Write the atomic number of the element present in the third period and seventeenth group of the periodic table.

Hint: Element is present in third period. This means the principal quantum number ($n$) is 3. The third period contains 8 elements, starts with the element having atomic number 11 and ends with element having atomic number 18. Also, you should know that the number of electrons in the ground state of an element is equal to its atomic number, for the neutral atom.

Complete step by step answer:
We are given that element present in the third period. Period number identifies the highest principle quantum number ($n$) of the element in which the last electron enters. Therefore, for the third period, $n = 3$. Also given that, the element is present in the seventeenth group. General configuration of the elements present in the seventeenth group is: $n{s^2}n{p^5}$.
Therefore, general electronic configuration of the required element is: $3{s^2}3{p^5}$ (since, $n = 3$).
Now, complete electronic configuration of the element is: $1{s^2}2{s^2}2{p^6}3{s^2}3{p^5}$.
Now, counting the total number of electrons in the ground state of the element: $1 + 2 + 6 + 2 + 5 = 17$ electrons.
We know that the total number of electrons in an element in its ground state is equal to its atomic number.
We calculated above that there are 17 electrons in the element in its ground state, hence the atomic number of the element is 17. The element with atomic number 17 is chlorine ($Cl$).

Note: The first period contains two elements and the subsequent periods consists of 8, 8, 18, 18, 32 elements respectively. The third period starts with the element sodium (atomic number = 11, contains 11 electrons) and ends with the element argon (atomic number 18, contains 18 electrons).