Question 20.1: Writing Electron Configurations of Transition Metals Write t......
Writing Electron Configurations of Transition Metals
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 transition metal ions, all the valence electrons occupy the d orbitals. For polyatomic ions, first determine the oxidation number of the transition metal, and then assign the valence electrons to the d orbitals as you would if the metal were a simple ion.
(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 4s² .
(b) A neutral Cr atom (Z = 24) has a total of 24 electrons; a Cr^{3+} ion has 24 – 3 = 21 electrons. Because all the valence electrons occupy the 3d orbitals, the electron configuration of Cr^{3+} is [Ar] 3d³ .
(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(VI) atom has a total of 20 electrons, 6 less than a neutral iron atom (Z = 26). Because the valence electrons occupy the 3d orbitals, the electron configuration of Fe(VI) is [Ar] 3d².