Example 6.11: Predicting Electron Configurations of Ions
Predicting Electron Configurations of Ions
What is the electron configuration and orbital diagram of:
(a)
\(\ce{Na^+}\)
(b)
\(\ce{P^3–}\)
(c)
\(\ce{Al^2+}\)
(d)
\(\ce{Fe^2+}\)
(e)
\(\ce{Sm^3+}\)
Solution
First, write out the electron configuration for each parent atom. We have chosen to show the full, unabbreviated configurations to provide more practice for students who want it, but listing the core-abbreviated electron configurations is also acceptable.
Next, determine whether an electron is gained or lost. Remember electrons are negatively charged, so ions with a positive charge have
lost an electron. For main group elements, the last orbital gains or loses the electron. For transition metals, the last s orbital loses an electron before the d orbitals.
\(\ce{ Na: 1s^2 2s^2 2p^6 3s^1}\)
Sodium cation loses one electron, so
\(\ce{Na+}\) :
[a]
\(\ce{ Na+: 1s^2 2s^2 2p^6}\)
\(\ce{ P: 1s^2 2s^2 2p^6 3s^2 3p^3}\)
Phosphorus trianion gains three electrons, so
\(\ce{P^3-}\) :
[b]
\(\ce{ P^31: 1s^22s^22p^63s^23p^6}\)
\(\ce{ Al: 1s^2 2s^2 2p^6 3s^2 3p^1 }\)
Aluminum dication loses two electrons
\(\ce{Al^2+:}\)
[c]
\(\ce{ Al^2+: 1s^22s^22p^63s^1}\)
\(\ce{ Fe: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6}\)
Iron(II) loses two electrons and, since it is a transition metal, they are removed from the 4s orbital
\(\ce{Fe^2+:}\)
[d]
\(\ce{ Fe^2+: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6}\)
\(\ce{ Sm: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4f^6}\)
Samarium trication loses three electrons. The first two will be lost from the 6s orbital, and the final one is removed from the 4f orbital.
[e]
\(\ce{ Sm^3+: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 4f^5}\)