/*! 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 84 Name the following coordination ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Name the following coordination compounds. a. \(\mathrm{Na}_{4}\left[\mathrm{Ni}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]\) b. \(\mathrm{K}_{2}\left[\mathrm{CoCl}_{4}\right]\) c. \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{SO}_{4}\) d. \(\left[\mathrm{Co}(\mathrm{en})_{2}(\mathrm{SCN}) \mathrm{Cl}\right] \mathrm{Cl}\)

Short Answer

Expert verified
a. Tetrakis(oxalato)nickelate(II) sodium b. Potassium tetrachlorocobaltate(II) c. Tetraamminecopper(II) sulfate d. Pentaamminethiocyanato-κN-chlorocobalt(III) chloride

Step by step solution

01

Identify the central metal ion

Find the metal atom in each coordination complex as it will be the central metal ion. a. Ni (Nickel) b. Co (Cobalt) c. Cu (Copper) d. Co (Cobalt)
02

Identify the ligands

Determine the ligands surrounding the central metal ion in each coordination complex. a. \(C_{2}O_{4}\) (Oxalate) b. Cl (Chloride) c. \(NH_{3}\) (Ammine) d. en (Ethylenediamine), SCN (Thiocyanate), Cl (Chloride)
03

Determine the charge on the coordination complex

Calculate the overall charge on each coordination complex. a. Charge on \(Na_{4}\) = +4, therefore charge on the complex = -4 b. Charge on \(K_{2}\) = +2, therefore charge on the complex = -2 c. Charge on \(SO_{4}\) = -2, therefore charge on the complex = +2 d. Charge on \(Cl\) = -1, therefore charge on the complex = +1
04

Name the coordination complex using the standard nomenclature rules

Name each coordination complex following IUPAC nomenclature rules. a. Tetrakis(oxalato)nickelate(II) sodium b. Potassium tetrachlorocobaltate(II) c. Tetraamminecopper(II) sulfate d. Pentaamminethiocyanato-κN-chlorocobalt(III) chloride So, the names of the coordination complexes are: a. Tetrakis(oxalato)nickelate(II) sodium b. Potassium tetrachlorocobaltate(II) c. Tetraamminecopper(II) sulfate d. Pentaamminethiocyanato-κN-chlorocobalt(III) chloride

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 91Ó°ÊÓ!

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

Use standard reduction potentials to calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ}\), and \(K\) (at \(298 \mathrm{~K}\) ) for the reaction that is used in production of gold: $$2 \mathrm{Au}(\mathrm{CN})_{2}^{-}(a q)+\mathrm{Zn}(s) \longrightarrow 2 \mathrm{Au}(s)+\mathrm{Zn}(\mathrm{CN})_{4}^{2-}(a q)$$ The relevant half-reactions are $$\begin{aligned}\mathrm{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} & \mathscr{C}^{\circ} &=-0.60 \mathrm{~V} \\ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} & \mathscr{E}^{\circ} &=-1.26 \mathrm{~V} \end{aligned}$$

Draw all geometrical and linkage isomers of square planar \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2}(\mathrm{SCN})_{2}\right]\).

Molybdenum is obtained as a by-product of copper mining or is mined directly (primary deposits are in the Rocky Mountains in Colorado). In both cases it is obtained as \(\mathrm{MoS}_{2}\), which is then converted to \(\mathrm{MoO}_{3}\). The \(\mathrm{MoO}_{3}\) can be used directly in the production of stainless steel for high-speed tools (which accounts for about \(85 \%\) of the molybdenum used). Molybdenum can be purified by dissolving \(\mathrm{MoO}_{3}\) in aqueous ammonia and crystallizing ammonium molybdate. Depending on conditions, either \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Mo}_{2} \mathrm{O}_{7}\) or \(\left(\mathrm{NH}_{4}\right)_{6} \mathrm{Mo}_{7} \mathrm{O}_{24} \cdot 4 \mathrm{H}_{2} \mathrm{O}\) is obtained. a. Give names for \(\mathrm{MoS}_{2}\) and \(\mathrm{MoO}_{3}\). b. What is the oxidation state of Mo in each of the compounds mentioned above?

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

There are three salts that contain complex ions of chromium and have the molecular formula \(\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}\). Treating \(0.27 \mathrm{~g}\) of the first salt with a strong dehydrating agent resulted in a mass loss of \(0.036 \mathrm{~g}\). Treating \(270 \mathrm{mg}\) of the second salt with the same dehydrating agent resulted in a mass loss of \(18 \mathrm{mg}\). The third salt did not lose any mass when treated with the same dehydrating agent. Addition of excess aqueous silver nitrate to \(100.0-\mathrm{mL}\) portions of \(0.100 M\) solutions of each salt resulted in the formation of different masses of silver chloride; one solution yielded 1430 \(\mathrm{mg} \mathrm{AgCl} ;\) another, \(2870 \mathrm{mg} \mathrm{AgCl}\); the third, \(4300 \mathrm{mg} \mathrm{AgCl}\). Two of the salts are green and one is violet. Suggest probable structural formulas for these salts, defending your answer on the basis of the preceding observations. State which salt is most likely to be violet. Would a study of the magnetic properties of the salts be helpful in determining the structural formulas? Explain.

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.