[Kr]5s1 would be the noble gas electron configuration for Rb.
[Kr] 5s1
Rb is likely to adopt a noble gas configuration of [Kr] when it forms ions. This means it will lose one electron to achieve a stable octet configuration, similar to the nearest noble gas element, krypton.
Rubidium has an atomic number of 37, making it an alkali metal. This means that its last shell is an s with only one electron. The full notation is [Kr] 5s1.
Rubidium (Rb) has a +1 ion, will have the same electron configuration as krypton (Kr) because the +1 status means it has lost an electron. The configuration is written 1s22s22p63s23p64s23d104p6.
1s2 2s2p6 3s2p6d10 4s2p6 5s1
[Kr] 5s1
Na, Rb, Al3+, and S2- do not have a noble gas configuration as they do not have the complete outer shell of electrons like a noble gas. O2, Br, Ca, and O2- have noble gas configurations as they either have a complete outer shell of electrons or have gained/lost electrons to achieve a stable noble gas configuration.
Rb is likely to adopt a noble gas configuration of [Kr] when it forms ions. This means it will lose one electron to achieve a stable octet configuration, similar to the nearest noble gas element, krypton.
Rubidium has an atomic number of 37, making it an alkali metal. This means that its last shell is an s with only one electron. The full notation is [Kr] 5s1.
Rubidium (Rb) has a +1 ion, will have the same electron configuration as krypton (Kr) because the +1 status means it has lost an electron. The configuration is written 1s22s22p63s23p64s23d104p6.
google it
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 Rb+1 or Kr
1s2 2s2p6 3s2p6d10 4s2p6 5s1
Rb has the electronic config. of = [Kr] 5s1
1s2 2s2p6 3s2p6d10 4s2p6 5s1
The lowest atomic number ion is Se^2-. The highest atomic number ion is Sr^2+. The ones in the middle are NOT Br^- or Rb^+. However, I do not know what the correct answer for those two is.
In the Rb atom, the electron configuration is [Kr]5s^1. This means there is 1 electron in the 5s sublevel of the Rb atom.