Classify the elements having atomic numbers `(9,12,16,34,53,56)` into three separate pairs on the basis of similar chemical properties.

Explanation

`2nd` period ends with `Z =10`
`3rd` period with `Z = 10+8 = 18`
`4th` period ens with `Z = 18+18 = 36`
`5th` period ends with `Z = 36+18 = 54`
Note: Magic numbers are `8,18` and `32`.
(a) `Z = 9(10-1) ("group" 18-1) = "group" 17`, `2nd` period element is `F`.
(b) `Z =12 (10+2) ("group" 2)`, `3rd` period, element is `Mg`.
( c) `Z =16 (18-2)("group" 18-2) = "group 16` , `3rd` period element is `S`.
(d) `Z = 34(36-2) ("group" 18-2) = "group" 16`, `4th` period, elements is `Se`.
(e) `Z = 53 (54-1) ("group" 18-1) = "group" 17`, `5th` period, elements is `I`.
(f) `Z = 56(54+2) ("group" 2)`, `5th` period and elements is `Ba`.
(i) Therefore, elements with atomic number `9(F)` and `53(I)` belong to group `17` i.e. halogens.
(ii) Elements with atomic number `12 (Mg)` and `56 (Ba)` belong to group `2`, i.e. alkaline earth metals.
(iii) Elements with atomic number `16(S)` and `34(Se)` belongs to group `16`, i.e. oxygen family.