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

Four-coordinate metals can have either a tetrahedral or a square-planar geometry; both possibilities are shown here for \(\left[\mathrm{Pt} \mathrm{Cl}_{2}\left(\mathrm{NH}_{3}\right)_{2}\right] .(\mathbf{a})\) What is the name of this molecule? (b) Would the tetrahedral molecule have a geometric isomer? (c) Would the tetrahedral molecule be diamagnetic or paramagnetic? (d) Would the square- planar molecule have a geometric isomer? (e) Would the square-planar molecule be diamagnetic or paramagnetic? (f) Would determining the number of geometric isomers help you distinguish between the tetrahedral and square-planar geometries? (g) Would measuring the molecule's response to a magnetic field help you distinguish between the two geometries? [Sections \(23.4-23.6]\)

Short Answer

Expert verified
The name of the complex \(\left[\mathrm{Pt}\mathrm{Cl}_{2}\left(\mathrm{NH}_{3}\right)_{2}\right]\) is diamminedichloroplatinum(II). The tetrahedral geometry does not have geometric isomers and is diamagnetic, while the square-planar geometry has two geometric isomers (cis and trans) and is also diamagnetic. The presence of geometric isomers can help distinguish between the two geometries, but magnetic properties cannot.

Step by step solution

01

(a) Name of the molecule

The complex under consideration is \(\left[\mathrm{Pt}\mathrm{Cl}_{2}\left(\mathrm{NH}_{3}\right)_{2}\right]\). To name this complex, we follow the IUPAC rules, where we list the ligands alphabetically, followed by the central metal atom. In this case, our ligands are two chlorides (Cl) and two ammonia molecules (NH3). Therefore, the name of the molecule is "diamminedichloroplatinum(II)."
02

(b) Geometric isomerism in tetrahedral geometry

Geometric isomers are molecules with the same chemical formula but different spatial arrangements of their atoms. In tetrahedral geometry, all four ligand positions are equivalent, so no geometric isomers are possible for this molecule.
03

(c) Magnetic properties in tetrahedral geometry

In the tetrahedral geometry, the molecule would be diamagnetic, as platinum (II) has a d8 configuration: 1s² 2s² 2p6 3s² 3p6 3d10 4s² 4p6 4d8. All the d-electrons are thus paired, and there are no unpaired electrons. Diamagnetic substances have no unpaired electrons and are not attracted to a magnetic field.
04

(d) Geometric isomerism in square-planar geometry

In square-planar geometry, different spatial arrangements of the ligands are possible. For the molecule \(\left[\mathrm{Pt}\mathrm{Cl}_{2}\left(\mathrm{NH}_{3}\right)_{2}\right]\), there are two possible geometric isomers: cis and trans. In the cis isomer, both ammonia ligands are adjacent to each other, while in the trans isomer, the ammonia ligands are opposite to each other. So, the square-planar molecule has geometric isomers.
05

(e) Magnetic properties in square-planar geometry

In the square-planar geometry, the molecule would also be diamagnetic. The reason is the same as for the tetrahedral geometry - platinum (II) has a d8 configuration with all its d-electrons paired, hence there are no unpaired electrons.
06

(f) Distinguishing between tetrahedral and square-planar geometries using geometric isomers

As we have discussed earlier, the tetrahedral molecule doesn't have geometric isomers, while the square-planar molecule does have geometric isomers (cis and trans). Therefore, the presence of geometric isomers can be used to distinguish between tetrahedral and square-planar geometries of the complex.
07

(g) Distinguishing between tetrahedral and square-planar geometries using magnetic properties

As previously discussed, both the tetrahedral and the square-planar molecules are diamagnetic because no unpaired electrons are present in the platinum (II) d8 configuration. Therefore, measuring the magnetic properties of the molecule would not help in distinguishing between these two geometries, as both of them would be diamagnetic.

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.

