Question

\text{Total number of ions from the following with noble gas configuration is \\_\\_\\_\\_\\_\\_.} \\\ \text{Sr}^{2+} \, (Z = 38), \, \text{Cs}^{+} \, (Z = 55), \, \text{La}^{2+} \, (Z = 57), \, \text{Pb}^{2+} \, (Z = 82), \, \text{Yb}^{2+} \, (Z = 70), \, \text{and} \, \text{Fe}^{2+} \, (Z = 26)

Answer 1

To determine if an ion has a noble gas configuration, we examine its electron configuration and compare it with that of a nearby noble gas:

- Sr²⁺ ($Z = 38$) loses two electrons, resulting in the electron configuration $[Kr]$, which matches the noble gas krypton.
- Cs⁺ ($Z = 55$) loses one electron, resulting in the electron configuration $[Xe]$, matching xenon.
- La³⁺ ($Z = 57$) loses three electrons, resulting in the electron configuration $[Xe]$, also matching xenon.
- Yb²⁺ ($Z = 70$) loses two electrons, resulting in the electron configuration $[Xe]$, matching xenon.

On the other hand:
- Pb²⁺ does not match any noble gas configuration due to its partially filled $d$-orbitals.
- Fe²⁺ does not match a noble gas configuration either, as it retains electrons in the $d$-orbital.

Thus, only *Sr²⁺, Cs⁺, La³⁺, and Yb²⁺ have noble gas configurations, totaling four ions.

The Correct answer is: 2