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Chapter 21: Problem 38

Name the following coordination compounds. a. $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Br}\right] \mathrm{Br}_{2}$ b. \(\mathrm{Na}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\) c. $\left[\mathrm{Fe}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{2}\left(\mathrm{NO}_{2}\right)_{2}\right] \mathrm{Cl}$ d. $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{2}\right]\left[\mathrm{PtI}_{4}\right]$

Short Answer

Expert verified
The short version answer for the given question is: a. Penta-aqua-bromo-chromium(III) tribromide b. Trisodium hexa-cyano-cobaltate(III) c. Di-ammine-di-nitrito-iron(III) chloride d. Tetra-ammine-di-iodido-platinum(IV) tetraiodidoplatinate(II)

Step by step solution

01

Naming Complex a: \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Br}\right] \mathrm{Br}_{2}\)

First, list the ligands alphabetically, which are water, \(\mathrm{H}_{2}\mathrm{O}\) (called aqua) and bromide, \(\mathrm{Br}\). Since there are five water molecules, we use the prefix 'penta.' The complex formula is then written as penta-aqua-bromo-chromium(III). The overall name of this compound is penta-aqua-bromo-chromium(III) tribromide.
02

Naming Complex b: \(\mathrm{Na}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\)

For this compound, the ligand is cyanide which is named as cyano. We have six cyano ligands, so we use the prefix 'hexa' and name the complex ion as hexa-cyano-cobalt(III). Since there are three sodium ions, the full name of the compound is trisodium hexa-cyano-cobaltate(III).
03

Naming Complex c: \(\left[\mathrm{Fe}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{2}\left(\mathrm{NO}_{2}\right)_{2}\right] \mathrm{Cl}\)

In this complex, we have the ligands en (ethylenediamine) and nitrito (NO₂). We have two of each ligand, so we use the prefix 'di' for both. Since Fe is iron, and the complex ion has a negative charge, we add the suffix '-ate' to the metallic element name. The complex ion becomes di-ammine-di-nitrito-iron(III). Finally, add the chloride counterion, and the compound's name is di-ammine-di-nitrito-iron(III) chloride.
04

Naming Complex d: \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{2}\right]\left[\mathrm{PtI}_{4}\right]\)

This compound has two complex ions. For the first one, we have ammine (NH₃) and iodido (I) as ligands with four ammines and two iodidos present, giving us tetra-ammine-di-iodido-platinum(IV). The second complex ion contains four iodido ligands, giving us tetraiodidoplatinum(II). The overall name of this compound is tetra-ammine-di-iodido-platinum(IV) tetraiodidoplatinate(II).

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Most popular questions from this chapter

When aqueous KI is added gradually to mercury(II) nitrate, an orange precipitate forms. Continued addition of KI causes the precipitate to dissolve. Write balanced equations to explain these observations. $\left(\text {Hint} : \mathrm{Hg}^{2+} \text { reacts with } \mathrm{I}^{-} \text { to form } \mathrm{HgI}_{4}^{2-} .\right)\( Would you expect \)\mathrm{Hg} \mathrm{I}_{4}^{2-}$ to form colored solutions? Explain.

Name the following coordination compounds. a. \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{2}\) b. $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{I}_{3}$ c. \(\mathrm{K}_{2}\left[\mathrm{PtC}_{4}\right]\) d. \(\mathrm{K}_{4}\left[\mathrm{Pt} \mathrm{C}_{6}\right]\) e. $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$ f. \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{NO}_{2}\right)_{3}\right]\)

Tetrahedral complexes of \(\mathrm{Co}^{2+}\) are quite common. Use a \(d\) -orbital splitting diagram to rationalize the stability of \(\mathrm{Co}^{2+}\) tetrahedral complex ions.

Why are \(\mathrm{CN}^{-}\) and \(\mathrm{CO}\) toxic to humans?

Draw the \(d\) -orbital splitting diagrams for the octahedral complex ions of each of the following. a. \(\mathrm{Fe}^{2+}\) (high and low spin) b. \(\mathrm{Fe}^{3+}(\text { high spin })\) c. \(\mathrm{Ni}^{2+}\)
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