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The quantum numbers for the outermost electron of rubidium with the configuration [Kr] 5s¹ and principal quantum number n = 5 are:

1. l = 0, corresponding to an s-orbital
2. m = 0, with spin quantum number +1/2
3. m = 0, with spin quantum number -1/2
4. l = 1, corresponding to a p-orbital

Correct answer: l = 0, corresponding to an s-orbital

Solution

The electronic configuration of Rb shows that the outermost electron is in the 5s orbital. For an s-orbital, the azimuthal quantum number (l) is 0.

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