Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Give the oxidation number of the metal ion in each of the following compounds. (a) \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{SO}_{4}\) (c) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}\) (b) \(\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\) (d) \(\mathrm{Cr}(\mathrm{en})_{2} \mathrm{Cl}_{2}\)

Short Answer

Expert verified
(a) Mn = +2; (b) Co = +3; (c) Co = +3; (d) Cr = +2.

Step by step solution

01

Understanding Coordination Compounds

Coordination compounds consist of a central metal ion bonded to ligands. The oxidation number of the metal ion can be determined by considering the charges of the ligands and the overall charge of the compound.
02

Calculate oxidation number for Mn in [Mn(NH3)6]SO4

The complex ion is \([\mathrm{Mn(NH_3)_6}]^{2+}\) since \(\mathrm{SO_4}^{2-}\) has a charge of \(-2\). Ammonia (\(\mathrm{NH_3}\)) carries no charge. The oxidation number of \(\mathrm{Mn}\) must balance the \(+2\) charge of the ion: \(\text{Oxidation number of Mn} = +2\).
03

Calculate oxidation number for Co in K3[Co(CN)6]

The complex ion is \([\mathrm{Co(CN)_6}]^{3-}\) because each \(\mathrm{K}^+\) has a \(+1\) charge and there are three \(\mathrm{K}\) ions. Cyanide (\(\mathrm{CN}^{-}\)) has a charge of \(-1\) each. The \(\mathrm{Co}\) oxidation number must balance the \(-6\) total from the cyanides to result in \(-3\): \(\text{Oxidation number of Co} = +3\).
04

Calculate oxidation number for Co in [Co(NH3)4Cl2]Cl

The complex ion is \([\mathrm{Co(NH_3)_4Cl_2}]^{+}\) since \(\mathrm{Cl}^{-}\) has a charge of \(-1\). Ammonia (\(\mathrm{NH_3}\)) is neutral while chloride (\(\mathrm{Cl}^{-}\)) has a \(-1\) charge, contributing \(-2\) total. The oxidation number of \(\mathrm{Co}\) must balance to result in \(+1\): \(\text{Oxidation number of Co} = +3\).
05

Calculate oxidation number for Cr in Cr(en)2Cl2

The compound has no net charge. Ethylenediamine (en) is neutral, contributing no charge, while each chloride (\(\mathrm{Cl}^{-}\)) contributes \(-1\), totaling \(-2\). The oxidation number of \(\mathrm{Cr}\) must balance the \(-2\) to zero: \(\text{Oxidation number of Cr} = +2\).

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!

Key Concepts

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

Oxidation Number Calculation
In coordination chemistry, calculating the oxidation number of a metal ion is crucial for understanding the nature of the compound. The oxidation number represents the total charge the metal would have if all ligands were removed along with the pairs of electrons they share with the metal. To determine this number, consider the following steps:
  • Identify the overall charge of the complex, if provided; this charge is often indicated at the end of the chemical formula inside square brackets.
  • Determine the charges on the ligands attached to the metal. Each ligand can either be neutral or charged. Common ligands include ammonia (\(\mathrm{NH_3}\)), which is neutral, and chloride (\(\mathrm{Cl^-}\)), which carries a \(-1\) charge.
  • Calculate the total charge contribution by the ligands. This is done by multiplying the individual ligand's charge by its quantity in the complex.
  • Use the formula: \[\text{Total Charge} = \text{Metal Oxidation Number} + \text{Total Ligand Charges}\] to find the oxidation number of the metal.
Understanding this process allows one to predict the chemical behavior of the compound and its possible reactions.
Transition Metal Complexes
Transition metal complexes play a vital role in coordination chemistry. These are compounds where a central metal ion binds to surrounding molecules or ions, termed ligands. The transition metals, found in the middle of the periodic table, possess partially filled d orbitals, which allow them to form various structures and participate in unique reactions.
Transition metal complexes can exhibit different coordination numbers which refer to the number of ligand bonds that the central metal can accommodate. For example, a coordination number of 6, as seen in \([\mathrm{Mn(NH_3)_6}]^{2+}\), denotes a typical octahedral geometry.
These complexes possess unique properties such as color and magnetism due to the electron transitions in their d orbitals. This quality makes them crucial in industrial applications and biological systems, like hemoglobin in the human body.
Ligands and Charges
Ligands are atoms, ions, or molecules that donate pairs of electrons to the central metal ion in a coordination complex. The effectiveness of a ligand is often determined by its charge and the availability of electrons to share.
  • Neutral ligands, like \(\mathrm{NH_3}\) (ammonia) and \(\mathrm{en}\) (ethylenediamine), carry no net charge and thus do not contribute to the net charge of the complex.
  • Anionic ligands, such as \(\mathrm{Cl^-}\) (chloride) and \(\mathrm{CN^-}\) (cyanide), carry negative charges and will increase the net charge of the complex negatively.
The role of these ligands extends beyond charge balance; they significantly affect the metal's oxidation state, geometry, and overall reactivity. In complex \([\mathrm{Co(NH_3)_4Cl_2}^+]\), the ammonia ligands are neutral, while chloride ions contribute negatively, influencing the cobalt's oxidation state. Thus, understanding ligands and their charges is crucial for predicting the stability and chemical behavior of coordination complexes.

