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Chapter 23: Problem 46

Give systematic names for the following formulas: (a) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{NO}_{2}\right)_{2}\right] \mathrm{Cl}\) (b) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]\) (c) \(\mathrm{K}_{2}\left[\mathrm{CuCl}_{4}\right]\)

Short Answer

Expert verified
tetraamminenitrocobalt(III) chloride

Step by step solution

01

Determine the central metal ion and its oxidation state for compound (a)

Identify the central metal ion Co (Cobalt). Four NH3 (ammine) ligands, two NO2 (nitro) ligands, and one Cl (chloride) counter-ion are attached to Cobalt. Since NH3 is neutral and each NO2 ligand has a charge of -1, let the oxidation state of Co be x. The equation for the oxidation state is: x + 4(0) + 2(-1) + (-1) = 0, solving for x, we get: x = +3.
02

Name the ligands for compound (a)

Identify and list the ligands in the coordination sphere. NH3 ligands are named as 'ammine', and NO2 ligands are named as 'nitro'. The overall compound contains four ammine and two nitro ligands.
03

Combine the names for compound (a)

The order of ligands is in alphabetical order: Ammine before nitro. The central metal with its oxidation state and the counter ion at the end:

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Key Concepts

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

Coordination Chemistry
Coordination chemistry studies compounds formed between metal ions and ligands. These compounds are called coordination compounds or complexes.
Metal ions in these complexes typically have empty orbitals that can accept electron pairs from ligands. Ligands are ions or molecules that donate these electron pairs to the metal ion.
  • The central metal ion: This is the metal atom that forms the core of the coordination complex.
  • Ligands: These are ions or molecules that bond to the central metal. They can be atoms, ions, or molecules with lone pairs of electrons.
  • Coordination sphere: This includes the central metal ion and its attached ligands, but not the counter ions.
Coordination compounds can display a rich variety of colors, magnetic properties, and reactivities due to the nature of the metal and the types of ligands attached to it.
Oxidation States
The oxidation state is crucial for naming coordination compounds. It represents the charge of the central metal ion if all ligands were removed along with electron pairs they had shared with the metal.
To calculate the oxidation state:
  • Identify the charges of the ligands in the complex.
  • Use an equation where the sum of the charge contributions from the ligands and the oxidation state of the metal equals the overall charge of the complex.
For example, in the compound \(\text{[Co(NH}_3\text{)}_4\text{(NO}_2\text{)}_2\text{]Cl}\):
\(x + 4(0) + 2(-1) + (-1) = 0\)
Solving for \(x\) (the oxidation state of Co), we get \(x = +3\).
Remember, knowing the oxidation state helps in accurately naming the complex and understanding its reactivity and properties.
Naming Ligands
Ligands determine how coordination compounds are named. Here’s a quick primer on naming ligands:
  • Neutral ligands like \(NH_3\) (ammine) and \(H_2O\) (aqua) are named based on their common names.

  • Anionic ligands, such as \(\text{NO}_2\) (nitro) and \(\text{Cl}^-\) (chloro), usually end in \'o\'.

When naming the entire coordination complex:
  • Name the ligands in alphabetical order before naming the central metal ion.
  • Use prefixes like 'tetra-' (4), 'hexa-' (6) to indicate the number of each type of ligand.
  • State the oxidation state of the metal in Roman numerals within parentheses.
For instance, \(\text{[CuCl}_4\text{]}^{2-}\) is named 'tetrachlorocuprate(II)'. This systematic, rule-based naming helps chemists easily identify and describe complex compounds accurately.

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Most popular questions from this chapter

Draw orbital-energy splitting diagrams and use the spectrochemical series to show the orbital occupancy for each of the following (assuming that \(\mathrm{H}_{2} \mathrm{O}\) is a weak-field ligand): (a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) (b) \(\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}\) (c) \(\left[\mathrm{FeF}_{6}\right]^{3-}\)

Palladium, like its group neighbor platinum, forms fourcoordinate \(\mathrm{Pd}(\mathrm{II})\) and \(\mathrm{six}\) -coordinate \(\mathrm{Pd}(\mathrm{IV})\) complexes. Write formulas for the complexes with these compositions: (a) \(\mathrm{PdK}\left(\mathrm{NH}_{3}\right) \mathrm{Cl}_{3}\) (b) \(\mathrm{PdCl}_{2}\left(\mathrm{NH}_{3}\right)_{2}\) (c) \(\mathrm{PdK}_{2} \mathrm{Cl}_{6}\) (d) \(\operatorname{Pd}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{4}\)

(a) What is the crystal field splitting energy \((\Delta) ?\) (b) How does it arise for an octahedral field of ligands? (c) How is it different for a tetrahedral field of ligands?

Why are there both high-spin and low-spin octahedral complexes but only high- spin tetrahedral complexes?

How many \(d\) electrons \(\left(n\right.\) of \(\left.d^{n}\right)\) are in the central metal ion in (a) \(\left[\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{SO}_{4}\) (b) \(\mathrm{Na}_{2}\left[\mathrm{Os}(\mathrm{CN})_{6}\right] ;\) (c) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{CO}_{3} \mathrm{I}\right] ?\)
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