Question

Which of the following ions is(are) expected to form colored octahedral aqueous complex ions? a. \(\mathrm{Zn}^{2+}\) b. \(\mathrm{Cu}^{2+}\) c. \(\mathrm{Mn}^{3+}\) d. \(\mathrm{Ti}^{4+}\)

Short answer

The ions that are expected to form colored octahedral aqueous complex ions are b. \(\mathrm{Cu}^{2+}\) (having one unpaired electron in the 3d orbitals) and c. \(\mathrm{Mn}^{3+}\) (having four unpaired electrons in the 3d orbitals).

Step-by-step solution

Write down the electronic configurations of the metal ions

We need to know the electronic configurations of the given ions, so let's write them down: a. \(\mathrm{Zn^{2+}}\): \([Ar] 3d^{10}\) b. \(\mathrm{Cu^{2+}}\): \([Ar] 3d^{9}\) c. \(\mathrm{Mn^{3+}}\): \([Ar] 3d^{4}\) d. \(\mathrm{Ti^{4+}}\): \([Ar] 3d^{0}\)

Analyze unpaired electrons in d-orbitals

Now, we will analyze each ion's electronic configuration to see if they have unpaired electrons in their d-orbitals: a. \(\mathrm{Zn^{2+}}\): There are no unpaired electrons in the 3d orbitals. b. \(\mathrm{Cu^{2+}}\): There is one unpaired electron in the 3d orbitals. c. \(\mathrm{Mn^{3+}}\): There are four unpaired electrons in the 3d orbitals. d. \(\mathrm{Ti^{4+}}\): There are no unpaired electrons in the 3d orbitals.

Determine which ions form colored octahedral complexes

Based on the presence of unpaired electrons in their d-orbitals, we can determine which ions are likely to form colored octahedral complexes: a. \(\mathrm{Zn^{2+}}\): No unpaired electrons, so it is not expected to form a colored complex. b. \(\mathrm{Cu^{2+}}\): One unpaired electron, so it is expected to form a colored complex. c. \(\mathrm{Mn^{3+}}\): Four unpaired electrons, so it is also expected to form a colored complex. d. \(\mathrm{Ti^{4+}}\): No unpaired electrons, so it is not expected to form a colored complex. So, the answer is: b. \(\mathrm{Cu}^{2+}\) and c. \(\mathrm{Mn}^{3+}\) are expected to form colored octahedral aqueous complex ions.