/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 Give formulas corresponding to t... [FREE SOLUTION] | 91影视

91影视

Give formulas corresponding to the following names: (a) Tetraamminezinc sulfate (b) Pentaamminechlorochromium(III) chloride (c) Sodium bis(thiosulfato)argentate(I)

Short Answer

Expert verified
(a) [Zn(NH3)4]SO4 (b) [Cr(NH3)5Cl]Cl2 (c) Na3[Ag(S2O3)2]

Step by step solution

01

- Identify the central metal and its oxidation state

For each compound, identify the central metal ion and its oxidation state from the given name.
02

- Determine the ligands

Identify all the ligands attached to the central metal ion as given in the compound name and note their charges.
03

- Write the formula with correct structure

Combine the central metal ion and ligands in a structured format, ensuring the correct number of ligands as indicated. Finally, write the remaining part of the chemical name as part of the formula.
04

- Summarize the formulas

Summarize all the formulas corresponding to the given names.

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 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

coordination compounds
Coordination compounds consist of a central metal ion bonded to surrounding molecules or ions known as ligands. The central metal ion usually comes from transition metals, which have the ability to form multiple bonds with various ligands. Coordination compounds are significant in many biological systems and industrial applications.
For example, in the case of Tetraamminezinc sulfate, the central metal ion is zinc, which forms coordination bonds with four ammonia (NH鈧) molecules. These compounds can vary widely in their structures and properties, but they always feature a metal ion at the center connected to ligands.
Understanding how to name and write the formulas for coordination compounds is crucial in chemistry. When dealing with these compounds, you must identify the following:
  • The central metal ion
  • Its oxidation state
  • The ligands attached
By recognizing these components, you can determine both the names and the structures of the compounds efficiently.
oxidation states
Oxidation states indicate the degree of oxidation (loss of electrons) of an atom within a compound. It鈥檚 crucial to determine the oxidation state of the central metal ion in coordination compounds.
For instance, in the compound Pentaamminechlorochromium(III) chloride, the central metal is chromium with an oxidation state of +3 (denoted by the Roman numeral III in the name).
The oxidation state helps to balance the overall charge of the compound and plays a pivotal role in defining the properties and reactivity of the metal ion. Here鈥檚 how you can determine the oxidation state:
  • Identify the total charge of the compound.
  • Sum the charges from all ligands and counter ions.
  • Solve for the metal鈥檚 charge using the net charge of the compound.
Understanding how to determine and apply oxidation states is essential for writing and naming coordination compounds correctly.
ligands
Ligands are molecules or ions that bind to the central metal ion in a coordination compound. They can vary in size, charge, and the number of donor atoms they possess. Ligands can act as either monodentate (binding through a single atom) or polydentate (binding through multiple atoms).
For example, in Sodium bis(thiosulfato)argentate(I), the ligand here is thiosulfate (S鈧侽鈧兟测伝), which binds to the silver ion. Identifying the correct ligands and knowing their structure helps in determining the correct formula for coordination compounds.
Here's how you can identify and work with ligands:
  • Determine the names and charges of the ligands from the compound name.
  • Understand the bonding pattern (how the ligand bonds to the metal ion).
  • Note the number of each type of ligand.
By analyzing these aspects, you can accurately write the formulas for various coordination compounds and understand their behavior in different chemical contexts.

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

A shortcut to finding optical isomers is to see if the complex has a plane of symmetry-a plane passing through the metal atom such that every atom on one side of the plane is matched by an identical one at the same distance from the plane on the other side. Any planar complex has a plane of symmetry, since all atoms lie in one plane. Use this approach to determine whether these exist as optical isomers: (a) \(\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) (tetrahedral); (b) \(\left[\mathrm{Pt}(\mathrm{en})_{2}\right]^{2+} ;\) (c) trans-[PtBr \(\left.\left._{4} \mathrm{Cl}_{2}\right]^{2-} ;\) (d) trans-[Co(en) \(\left._{2} \mathrm{~F}_{2}\right]^{+}\) (e) \(c i s-\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{~F}_{2}\right]^{+}\)

\(\mathrm{VF}_{5}\) is a liquid that boils at \(48^{\circ} \mathrm{C},\) whereas \(\mathrm{VF}_{3}\) is a solid that melts above \(800^{\circ} \mathrm{C}\). Explain this difference in properties.

Octahedral [Ni(NH \(_{3}\) ) \(\left._{6}\right]^{2+}\) is paramagnetic, whereas planar \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) is diamagnetic, even though both metal ions are \(d^{8}\) species. Explain.

Give the electron configuration of (a) \(\mathrm{Pm} ;\) (b) \(\mathrm{Lu}^{3+} ;\) (c) \(\mathrm{Th}\) (d) \(\mathrm{Fm}^{3+}\).

Which of these ions cannot form both high- and low-spin octahedral complexes: (a) \(\mathrm{Mn}^{3+} ;\) (b) \(\mathrm{Nb}^{3+} ;\) (c) \(\mathrm{Ru}^{3+}\) (d) \(\mathrm{Ni}^{2+} ?\)

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.