Question 7.PS.10: Electron Configurations for Transition Elements and Ions (a)......
Electron Configurations for Transition Elements and Ions
(a) Write the electron configuration for the Co atom using the noble gas notation. Then draw the orbital box diagram for the electrons beyond the preceding noble gas configuration.
(b) Cobalt commonly exists as 2+ and 3+ ions. How does the orbital box diagram given in part (a) have to be changed to represent the outer electrons of Co^{2+} and Co^{3+}?
(c) How many unpaired electrons do Co, Co^{2+}, and Co^{3+} each have?
(a)
[Ar]3d^74s^2; [Ar] \boxed{↑↓}\quad\boxed{↑↓}\quad\overset{3d}{\boxed{↑ }}\quad\boxed{↑ }\quad\boxed{↑ } \quad\quad\overset{4s}{\boxed{↑↓}}
(b) For Co^{2+}, remove the two 4s electrons from Co to give
Co^{2+} [Ar] \boxed{↑↓}\quad\boxed{↑↓}\quad\overset{3d}{\boxed{↑ }}\quad\boxed{↑ }\quad\boxed{↑ }For Co^{3+}, remove the two 4s electrons and one of the paired 3d electrons from Co to give
Co^{3+} [Ar] \boxed{↑↓}\quad\boxed{↑}\quad\overset{3d}{\boxed{↑ }}\quad\boxed{↑ }\quad\boxed{↑ }(c) Co has three unpaired electrons; Co^{2+} has three unpaired electrons; Co^{3+} has four unpaired electrons.
Strategy and Explanation
(a) Co, with an atomic number of 27, is in the fourth period. It has nine more electrons than Ar, with two of the nine in the 4s subshell and seven in the 3d subshell, so its electron configuration is [Ar] 3d^74s^2. For the orbital box diagram, all d orbitals get one electron before pairing occurs (Hund’s rule).
(b) To form Co^{2+}, two electrons are removed from the outermost shell (4s subshell). To form Co^{3+} from Co^{2+}, one of the paired electrons is removed from a 3d orbital.
(c) Looking at the orbital box diagrams, you should see that the numbers of unpaired electrons are three for Co, three for Co^{2+}, and four for Co^{3+}.