Question

write the complete ground - state electron configuration of rb+.for multi - digit superscripts or coefficients, use each number in succession.

Explanation:

Step1: Determine the atomic number of Rb

The atomic number of Rb (rubidium) is 37. So, a neutral Rb atom has 37 electrons.

Step2: Consider the ion formation

\(Rb^{+}\) has lost one electron. So, it has \(37 - 1=36\) electrons.

Step3: Write the electron configuration

Using the Aufbau principle, the electron configuration of an atom/ion with 36 electrons is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}\)

Answer:

\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}\)