Question

If periodic table would have contained 10 period, maximum elements in that periodic table would be:

• A. 72
• B. 190
• C. 144
• D. 290

Answer 1

The correct option is A 72

For, even no. of period $=\frac{(n+2{)}^2}{2}$

So, $10$ periods $=\frac{(10+2{)}^2}{2}$

$=\frac{144}{2}=72$

$\therefore$ Number of elements $=72$