Geometric Isomers
Geometric isomerism is a fascinating topic in coordination chemistry. It refers to the existence of two or more compounds with the same type and number of atoms but in different spatial arrangements. An analogy is having a set of the same LEGO blocks that can be put together in various ways. In coordination complexes, the arrangement of ligands around the central metal atom can lead to different geometric isomers. A classic example is seen in the square-planar complex, \(\left[\mathrm{Pt}\mathrm{Cl}_{2}\left(\mathrm{NH}_{3}\right)_{2}\right]\), which can exist as either cis or trans isomers:
  • In the **cis isomer**, similar ligands are adjacent to each other, forming a 'V' shape.
  • In the **trans isomer**, similar ligands are opposite each other, forming a '+' shape.
This isomerism is significant because each isomer can have distinct chemical and physical properties, such as different reactivities and colors.
Tetrahedral Geometry
Tetrahedral geometry is one of the simplest and most common types of geometries found in coordination compounds, particularly those with a coordination number of four. Think of it like a pyramid, with the metal atom at the center and four ligands at the corners. \(\left[\text{AB}_4\right]\) is a typical stoichiometry for such complexes.A key characteristic of tetrahedral complexes is their lack of geometric isomers. This is because all four positions around the central atom are equivalent, so rearranging them doesn't lead to new structures. In terms of bond angles, the ideal angle is 109.5 degrees, providing a symmetrical and three-dimensional structure. This geometry is often seen in high-spin complexes where the ligand field is weak, allowing electron pairing.
Square-Planar Geometry
Square-planar geometry is particularly characteristic of d8 metal ion complexes, like our platinum (II) complex here. Imagine a flat, square tabletop with the metal atom at the center and ligands at each corner. This geometry is common in transition metals, where the ligand field stabilizes the squared planar arrangement.An intriguing aspect of square-planar complexes is their ability to form geometric isomers. As mentioned earlier, \(\left[\mathrm{Pt}\mathrm{Cl}_{2}\left(\mathrm{NH}_{3}\right)_{2}\right]\) can exist in both cis and trans forms because the positions around the metal are not equivalent, allowing for structural variations. In terms of bond angles, the **90 degrees** between adjacent ligands leads to potential differences in chemical behavior and reactivity, depending on the isomerism.
Magnetic Properties
Understanding the magnetic properties of coordination complexes provides insight into their electronic structures. These properties depend on the presence or absence of unpaired electrons.For both tetrahedral and square-planar geometries involving platinum (II), the d8 electronic configuration plays a critical role. All d-electrons are paired, leading to a state known as **diamagnetism**. Diamagnetic substances are repelled by magnetic fields because they lack unpaired electrons.Consequently, regardless of the precise geometry, platinum (II) complexes like \(\left[\mathrm{Pt}\mathrm{Cl}_{2}\left(\mathrm{NH}_{3}\right)_{2}\right]\) would not exhibit attraction to a magnetic field. This uniformity in magnetic behavior makes it challenging to use magnetic properties alone to distinguish between these geometries, reaffirming why geometric isomerism may offer a clearer distinction.

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

The most important oxides of iron are magnetite, \(\mathrm{Fe}_{3} \mathrm{O}_{4}\), and hematite, \(\mathrm{Fe}_{2} \mathrm{O}_{3} .\) (a) What are the oxidation states of iron in these compounds? (b) One of these iron oxides is ferrimagnetic, and the other is antiferromagnetic. Which iron oxide is more likely to be ferrimagnetic? Explain.

Carbon monoxide is toxic because it binds more strongly to the iron in hemoglobin (Hb) than does \(\mathrm{O}_{2}\), as indicated by these approximate standard free-energy changes in blood: $$ \begin{aligned} \mathrm{Hb}+\mathrm{O}_{2} & \longrightarrow \mathrm{HbO}_{2} & \Delta G^{\circ}=-70 \mathrm{~kJ} \\ \mathrm{Hb}+\mathrm{CO} & \longrightarrow \mathrm{HbCO} & \Delta G^{\circ}=-80 \mathrm{~kJ} \end{aligned} $$ Using these data, estimate the equilibrium constant at 298 K for the equilibrium $$ \mathrm{HbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HbCO}+\mathrm{O}_{2} $$

Explain why the transition metals in periods 5 and 6 have nearly identical radii in each group.

Pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\), abbreviated py, is the molecule (a) Would you expect pyridine to act as a monodentate or bidentate ligand? (b) For the equilibrium reaction $$ \left[\mathrm{Ru}(\mathrm{py})_{4}(\mathrm{bipy})\right]^{2+}+2 \mathrm{py} \rightleftharpoons\left[\mathrm{Ru}(\mathrm{py})_{6}\right]^{2+}+\mathrm{bipy} $$ would you predict the equilibrium constant to be larger or smaller than one?

Solutions of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+},\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (both octahedral), and \(\left[\mathrm{CoCl}_{4}\right]^{2-}(\) tetrahedral) are colored. One is pink, one is blue, and one is yellow. Based on the spectrochemical series and remembering that the energy splitting in tetrahedral complexes is normally much less than that in octahedral ones, assign a color to each complex.

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