# What is the electron configuration of the P3- and Mo3+ ions?

Nov 3, 2015

See explanation.

#### Explanation:

The electron configuration of ""_15P is:

""_15P: 1s^(2)2s^(2)2p^(6)3s^(2)3p^(3)

When phosphorous gains $3$ electrons to form the ion ${P}^{3 -}$ the electron configuration becomes:

""_15P^(3-): 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)

The electron configuration of ""_42Mo is:

""_42Mo: 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(2)3d^(10)4p^(6)5s^(color(green)(2))4d^(color(green)(4))

When Molybdenum loses $3$ electrons to form the ion $M {o}^{3 +}$ the electron configuration becomes:
""_42Mo^(3+): 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(2)3d^(10)4p^(6)5s^(color(red)(0))4d^(color(red)(3))

Some people they get confused and remove the $3$ electrons from $4 {d}^{4}$ first, however, the electrons will be removed from $5 {s}^{2}$ first and the remaining $1$ electron will be removed from $4 {d}^{4}$.

You should keep in mind that the $5 s$ orbital gets filled before the $4 d$ orbital, however, when removing electrons; the electrons will be removed from $5 s$ first.