Question 20.1: Write the electron configuration of the metal in each of the......
Write the electron configuration of the metal in each of the following atoms or ions:
(a) Ni (b) Cr^{3+} (c) FeO_{4} ^{2-} (ferrate ion)
STRATEGY
In neutral atoms of the first transition series, the 4s orbital is usually filled with 2 electrons and the remaining electrons occupy the 3d orbitals. In writing electron configurations for transition metal ions, remove electrons from valence ns orbitals before valence (n – 1) d orbitals. For polyatomic ions, first determine the oxidation number of the transition metal.
(a) Nickel (Z = 28) has a total of 28 electrons, including the argon core of 18. Two of the 10 valence electrons occupy the 4s orbital, and the remaining eight are assigned to the 3d orbitals. The electron configuration is therefore [Ar] 3d^{8}4s^{2}.
Ni: \overset{\uparrow\downarrow}{—} \overset{\uparrow\downarrow}{—} \overset{\uparrow\downarrow}{\underset{3d}{—}} \overset{\uparrow}{—} \overset{\uparrow}{—} \overset{\uparrow\downarrow}{\underset{4s}{—}}(b) A neutral Cr atom (Z = 24) has a 6 valence electrons and the electron configuration is [Ar] 3d^{5}4s^{1}. A Cr^{3+} ion has lost 3 valence electrons and has the configuration [Ar] 3d^{3}.
Cr: \overset{\uparrow}{—} \overset{\uparrow}{—} \overset{\uparrow }{\underset{3d}{—}} \overset{\uparrow}{—} \overset{\uparrow}{—} \overset{\uparrow }{\underset{4s}{—}} Cr^{3+}: \overset{\uparrow}{—} \overset{\uparrow}{—} \overset{\uparrow}{\underset{3d}{—}} \overset{}{—} \overset{}{—} \underset{4s}{—}
(c) The oxidation number of each of the four oxygens in FeO_{4} ^{2-} is -2 and the overall charge on the oxoanion is -2, so the oxidation number of the iron must be +6. An iron atom has 8 valence electrons and the electron configuration [Ar] 3d^{6}4s^{2}. An iron(VI) atom has lost 6 valence electrons and the electron configuration is [Ar] 3d^{2}.
Fe: \overset{\uparrow\downarrow }{—} \overset{\uparrow}{—} \overset{\uparrow }{\underset{3d}{—}} \overset{\uparrow}{—} \overset{\uparrow}{—} \overset{\uparrow\downarrow }{\underset{4s}{—}} Fe^{6+}: \overset{\uparrow}{—} \overset{\uparrow}{—} \underset{3d}{—} — — \underset{4s}{—}