Question

Atomic number of an element is 37. The
valu e of (n + l) for the orbital from which
the electron is removed to form +1 ion of
its is

Answer 1

Answer:

5

Explanation:

The Electronic Configuration of Rubidium(Atomic no:37) 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 5s^1

so the one electron should be removed from outermost shell( 5s^1) the value of (n+l) for this is

= 5+0=0

As value of l for s orbital is 0.