the Periodic Law of the Chemical Elements. 



101 



10 are higher than the mean, whilst of the 9 remaining 

 values, 5 occur in groups 1, 2, and 3, and of these five, four 

 are lower than the mean, so that the high values of c occur 

 chiefly with elements belonging to the higher groups (viz. 5, 

 6, and 7), while the low values occur chiefly with elements 

 belonging to the lower groups (viz. 1, 2, and 3). Four of 

 the values furthest from the mean occur in group 4 ; of these 

 two (Ti and Gre) are high, and two (0 and Si) are low values 

 of c. 



It will be observed that the elements of group 8 have 

 not been included in the above table on account of the dif- 

 ficulty of assigning a value to v in their case. This difficulty, 

 however, is removed if v represents, not the valency, but the 

 numerical order of the elements in the series to which they 

 respectively belong, in which case c has the following values 

 in group 8 : — 



m 



=5. 



m- 



=12. 



w=19. 



m- 



=26. 



v= 8. 



Fe. 



.7*14 



Eu. 



..6-98 





Os. 



..6-69 



Co. 



.7-34 



Eh. 



..6-94 



Wanting. 



Ir. 



..6-64 



t/= 9. 



m. 



.7-18 



Pd. 



.7-00 





Pt. 



..6-66 



^=10. 



For potassium m = 5, so that the equation A=6'6(m + \/v) 

 becomes A =6*6(5 +\/v) for the elements of the potassium 

 series ; and since the common difference in the arithmetical 

 series 5, 8J, 12, 15J &c. is 3^, therefore for any element in 

 series IV. or upwards 



A=6-6[5 + 3'5(a-4)+v / ^] = 6-6(3-5a-9 + v / ^) J 



where a=the number of the series to which the element 

 belongs, and v the numerical order of the element in its own 

 series. 



The equation A= c{m+ s/v) becomes A=c(m+1) in the 

 case of elements belonging to the first group. So that for 

 potassium and silver, 



(1) cK + l) = 39, and (2) c(m 7 + l) = 107'7. 

 Let x represent the common difference in the arithmetical 

 series m 4 , m 5 , ra 6 , m 7 , &c. corresponding to the 4th, 5th, 6th, 

 7th, &c. series of elements. 



Then m 7 = m 4 + 3x (3). 



