Question

Give the electronic configuration of the given metal (atom/ ions).
    \[FeO_{4}^{2-}\] (ferrate ions)
a.) \[\left[ Ar \right]3{{s}^{2}}\]
b.) \[\left[ Ar \right]4{{s}^{2}}\]
c.) \[\left[ Ar \right]3{{d}^{2}}\]
d.) None of these

Answer 1

Hint: The electron configuration of an element can be defined as how electrons are distributed in its atomic orbitals. Electron configurations of atoms follow a standard notation in which all electron-containing atomic subshells which are present with the number of electrons they hold written in superscript are placed in a sequence.

Complete step by step answer:
Atomic number of iron is 26 and its electronic configuration in its elemental state is \[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{6}}4{{s}^{2}}\]. The oxidation number of iron in the given compound that is ferrate ion is +6.

Electron configuration of Fe(IV) cation in \[FeO_{4}^{2-}\]is \[\left[ Ar \right]3{{d}^{2}}\]or
\[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{2}}4{{s}^{0}}\]
So, the correct answer is “Option C”.

Note: A molecular orbital diagram or commonly known as MO diagram, is a qualitative descriptive tool explaining chemical bonding in molecules in terms of molecular orbital theory in general and the linear combination of atomic orbitals method in particular.
Oxidation number or state of an atom/ion is the number of electrons an atom/ion that the molecule has either gained or lost compared to the neutral atom. Electropositive metal atoms, of group I, 2 and 3 lose a specific number of electrons and always have constant positive oxidation numbers.