Question 7.5.1: Write electron configurations for cations. Write electron co......
Write electron configurations for cations.
Write electron configurations for the following ions in spdf and orbital box notation.
a. Al^{3+} (Do not use noble gas notation.)
b. Cr^{2+} (Use noble gas notation.)
You are asked to write an electron configuration for a cation using spdf and orbital box notation.
You are given the identity of the cation.
a. Aluminum is element 13. The element loses three electrons from its highest-energy orbitals to form the Al^{3+} ion.
\text{Al :}1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}\underset{1s}{ \boxed{\uparrow \downarrow } }\,\underset{2s}{ \boxed{\uparrow \downarrow } }\,\underset{2p}{ \boxed{\uparrow \downarrow } \boxed{\uparrow \downarrow } \boxed{\uparrow \downarrow } }\,\underset{3s}{ \boxed{\uparrow \downarrow } }\,\underset{3p}{ \boxed{\uparrow \, \, }\boxed{ \,\,}\boxed{\,\,}}
\text{Al}^{3+} : 1s^{2}2s^{2}2p^{6}\underset{1s}{ \boxed{\uparrow \downarrow } }\,\underset{2s}{ \boxed{\uparrow \downarrow } }\,\underset{2p}{ \boxed{\uparrow \downarrow } \boxed{\uparrow \downarrow } \boxed{\uparrow \downarrow } }
b. Chromium is element 24. The element loses two electrons from its highest-energy orbitals to form the Cr^{2+} ion.
Cr: [Ar]4s^13d^{5} [Ar] \underset{4s}{ \boxed{\uparrow \,\,} }\,\underset{3d}{ \boxed{\uparrow \,\,} \boxed{\uparrow \,\,} \boxed{\uparrow \,\,} \boxed{\uparrow \,\,} \boxed{\uparrow \,\,} }
Cr^{2+}: [Ar]3d^{4} [Ar] \underset{4s}{\boxed{\,\,} }\,\underset{3d}{ \boxed{\uparrow \,\,} \boxed{\uparrow \,\,} \boxed{\uparrow \,\,} \boxed{\uparrow \,\,} \boxed{ \,\,} }