Give formulas for the following complex ions. a. tetrachloroferrate(III) ion b. pentaammineaquaruthenium(III) ion c. tetracarbonyldihydroxochromium(III) ion d. amminetrichloroplatinate(II) ion

Short Answer

Expert verified
a. \(\textrm{[FeCl}_{4}\textrm{]}^{3-}\) b. \(\textrm{[Ru(NH}_{3}\textrm{)}_{5}\textrm{(H}_{2}\textrm{O)}]^{3+}\) c. \(\textrm{[Cr(CO)}_{4}\textrm{(OH)}_{2\textrm{]}^{3-}\) d. \(\textrm{[Pt(NH}_{3}\textrm{)Cl}_{3\textrm{]}^{-}\)

Step by step solution

01

a. Tetrachloroferrate(III) ion

Tetrachloroferrate(III) ion consists of a central iron (Fe) atom with an oxidation state of +3 and four chloride (Cl) ligands surrounding it. The formula for the tetrachloroferrate(III) ion can be written as: \[ \textrm{[FeCl}_{4}\textrm{]}^{3-} \]
02

b. Pentaammineaquaruthenium(III) ion

Pentaammineaquaruthenium(III) ion consists of a central ruthenium (Ru) atom with an oxidation state of +3, five ammine (NH3) ligands, and one aqua (H2O) ligand. The formula for the pentaammineaquaruthenium(III) ion can be written as: \[ \textrm{[Ru(NH}_{3}\textrm{)}_{5}\textrm{(H}_{2}\textrm{O)}]^{3+} \]
03

c. Tetracarbonyldihydroxochromium(III) ion

Tetracarbonyldihydroxochromium(III) ion consists of a central chromium (Cr) atom with an oxidation state of +3, four carbonyl (CO) ligands, and two hydroxo (OH) ligands. The formula for the tetracarbonyldihydroxochromium(III) ion can be written as: \[ \textrm{[Cr(CO)}_{4}\textrm{(OH)}_{2\textrm{]}^{3-} \]
04

d. Amminetrichloroplatinate(II) ion

Amminetrichloroplatinate(II) ion consists of a central platinum (Pt) atom with an oxidation state of +2, one ammine (NH3) ligand, and three chloride (Cl) ligands. The formula for the amminetrichloroplatinate(II) ion can be written as: \[ \textrm{[Pt(NH}_{3}\textrm{)Cl}_{3\textrm{]}^{-} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What causes high-altitude sickness, and what is high-altitude acclimatization?

Qualitatively draw the crystal field splitting for a trigonal bipyramidal complex ion. (Let the \(z\) axis be perpendicular to the trigonal plane.)

How many bonds could each of the following chelating ligands form with a metal ion? a. acetylacetone (acacH), a common ligand in organometal:atalysts: b. diethylenetriamine, used in a variety of industrial processes: $$ \mathrm{NH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2} $$ c. salen, a common ligand for chiral organometallic catalysts: d. porphine, often used in supermolecular chemistry as well as catalysis; biologically, porphine is the basis for many different types of porphyrin- containing proteins, including heme proteins:

The complex ion \(\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) has an absorption maximum at around \(800 \mathrm{nm} .\) When four ammonias replace water, \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}^{2+},\) 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.

How many unpaired electrons are in the following complex ions? a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) (low-spin case) b. \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) c. \(\mathrm{V}(\mathrm{en})_{3}^{3+}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free