In noble-gas shorthand notation, what is the ground-state electron configuration of rubidium (Rb, atomic number 37)?

• A. $\left[Kr\right]s^1$
• B. $\left[Kr\right]d^1$
• C. $\left[Ar\right]s^2{p}^6{s}^1$
• D. $\left[Kr\right]s^2$

Answer: (A)

Verified step by step guidance

1. Identify the atomic number of rubidium (Rb), which is 37. This tells us rubidium has 37 electrons in its neutral ground state.
2. Recall that noble-gas shorthand notation uses the electron configuration of the nearest noble gas with a lower atomic number as a starting point. Find the noble gas with atomic number less than 37, which is krypton (Kr) with atomic number 36.
3. Write the electron configuration of krypton (Kr) in brackets: \([\mathrm{Kr}]\). This represents the first 36 electrons.
4. Determine the electron configuration for the 37th electron, which will be placed in the next available orbital after krypton. Since rubidium is in group 1 of the periodic table, the next electron goes into the 5s orbital.
5. Combine the noble gas core with the additional electron: \([\mathrm{Kr}]\,5s^{1}\). This is the ground-state electron configuration of rubidium in noble-gas shorthand notation.