CBSE Class 10 Science 2022 · Paper 31/6/1 · Question 8

(a) Döbereiner arranged elements with similar properties in groups of three, called triads. When the three are written in order of increasing atomic mass, the atomic mass of the middle element is roughly the average of the other two; for Li ($6.9$), Na ($23.0$) and K ($39.0$), $\frac{6.9+39.0}{2}=22.95\approx 23.0$. (b) In NCERT's account, Döbereiner could identify only three triads from the elements known at that time, so most elements could not be classified in this way and the system was not found useful. (c)(i) X has valency $3$ and its oxide matches the general formula ${\text{R}}_2{\text{O}}_3$, so X is placed in Group III. (c)(ii) Y has valency $2$ towards hydrogen and its hydride matches ${\text{RH}}_2$, which heads both Group II and Group VI; so Y is placed in Group II or Group VI, and the hydride formula alone cannot separate the two.