What is the group and period of an element with the outer electronic configuration $4s^24p^4$?
The highest principal quantum number 'n' is 4, so the element is in Period 4. The configuration $ns^2np^4$ indicates a p-block element. For p-block elements, the group number is $10 + ext{number of s-electrons} + ext{number of p-electrons}$. So, Group = $10 + 2 + 4 = 16$. This element is Selenium (Se, Z=34).