Let's determine the ion formed when potassium (K) achieves a noble-gas electron configuration.
1. Identify the atomic number:
- Potassium (K) has an atomic number of 19. This means that a neutral potassium atom has 19 protons and 19 electrons.
2. Determine the electronic configuration:
- The electronic configuration of potassium is [tex]\( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 \)[/tex]. This shows that potassium has one electron in its outermost shell (4s^1).
3. Achieving a noble-gas electron configuration:
- Noble gases have full outer electron shells. The closest noble gas before potassium in the periodic table is argon (Ar), which has an electronic configuration of [tex]\( 1s^2 2s^2 2p^6 3s^2 3p^6 \)[/tex].
- For potassium to achieve the noble-gas configuration of argon, it needs to lose one electron.
4. Forming the ion:
- When potassium loses one electron, the electron configuration becomes [tex]\( 1s^2 2s^2 2p^6 3s^2 3p^6 \)[/tex], matching that of argon.
- By losing one electron, potassium forms an ion with a +1 charge because it now has 19 protons (positively charged) and 18 electrons (negatively charged), resulting in a net positive charge.
5. Conclusion:
- The formula of the ion formed when potassium loses one electron to achieve a noble-gas electron configuration is [tex]\( K^+ \)[/tex].
Hence, the correct option is [tex]\( K^+ \)[/tex].
The numerical result indicating the correct option is:
- 2
This corresponds to:
- [tex]\( K^+ \)[/tex]
So, the ion formed when potassium achieves a noble-gas electron configuration is [tex]\( K^+ \)[/tex], matching the correct answer which is option 2.
