Question

According to Doberiener, in triads, the atomic weight of central element is ____ that of first and third elements in that triads
A. Geometric mean
B. Arithmetic mean
C. Harmonic mean
D. None

Answer 1

Hint: In the early times, Dobereiner made an attempt to group elements that shared similar properties. He arranged the then known elements into four to five groups. Each group consisted of three elements resembling properties.

Complete answer
Dobereiner showed that when the three elements in a triad were written in the order of increasing atomic masses; the atomic mass of the middle element was roughly the average of the atomic masses of the other two elements, that is, the first and third elements. For example, take the triad consisting of lithium (Li), sodium (Na) and potassium (K) with the respective atomic masses \[6.9,{\text{ }}23.0{\text{ }}and{\text{ }}39.0.\] the sum of atomic masses of lithium and potassium is 46. If the average of these two is taken, it is the atomic mass of sodium, which is the middle element.
Similarly, calcium (\[atomic{\text{ }}mass{\text{ }} = {\text{ }}40.1\]), strontium (\[atomic{\text{ }}mass{\text{ }} = {\text{ }}87.6\]) and barium\[\left( {atomic{\text{ }}mass{\text{ }} = {\text{ }}137.3} \right)\] form Dobereiner’s triad. The atomic mass of strontium is the roughly equal to arithmetic mean of the masses of calcium and barium, which sum up to \[177.4.\]
Another example is chlorine (\[35.5\]), bromine (\[80\]) and iodine (\[126.9\]). They form a set of Dobereiner’s triads. The sum of atomic masses of chlorine and iodine is \[162.4\], half of which is approximately equal to atomic mass of bromine.

Hence, option “B” is correct

Note:
Döbereiner could identify only three triads from the elements known at that time. Hence, this system of classification into triads was not found to be useful.