Atom with the electron configuration of 1S$^2$2S$^2$2P$^6$3S$^2$ belongs to

To determine the position of an element in the Periodic Table based on its electron configuration, we follow these two simple rules:

1. Analyze the Question

We are given the electron configuration: 1s² 2s² 2p⁶ 3s². We need to identify the Group and Period of this atom.

2. Explain the Concept

• Period: The period of an element is determined by the highest principal quantum number (n) in its electron configuration. This represents the outermost shell (energy level) that contains electrons.
• Group: The group of an element is determined by the number of valence electrons (electrons in the outermost shell). For elements in the s-block, the number of valence electrons directly corresponds to the Group number.

3. Show the Solution

• Step 1: Identify the Period. Look at the highest number appearing before the orbitals (s, p, d, f). In 1s² 2s² 2p⁶ 3s², the highest number is 3. Therefore, the element is in Period 3.

• Step 2: Identify the Group. Look at the outermost shell (the shell with the highest number, which is 3). The configuration shows 3s². There are 2 electrons in this outermost shell. Therefore, the element is in Group 2.

• Step 3: Conclusion. The element has 2 valence electrons in the 3rd energy level, placing it in Group 2, Period 3. (This element is Magnesium, Mg).

4. Identify the Answer

Comparing our result with the options provided: A. Group 3 and Period 2 B. Group 8 and Period 3 C. Group 2 and Period 3 D. Group 2 and Period 6

The correct option is C.

5. Exam Tip

To quickly find the group for s-block elements, just look at the exponent of the last 's' orbital. If it ends in 's¹', it's Group 1; if it ends in 's²', it's Group 2. For the period, just look at the largest number in front of the orbitals!