How can I determine the quantum number of a electron lost during ion formation?

How do quantum numbers change for ions?

A sample question: Give the set of four quantum numbers that could represent the electron lost to form the Rb ION from the Rb atom.

The answer given is $n=5; l=0; m_l=0; m_s= +\frac12$. I chose the answer: $n=4; l=1; m_l=1; m_s= -\frac12$.

I am looking for an explanation. I am confused because I thought the question asked me to describe $\ce{Rb+}$ and the answer given is the same that I would choose for $\ce{Rb}$ at ground state. If it is asking me to describe the electron, how is that technically possible if I don't know where the electron went? Sounds silly and I know I'm over thinking it. Some insight would be appreciated.

• It seems you have the correct mindset and required knowledge to solve this problem, but just made a tiny slip-up in interpretation. You were asked to describe the electron lost from the rubidium atom. Write out the electronic configurations for the neutral rubidium atom and the rubidium cation, and compare. Is it clear now? – Nicolau Saker Neto Apr 5 '15 at 23:00
• Yes, it is clear now. Representing the electron lost from the Rb atom as n=4; l=1; ml=1; ms= -1/2 was really describing Kr. Thanks for helping me reach my 'aha' moment. – Lesleann Apr 5 '15 at 23:24
• If you feel up for it, we encourage users to answer their own questions when possible! Describing what was on your mind and how you came to understand the correct solution may help future readers. – Nicolau Saker Neto Apr 6 '15 at 0:07

The question you were asked was in reference to the quantum numbers that correspond to the electron lost when $\ce{Rb}$ goes to $\ce{Rb+}$. The electronic configuration of $\ce{Rb}$ is $[\ce{Kr}]\ 5\mathrm{s}^{1}$; thus, you need the set of quantum numbers for a $5\mathrm{s}^{1}$ electron. That set of quantum numbers is:
$$n = 5;\ l= 0;\ m_{l} = 0;\ m_{s} = +{1\over2}$$