QuestionWrite ground-state electron configurations for these ions: (a) Sc $^{3+}$ , (b) Ti $^{4+}$ , (c) V $^{5+}$ , (d) Cr $^{3+}$ , (e) Mn $^{2+}$ , (f) Fe $^{2+}$ , (g) Fe $^{3+}$ , (h) Co $^{2+}$ , (i) Ni $^{2+}$ , (j) Cu $^{+}$ , (k) Cu $^{2+}$ , (l) Ag $^{+}$ , (m) Au $^{+}$ , (n) Au $^{3+}$ , (o) Pt $^{2+}$ .

Studdy Solution

STEP 1

Assumptions1. The electron configurations are written for ground-state atoms. . The electron configurations are written in the order of increasing atomic number.
3. The electron configurations are written for transition metal ions.
4. The electron configurations are written in the format of [previous noble gas] followed by the electron configuration of the valence electrons.
5. The ions have lost electrons from the highest energy level first.

STEP 2

We will start with $\mathrm{Sc}^{+}$ . The atomic number of Scandium (Sc) is21, so the electron configuration of a neutral Sc atom would be [Ar]4s2d1. However, we are dealing with $\mathrm{Sc}^{+}$ , which means it has lost electrons.

STEP 3

Since thes orbital is higher in energy than the3d orbital, the electrons are removed from thes orbital first. Thus, all three electrons (two from thes orbital and one from the3d orbital) are removed, leaving us with the electron configuration of $\mathrm{Sc}^{3+}$ .
$\mathrm{Sc}^{3+}: [Ar]$

STEP 4

Next, we will find the electron configuration for $\mathrm{Ti}^{4+}$ . The atomic number of Titanium (Ti) is22, so the electron configuration of a neutral Ti atom would be [Ar]4s23d2. However, we are dealing with $\mathrm{Ti}^{4+}$ , which means it has lost4 electrons.

STEP 5

Again, since the4s orbital is higher in energy than the3d orbital, the electrons are removed from the4s orbital first. Thus, all four electrons (two from the4s orbital and two from the3d orbital) are removed, leaving us with the electron configuration of $\mathrm{Ti}^{4+}$ .
$\mathrm{Ti}^{4+}: [Ar]$

STEP 6

We will continue this process for the remaining transition metal ions. The electron configurations for the remaining ions are as follows $\mathrm{V}^{5+}: [Ar]$ $\mathrm{Cr}^{3+}: [Ar]3d3$ $\mathrm{Mn}^{2+}: [Ar]3d5$ $\mathrm{Fe}^{2+}: [Ar]3d6$ $\mathrm{Fe}^{3+}: [Ar]3d5$ $\mathrm{Co}^{2+}: [Ar]3d$ $\mathrm{Ni}^{2+}: [Ar]3d8$ $\mathrm{Cu}^{+}: [Ar]3d10$ $\mathrm{Cu}^{2+}: [Ar]3d9$ $\mathrm{Ag}^{+}: [Kr]4d10$ $\mathrm{Au}^{+}: [Xe]4f145d10$ $\mathrm{Au}^{3+}: [Xe]4f145d$ $\mathrm{Pt}^{2+}: [Xe]4f145d8$

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