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

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

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Answer 1 · 2015-11-03T17:38:23

See explanation.

Explanation:

The electron configuration of $_{15} P$ $""_15P$ is:

$_{15} P : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{3}$ $""_15P: 1s^(2)2s^(2)2p^(6)3s^(2)3p^(3)$

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

$_{15} P^{3 -} : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$ $""_15P^(3-): 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)$

The electron configuration of $_{42} M o$ $""_42Mo$ is:

$_{42} M o : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2} 4 d^{4}$ $""_42Mo: 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(2)3d^(10)4p^(6)5s^(color(green)(2))4d^(color(green)(4))$

When Molybdenum loses $3$ $3$ electrons to form the ion $M o^{3 +}$ $Mo^(3+)$ the electron configuration becomes:
$_{42} M o^{3 +} : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{0} 4 d^{3}$ $""_42Mo^(3+): 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(2)3d^(10)4p^(6)5s^(color(red)(0))4d^(color(red)(3))$

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

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

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

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