Final answer:
The outer electronic configuration ns²np⁴ corresponds to the element sulfur (S) in its ground state, which belongs to group 16 of the periodic table and has six electrons in its outer p subshell.
Step-by-step explanation:
The outer electronic configuration ns²np⁴ refers to an element with six electrons in its outermost p subshell (since the p subshell can hold a total of six electrons). To find which element in its ground state corresponds to this configuration, we refer to the periodic table and seek elements in the oxygen group (group 16), as they have six electrons in their outer p subshell. The elements within this group include oxygen (O), sulfur (S), selenium (Se), tellurium (Te), and polonium (Po).
Considering the periodic trends and the fact that 'n' represents the principal quantum number which increases with the period number, we need to find an element with the principal quantum number that matches the electron configuration. Sulfur (S), with the atomic number 16, has the ground state electron configuration of [Ne]3s²3p⁴. This matches the given ns²np⁴ configuration, where 'n' is equal to 3 for sulfur when considering the highest energy level in its electron distribution.
Thus, the element with the outer electronic configuration ns²np⁴ in its ground state is sulfur (S).