Question

How many elements would a period have maximum comprise of there are ten periods in a periodic table?

Answer 1

Hint: As we know that in the periodic table, the elements are arranged in an increasing order of atomic number. Moving down from up to down represents the group and moving from left to right represents the period. The elements are placed in these groups and periods which together comprise a periodic table.

Complete answer:
There is a formula to solve this question. Let's start solving by using this formula.
\[n\]= number of period
So in this case \[n = 10\] (given)
No. of sub shells \[ = {\left[ {\dfrac{{(n + 2)}}{2}} \right]^2}\]here n=even number
No. of sub shells \[ = {\left[ {\dfrac{{(n + 1)}}{2}} \right]^2}\]here n=odd number
Electron in one subshell is 2 i.e. spin (\[\dfrac{{ + 1}}{2},\dfrac{{ - 1}}{2}\])
Thus the formula for total number of element in the nth period: \[2 \times {\left[ {\dfrac{{n + 2}}{2}} \right]^2}\]
Here $n = 10$, so it can solved as given below
\[ = 2 \times {\left[ {\dfrac{{n + 2}}{2}} \right]^2}\]
Now we can substitute the given value we get,
\[ = 2 \times {\left[ {\dfrac{{10 + 2}}{2}} \right]^2}\]
On simplification we get,
$ = 72$
In a periodic table there are a total of seven rows which represent periods and a total of $18$ columns which represent groups. Generally metals are placed on the left while non-metals are placed on the right. An example of group $17$ elements are the halogens; and group $18$ are the noble gases.

Note:
We need to remember that the metal and nonmetals can be further classified into subcategories that show a graduation from metallic to non-metallic properties, when going left to right in the rows. The metals may be subdivided into the highly reactive alkali metals, less reactive alkaline earth metals, lanthanides and actinides and transition metals that also have its sub class which has inner transition metals.