Atom with the electron configuration of 1S\(^2\)2S\(^2\)2P\(^6\)3S\(^2\) belongs to

Answer Details

To determine the group and period of an element from its electron configuration, two pieces of information are needed:

• The period is the highest principal quantum number (the largest shell number) occupied by electrons.
• The group (for s-block and p-block elements) is determined by the number of electrons in the outermost shell.

The electron configuration given is \(1s^2\,2s^2\,2p^6\,3s^2\). The total number of electrons is \(2 + 2 + 6 + 2 = 12\), which identifies the element as magnesium (Mg).

The highest principal quantum number is 3 (from the \(3s^2\) subshell), so the element is in Period 3.

The outermost shell (\(n = 3\)) contains only 2 electrons (both in the \(3s\) subshell). Since these are s-block electrons, the element is in Group 2.

Therefore, the element belongs to Group 2 and Period 3.