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

In this question, we explore the differences between metal coordination by monodentate and bidentate ligands. Formation constants, \(K_{t}\), for \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}(\mathrm{aq})\) and \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}(\mathrm{aq})\) are as follows: \(\mathrm{Ni}^{2+}(\mathrm{aq})+6 \mathrm{NH}_{3}(\mathrm{aq}) \longrightarrow\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}(\mathrm{aq}) \quad K_{\mathrm{f}}=10^{8}\) \(\mathrm{Ni}^{2+}(\mathrm{aq})+3 \mathrm{en}(\mathrm{aq}) \longrightarrow\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}(\mathrm{aq})\) \(K_{f}=10^{18}\) The difference in \(K_{f}\) between these complexes indicates a higher thermodynamic stability for the chelated complex, caused by the chelate effect. Recall that \(K\) is related to the standard free energy of the reaction by \(\Delta_{r} G^{\circ}=-R T \ln K\) and \(\Delta_{r} G^{\circ}=\) \(\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ} .\) We know from experiment that \(\Delta_{t} H^{\circ}\) for the \(\mathrm{NH}_{3}\) reaction is \(-109 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn}\) and \(\Delta_{i} H^{\circ}\) for the ethylenediamine reaction is \(-117 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn} .\) Is the difference in \(\Delta_{r} H^{\circ}\) suffi- cient to account for the \(10^{10}\) difference in \(K_{f} ?\) Comment on the role of entropy in the second reaction.

Determine the number of unpaired electrons in the following tetrahedral complexes. All tetrahedral complexes are high spin. (a) \(\left[\mathrm{FeCl}_{4}\right]^{2-}\) (c) \(\left[\mathrm{MnCl}_{4}\right]^{2}\) (b) \(\mathrm{Na}_{2}\left[\mathrm{CoCl}_{4}\right]\) \((\mathrm{d})\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{ZnCl}_{4}\right]\)

For the high-spin coordination compound \(\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2},\) identify the following. (a) the oxidation number of manganese (b) the coordination number for manganese (c) the coordination geometry for manganese (d) the number of unpaired electrons per metal atom (e) whether the complex is diamagnetic or paramagnetic (f) the number of geometric isomers

From experiment, we know that \(\left[\mathrm{CoF}_{6}\right]^{3-}\) is paramagnetic and \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) is diamagnetic. Using the ligand field model, depict the electron configuration for each ion, and use this model to explain the magnetic property. What can you conclude about the effect of these ligands on the magnitude of \(\Delta_{0} ?\)

In water, the chromium(II) ion, \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) absorbs light with a wavelength of about \(700 \mathrm{nm}\). What color is the solution?

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