Answer:
Mn(IV): 3 outer-shell d electrons
Co(III): 6 outer-shell d electrons
Fe(II): 6 outer-shell d electrons
Ni(II): 8 outer-shell d electrons
Explanation:
Mn(IV):
The ion is Mn⁴⁺, so it has lost 4 valence electrons. The ground state electronic configuration of Mn is [Ar]3d ⁵ 4s ². When the neutral atom loses 4 electrons, the electronic configuration is [Ar]3d ³.
outer-shell d electrons = 3
Co(III):
The ion is Co³⁺, so it has lost 3 valence electrons. The ground state electronic configuration of Co is [Ar]3d ⁷4s ². When the neutral atom loses 3 electrons, the electronic configuration is [Ar]3d ⁶.
outer-shell d electrons = 6
Fe(II):
The ion is Fe²⁺, so it has lost 2 valence electrons. The ground state electronic configuration of Fe is [Ar]3d ⁶4s ². When the neutral atom loses 2 electrons, the electronic configuration is [Ar]3d ⁶.
outer-shell d electrons = 6
Ni(II):
The ion is Ni²⁺, so it has lost 2 valence electrons. The ground state electronic configuration of Ni is [Ar]3d ⁸4s ². When the neutral atom loses 2 electrons, the electronic configuration is [Ar]3d ⁸.
outer-shell d electrons = 8
The d electron in the outer shell of each of the transition metal are:
Mn(IV): 3 d electrons in the outer shell.
Co(III): 6 d electrons.
Fe(II): 6 d electrons.
Ni(II): 8 d electrons.
Electronic configuration is the arrangement of the electrons of an atom.
There are subshell s, p, d, f.
The electrons are arranged in these subshells.
Each element has different arrangement of electrons.
Thus, the number of d electron in each of the element is: Mn(IV) :3, Co(III) :6, Fe(II) :6, Ni(II) :8.
Learn more about electronic configuration
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