The way to periodic Table . . .

[Hadi Bakhsh]

Asslam O Alikum
Sir , this is my little effort about periodic table , i know there will be mistakes and that can be rectified by you . . . Positive criticism will be welcomed . . .

The way to Periodic Table
This is the way to periodic table, that you can remember the whole periodic table along with the periods and groups.
• First of all you should remember that how many elements are present in each group.
Example. First period contain two elements, second and third contain eight elements.
• Where from period start and where it ends
Example . first period starts from Hydrogen with atomic no. 1 and ends at He with atomic no. 2 and second period start from Li with Atomic No. 3 and ends at Ne with atomic No.
10
• Remember the first or second or both famous elements of all groups.
Example. Lithium and Sodium (Natrium) is of IA, Carbon of IVA, Fluorine and Chlorine of VIIA and Helium of VIIIA.
• Daily remember just one or two elements with their atomic No. and their mass.
• If you are asked atomic no. with its group and period then first of all you guess where that element lies in period and how many elements are there in that period , then subtract the no. total no. of elements in that period from the atomic no. of that element .
Example . Tellurium, you want to know its period and group.
Tellurium’s atomic no. is 52 which is in 5th period because 5th period starts from Rb 37 and ends at Xe 54. We came to know period is 5th and there are 18 elements in group five so its very easy to know its group. Now subtract 18 (total no. of elements in 5th period) from 52 atomic no. of Tellurium 52-18=34 (atomic no. of Selenium) again subtracting to reach famous element of group, 34 (atomic no. of Se) – 18 (total elements in 4th period) = 16 atomic no. of Sulfur we know sulfur is in VIA group for more confirmation you can subtract 16 (at. No. of S) – 8 (total elements in 3rd period) = 8 (atomic no. of well known atom Oxygen.
This point is applicable to all except, Helium because He is S block and rests of all are P block and Lanthanides and Actinides.
Example2.Thallium, you want to know its period and group. It is very easy if you have remembered atomic no. and in which period lies.
Its atomic No. is 81 and it lies in 6th period which contains 32 elements. Now subtract
81(at. No. of Tl)-32(total no. of elements in 6th period) = 49(at. No of Indium) again subtracting 49(at. No of In) - 18(total no. of elements in 5th period) = 31(at. No of Gallium) again subtracting 31(at. No. of Ga) – 18 (total No. of elements in 4th period) = 13 is atomic no. of Aluminum and Al belongs to IIIA group.
Example. Niobium its at. No. is 41 and lies in 5th period with at. Mass 93 and 41 lies in 5th period which contains 18 elements. Now subtracting 41(at. No. of Nb) -18(total no. of elements in 5th period) = 23(at. No of Vanadium) again subtracting because it lies in Transition elements 23(at. No of V) – 18(total no. of elements in 4th period) = 5 atomic no. of Boron which is member of IIIA but Nb will be period of IIIB.