IF THERE WERE 10 PERIODS IN THE PERIODIC TABLE THEN HOW MANY ELEMENTS WOULD THIS PERIOD CAN MAXIMUM COMPRISE OF?

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$$\begin{gathered} No. of elements in a period is calculated by the formula: \\ If the period no. is even, no. of elements = 2\times {\left(\frac{n+2}{2}\right)}^2 \\ If the period no. is odd, no. of elements = 2 \times {\left(\frac{n+1}{2}\right)}^2 \\ Here, period number = 10 \\ So, n = 10 and the formula used will be of that of even number of period. \\ No. of elements in the 10th period = 2\times {\left(\frac{10+2}{2}\right)}^2 = 2\times 6^2 = 72 elements \end{gathered}$$

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