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Chapter 20: Problem 85

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

Expert verified
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

01

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. \(\mathrm{Ti^{4+}}\): \([Ar] 3d^{0}\)
02

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.
03

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.

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Most popular questions from this chapter

a. In the absorption spectrum of the complex ion \(\mathrm{Cr}(\mathrm{NCS})_{6}^{3-}\) there is a band corresponding to the absorption of a photon of light with an energy of \(1.75 \times 10^{4} \mathrm{cm}^{-1}\). Given 1 \(\mathrm{cm}^{-1}=1.986 \times 10^{-23} \mathrm{J},\) what is the wavelength of this photon? b. The \(\mathrm{Cr}-\mathrm{N}-\mathrm{C}\) bond angle in \(\mathrm{Cr}(\mathrm{NCS})_{6}^{3-}\) is predicted to be \(180^{\circ} .\) What is the hybridization of the \(\mathrm{N}\) atom in the NCS - ligand when a Lewis acid-base reaction occurs between \(\mathrm{Cr}^{3+}\) and \(\mathrm{NCS}^{-}\) that would give a \(180^{\circ}\) \(\mathrm{Cr}-\mathrm{N}-\mathrm{C}\) bond angle? \(\mathrm{Cr}(\mathrm{NCS})_{6}^{3-}\) undergoes substitution by ethylenediamine (en) according to the equation \(\mathrm{Cr}(\mathrm{NCS})_{6}^{3-}+2 \mathrm{en} \longrightarrow \mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}^{+}+4 \mathrm{NCS}^{-}\) Does \(\operatorname{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}^{+}\) exhibit geometric isomerism? Does \(\mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}^{+}\) exhibit optical isomerism?

Consider the complex ions \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}^{3+}, \mathrm{Co}(\mathrm{CN})_{6}^{3-},\) and \(\mathrm{CoF}_{6}^{3-} .\) The wavelengths of absorbed electromagnetic radiation for these compounds (in no specific order) are \(770 \mathrm{nm}\) \(440 \mathrm{nm},\) and \(290 \mathrm{nm} .\) Match the complex ion to the wavelength of absorbed electromagnetic radiation.

The complex ion \(\mathrm{Ru}\) (phen) \(_{3}^{2+}\) has been used as a probe for the structure of DNA. (Phen is a bidentate ligand.) a. What type of isomerism is found in \(\mathrm{Ru}(\text { phen })_{3}^{2+} ?\) b. \(\mathrm{Ru}(\text { phen })_{3}^{2+}\) is diamagnetic (as are all complex ions of \(\mathrm{Ru}^{2+}\) ). Draw the crystal field diagram for the \(d\) orbitals in this complex ion.$

A compound related to acetylacetone is 1,1,1 -trifluoroacetylacetone (abbreviated Htfa): Htfa forms complexes in a manner similar to acetylacetone. (See Exercise \(47 .\) Both \(\mathrm{Be}^{2+}\) and \(\mathrm{Cu}^{2+}\) form complexes with tfa \(^{-}\) having the formula \(\mathrm{M}\) (tfa) \(_{2}\). Two isomers are formed for each metal complex. a. The \(\mathrm{Be}^{2+}\) complexes are tetrahedral. Draw the two isomers of Be(tfa)_2. What type of isomerism is exhibited by \(\mathrm{Be}(\mathrm{tfa})_{2} ?\) b. The \(\mathrm{Cu}^{2+}\) complexes are square planar. Draw the two isomers of \(\mathrm{Cu}(\mathrm{tfa})_{2} .\) What type of isomerism is exhibited by $\mathrm{Cu}(\mathrm{tfa})_{2} ?

Give formulas for the following. a. potassium tetrachlorocobaltate(II) b. aquatricarbonylplatinum(II) bromide c. sodium dicyanobis(oxalato)ferrate(III) d. triamminechloroethylenediaminechromium(III) iodide
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