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

The glycinate ion, \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2}^{-},\) formed by deprotonation of the amino acid glycine, can function as a bidentate ligand, coordinating to a metal through the nitrogen of the amino group and one of the oxygen atoms. (equation can't copy) A copper complex of this ligand has the formula \(\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} .\) For this complex, determine the following. (a) the oxidation state of copper (b) the coordination number of copper (c) the number of unpaired electrons (d) whether the complex is diamagnetic or paramagnetic

Short Answer

Expert verified
(a) +2; (b) 6; (c) 1 unpaired electron; (d) Paramagnetic.

Step by step solution

01

Determine the oxidation state of copper

In the complex \( \text{Cu}\left(\text{H}_2\text{NCH}_2\text{CO}_2\right)_2\left(\text{H}_2\text{O}\right)_2 \), the glycinate ion \( \text{H}_2\text{NCH}_2\text{CO}_2^- \) acts as a bidentate ligand with a -1 charge per ligand. Since there are two glycinate ligands, the total charge contributed by these ligands is -2. The \( \text{H}_2\text{O} \) molecules are neutral. The whole complex is neutral, thus, the oxidation state of Cu must balance the -2 charge of the ligands, which makes the oxidation state of Cu \(+2\).
02

Determine the coordination number of copper

The glycinate ion acts as a bidentate ligand, coordinating to the metal at two points (one with the nitrogen and one with the oxygen atom). Since there are two glycinate ions, they provide four coordination sites. Additionally, there are two \( \text{H}_2\text{O} \) molecules, each coordinated with one point. Thus, the total coordination number of copper is \(6\).
03

Determine the number of unpaired electrons

Copper in the oxidation state \(+2\) has the electron configuration \([\text{Ar}] \, 3d^9\). For a \(3d^9\) configuration, there is one unpaired electron present.
04

Determine the magnetic property of the complex

Since the copper complex has one unpaired electron, it exhibits paramagnetic properties, as paramagnetism is associated with unpaired electrons.

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 State
In coordination chemistry, understanding the oxidation state of a metal in a complex is crucial. It helps determine the element's overall charge and its interaction with the ligands bound to it. In our copper complex, \[ \mathrm{Cu}\left(\mathrm{H}_2 \mathrm{NCH}_2 \mathrm{CO}_2\right)_2\left(\mathrm{H}_2 \mathrm{O}\right)_2, \]the oxidation state of copper is deduced based on the charges of the ligands and the overall charge of the complex.
  • The glycinate ion, functioning as a bidentate ligand, carries a \(-1\) charge.
  • There are two such ligands, providing a total charge of \(-2\).
  • The water molecules are neutral and do not add any charge.
The whole complex is neutral, meaning the oxidation state of the copper has to balance the \(-2\) from the ligands, making the copper's oxidation state \(+2\). Recognizing the oxidation state allows us to understand the electron configuration and magnetic properties of the metal center.
Coordination Number
The coordination number in a complex is the count of how many atoms are directly bonded to the central metal ion. Knowing this helps in predicting the geometry and structure of the complex. In our copper complex:
The glycinate ion acts as a bidentate ligand.
  • Each glycinate ligand binds at two points through a nitrogen atom and an oxygen atom from the carboxylate group.
Since there are two glycinate ions, they provide four coordination sites to the copper,
In addition to two coordination sites from the water molecules,
Thus, the total coordination number of copper is 6, typically resulting in an octahedral geometry.
Bidentate Ligand
Bidentate ligands are unique because they can attach to a metal ion at two separate points. This creates a more stable complex by forming a chelate ring with the metal.
In the given exercise, the glycinate ion is bidentate.
  • It uses the nitrogen atom from the amino group.
  • And one oxygen atom from the carboxylate group to coordinate with the metal ion.
This dual attachment not only gives rise to its bidentate nature but also majorly contributes to the coordination number and overall stability of the complex. Bidentate ligands are especially significant in coordination chemistry for their tendency to form chelate rings, increasing the stability of the complex.
Paramagnetism
The magnetic properties of a coordination complex are largely dictated by the presence of unpaired electrons. Complexes with unpaired electrons exhibit paramagnetism,
which means they are attracted to a magnetic field. In our copper complex,
the copper ion has an oxidation state of \(+2\), leading to an electron configuration of \[ \text{[Ar]} \, 3d^9. \]With one unpaired electron present in this configuration, the complex exhibits paramagnetic behavior. This property can be measured using magnetic susceptibility techniques and provides insight into the electronic structure of the metal center.

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

Which of the following complexes is (are) square planar? (a) \(\left[\mathrm{Ti}(\mathrm{CN})_{4}\right]^{2-}\) (b) \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) (c) \(\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{2-}\) (d) \(\left[\mathrm{Pt}(\mathrm{CN})_{4}\right]^{2-}\)

A The transition metals form a class of compounds called metal carbonyls, an example of which is the teurahedral complex \(\mathrm{Ni}(\mathrm{CO})_{4}\). Given the following thermodynamic data: $$\begin{array}{lcc} \hline & \Delta H_{f}^{\circ}(\mathrm{kJ} / \mathrm{mol}) & \mathrm{S}^{\circ}(\mathrm{J} / \mathrm{K} \cdot \mathrm{mol}) \\ \hline \mathrm{Ni} & 0 & 29.87 \\ \mathrm{CO}(\mathrm{g}) & -110.525 & +197.674 \\ \mathrm{Ni}(\mathrm{CO})_{4}(\mathrm{g}) & -602.9 & +410.6 \\ \hline \end{array}$$ (a) Calculate the equilibrium constant for the formation of \(\mathrm{Ni}(\mathrm{CO})_{4}(\mathrm{g})\) from nickel metal and CO gas. (b) Is the reaction of \(\mathrm{Ni}(\mathrm{s})\) and \(\mathrm{CO}(\mathrm{g})\) product-or reactant-favored? (c) Is the reaction more or less product-favored at higher temperatures? How could this reaction be used in the purification of nickel metal?

A The complex ion \(\left[\mathrm{Co}\left(\mathrm{CO}_{3}\right)_{3}\right]^{3-},\) an octahedral complex with bidentate carbonate ions as ligands, has one absorption in the visible region of the spectrum at \(640 \mathrm{nm}\). From this information: (a) Predict the color of this complex, and explain your reasoning. (b) Is the carbonate ion as weak- or strong-field ligand? (c) Predict whether \(\left[\mathrm{Co}\left(\mathrm{CO}_{3}\right)_{3}\right]^{3-}\) will be paramagnetic or diamagnetic.

Give the formula of a complex constructed from one \(\mathrm{Cr}^{3+}\) ion, two ethylenediamine ligands, and two ammonia molecules. Is the complex neutral or is it charged? If charged, give the charge.

Four isomers are possible for \(\left[\mathrm{Co (\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{+}\) Draw the structures of all four. (Two of the isomers are chiral, meaning that each has a non-superimposable mirror image.)

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