Question

Determine the quantum numbers n and l of the electron removed when bi is ionized to bi+

Answer 1

Electronic configuration of bismuth is given as:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^{6}4f^{14}5d^{10}6s^{2}6p^{3}$

When electron is removed from bismuth then bismuth is ionized to $Bi^{+}$, the electronic configuration becomes:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^{6}4f^{14}5d^{10}6s^{2}6p^{2}$

Now, n represents the principal quantum number which determines the distance from the orbital of the electrons i.e. size of the orbital and its energy.

Thus, outer most electron is present in shell number 6. Thus, principal quantum number is (n=6).

Now, l represents the azimuthual quantum number which determines the shape of an orbital.

Now, if the outer most electron is present in s orbital then the l =0, if the outer most electron is present in p orbital then the l =1, if the outer most electron is present in d orbital then the l =2, if the outer most electron is present in f orbital then the l =4 and if the outer most electron is present in d orbital then the l =2, if the outer most electron is present in h orbital then the l =5.

l 0 1 2 3 4 5

Letter s p d f h

Thus, azimuthual quantum number is 1 as the outer most electron is present in the p orbital.