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Chapter 21: Problem 85

Give formulas for the following. a. Hexakis(pyridine)cobalt(III) chloride b. Pentaammineiodochromium(III) iodide c. Tris(ethylenediamine)nickel(II) bromide d. Potassium tetracyanonickelate(II) e. Tetraamminedichloroplatinum(IV) tetrachloroplatinate(II)

Short Answer

Expert verified
The formulas of the given complex compounds are as follows: a. Hexakis(pyridine)cobalt(III) chloride: \[ [Co(C_5H_5N)_6]Cl_3 \] b. Pentaammineiodochromium(III) iodide: \[ [Cr(NH_3)_5I]I_2 \] c. Tris(ethylenediamine)nickel(II) bromide: \[ [Ni(C_2H_8N_2)_3]Br_2 \] d. Potassium tetracyanonickelate(II): \[ K_2[Ni(CN)_4] \] e. Tetraamminedichloroplatinum(IV) tetrachloroplatinate(II): \[ [Pt(NH_3)_4Cl_2][PtCl_4] \]

Step by step solution

01

Identify the metal, ligands, and their charges

The metal is cobalt with an oxidation state of +3, the ligands are pyridine (C5H5N, a neutral monodentate ligand), and the counterions are chloride.
02

Write the formula

Using the information above, we can write down the formula for hexakis(pyridine)cobalt(III) chloride as follows: \[ [Co(C_5H_5N)_6]Cl_3 \] b. Pentaammineiodochromium(III) iodide
03

Identify the metal, ligands, and their charges

The metal is chromium with an oxidation state of +3, the ligands are 5 molecules of ammonia (NH3, a neutral monodentate ligand) and 1 iodide ion (a monodentate ligand with a charge of -1), the counterions are iodide.
04

Write the formula

Using the information above, we can write down the formula for pentaammineiodochromium(III) iodide as follows: \[ [Cr(NH_3)_5I]I_2 \] c. Tris(ethylenediamine)nickel(II) bromide
05

Identify the metal, ligands, and their charges

The metal is nickel with an oxidation state of +2, the ligands are 3 molecules of ethylenediamine (C2H8N2, a neutral bidentate ligand), and the counterions are bromide.
06

Write the formula

Using the information above, we can write down the formula for tris(ethylenediamine)nickel(II) bromide as follows: \[ [Ni(C_2H_8N_2)_3]Br_2 \] d. Potassium tetracyanonickelate(II)
07

Identify the metal, ligands, and their charges

The metal is nickel with an oxidation state of +2, the ligands are 4 cyanide ions (CN-, a monodentate ligand with a charge of -1), and the counterions are potassium.
08

Write the formula

Using the information above, we can write down the formula for potassium tetracyanonickelate(II) as follows: \[ K_2[Ni(CN)_4] \] e. Tetraamminedichloroplatinum(IV) tetrachloroplatinate(II)
09

Identify the metal, ligands, and their charges for both cations and anions

There are two complex ions in this compound. For the cation: - The metal is platinum with an oxidation state of +4. - The ligands are 4 molecules of ammonia (NH3, a neutral monodentate ligand) and 2 chloride ions (Cl-, a monodentate ligand with a charge of -1). For the anion: - The metal is also platinum, but this time with an oxidation state of +2. - The ligands are 4 chloride ions (Cl-, a monodentate ligand with a charge of -1).
10

Write the formula

Using the information above, we can write down the formula for tetraamminedichloroplatinum(IV) tetrachloroplatinate(II) as follows: \[ [Pt(NH_3)_4Cl_2][PtCl_4] \]

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Most popular questions from this chapter

The complex ion \(\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) has an absorption maximum at around \(800 \mathrm{~nm}\). When four ammonias replace water, \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]^{2+}\), the absorption maximum shifts to around \(600 \mathrm{~nm} .\) What do these results signify in terms of the relative field splittings of \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\) ? Explain.

Write electron configurations for the following metals. a. \(\mathrm{Ni}\) b. \(\mathrm{Cd}\) c. \(\mathrm{Zr}\) d. \(\mathrm{Os}\)

Compounds of \(\mathrm{Sc}^{3+}\) are not colored, but those of \(\mathrm{Ti}^{3+}\) and \(\mathrm{V}^{3+}\) are. Why?

Consider the following data: $$\begin{aligned} \mathrm{Co}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}^{2+} & & \mathscr{E}^{\circ}=1.82 \mathrm{~V} \\ \mathrm{Co}^{2+}+3 \mathrm{en} \longrightarrow \mathrm{Co}(\mathrm{en})_{3}^{2+} & K &=1.5 \times 10^{12} \\ \mathrm{Co}^{3+}+3 \mathrm{en} \longrightarrow \mathrm{Co}(\mathrm{en}){ }^{3+} & K &=2.0 \times 10^{47} \end{aligned}$$ where en \(=\) ethylenediamine. a. Calculate \(\mathscr{E}^{\circ}\) for the half-reaction $$\mathrm{Co}(\mathrm{en})_{3}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}(\mathrm{en})_{3}^{2+}$$ b. Based on your answer to part a, which is the stronger oxidizing agent, \(\mathrm{Co}^{3+}\) or \(\mathrm{Co}(\mathrm{en})_{3}{ }^{3+}\) ? c. Use the crystal field model to rationalize the result in part b.

The \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\) ion is diamagnetic, but \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) is paramagnetic. Explain.
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