads : "The 1-sugars are the mirror images of the d-sugars (but not necessarily of opposite rotation)." Apparently the author is confusing the point that all members of the d-series are not necessarily dextrorotatory. Enantiomorphs, even in sugar chemistry, still have rotations that are equal in magnitude but of opposite sign. M. L. WOLFROM STAT=UNIYSPSIN C o ~ u r n u sOHIO ,

THB OHIO