Question 7.5.1: Write electron configurations for cations. Write electron co......

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]  $\underset{4s}{\boxed{\,\,} }\,\underset{3d}{ \boxed{\uparrow \,\,} \boxed{\uparrow \,\,} \boxed{\uparrow \,\,} \boxed{\uparrow \,\,} \boxed{ \,\,} }$