Write the formula for each of the following compounds, being sure to use brackets to indicate the coordination sphere: (a) hexaamminechromium(III) nitrate (b) tetraamminecarbonatocobalt(III) sulfate (c) dichlorobis(ethylenediamine) platinum(IV) bromide (d) potassium diaquatetrabromovanadate(III) (e) bis(ethylenediamine) zinc(II) tetraiodomercurate(II)

Short Answer

Expert verified
\[[\text{Cr}(\text{NH}_3)_6] (\text{NO}_3)_3\], \[[\text{Co}(\text{NH}_3)_4 (\text{CO}_3)] (\text{SO}_4)_2\], \[[\text{Pt}(\text{en})_2 \text{Cl}_2] (\text{Br})_4\], \[\text{K}[\text{V}(\text{H}_2\text{O})_2 \text{Br}_4]\], \[[\text{Zn}(\text{en})_2] [\text{HgI}_4]\]

Step by step solution

01

(a) hexaamminechromium(III) nitrate

The compound has a chromium(III) ion in the coordination sphere, bound to six ammine ligands. The nitrate ion is the counter ion. The overall charge on the coordination sphere is +3 since chromium has a charge of +3. Thus, the formula is: \[ [\text{Cr}(\text{NH}_3)_6] (\text{NO}_3)_3 \]
02

(b) tetraamminecarbonatocobalt(III) sulfate

The compound has a cobalt(III) ion in the coordination sphere, bound to four ammine ligands and one carbonate ligand. The sulfate ion is the counter ion. The overall charge on the coordination sphere is +3 since cobalt has a charge of +3. Thus, the formula is: \[ [\text{Co}(\text{NH}_3)_4 (\text{CO}_3)] (\text{SO}_4)_2 \]
03

(c) dichlorobis(ethylenediamine) platinum(IV) bromide

The compound has a platinum(IV) ion in the coordination sphere, bound to two ethylenediamine ligands and two chloride ligands. The bromide ion is the counter ion. The overall charge on the coordination sphere is +4 since platinum has a charge of +4. Thus, the formula is: \[ [\text{Pt}(\text{en})_2 \text{Cl}_2] (\text{Br})_4 \]
04

(d) potassium diaquatetrabromovanadate(III)

The compound has a vanadate(III) ion in the coordination sphere, bound to two aqua (water) ligands and four bromide ligands. The overall charge on the coordination sphere is -1 since vanadate has a charge of +3 and there are four bromide ligands with a charge of -1 each. The counter ion is the potassium ion. Thus, the formula is: \[ \text{K}[\text{V}(\text{H}_2\text{O})_2 \text{Br}_4] \]
05

(e) bis(ethylenediamine) zinc(II) tetraiodomercurate(II)

The compound has a zinc(II) ion in the coordination sphere, bound to two ethylenediamine ligands. The overall charge on the coordination sphere is +2 since zinc has a charge of +2. The counter ion is the tetraiodomercurate(II) ion, which has a -2 charge since mercury has a charge of +2 and there are four iodide ligands with a charge of -1 each. Thus, the formula is: \[ [\text{Zn}(\text{en})_2] [\text{HgI}_4] \]

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Most popular questions from this chapter

Write the names of the following compounds, using the standard nomenclature rules for coordination complexes: (a) \(\left[\mathrm{Rh}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}\) (b) \(\mathrm{K}_{2}\left[\mathrm{TiCl}_{6}\right]\) (c) \(\mathrm{MoOCl}_{4}\) (d) \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right] \mathrm{Br}_{2}\)

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