Question

Classify the elements having atomic numbers as given below into three separate pairs on the basis of similar chemical properties. Give brief electronic explanation:$9,12,16,34,53,56$.

Answer 1

The second period ends at atomic number $10$ while the third period ends at atomic number $18$. Therefore, $9,12$and $16$ are the first elements in their respective groups. The atomic numbers of the other elements of the same group can be deduced by adding magic numbers of $8,18,18$ and $32$ to elements of $2^{nd}$ period and by adding magic numbers of $18,18$ and $32$ to the elements of $3^{rd}$ period. Thus,
$9+8+18+18=53$
$12+8+18+18=56$
$16+18=34$
elements with atomic numbers $9\left(F\right)$ and $53\left(I\right)$ belong to halogen family $\left(group17\right)$; elements with atomic numbers $12\left(Mg\right)$ and $56\left(Ba\right)$ belong to alkaline earth metals $\left(Group2\right)$ while elements with atomic numbers $16\left(S\right)$ and $34\left(Se\right)$ belong to oxygen family $\left(Group16\right)$.