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

Chemists tried for a long time to make molecular compounds containing silicon- silicon double bonds, they finally succeeded in 1981 . The trick is having large, bulky R groups on the silicon atoms to make \(\mathrm{R}_{2} \mathrm{Si}=\mathrm{SiR}_{2}\) compounds. What experiments could you do to prove that a new compound has a silicon-silicon double bond rather than a silicon-silicon single bond?

Short Answer

Expert verified
To prove that a new compound has a silicon-silicon double bond rather than a single bond, a combination of X-Ray crystallography, infrared (IR) spectroscopy, and nuclear magnetic resonance (NMR) spectroscopy can be used. X-Ray crystallography can show differences in bond lengths, with double bonds being shorter than single bonds. IR spectroscopy can reveal unique absorption frequencies associated with double bonds compared to single bonds. NMR spectroscopy can provide information about the bonding environment of the silicon atoms, and signals consistent with a double bond support the presence of a silicon-silicon double bond in the compound.

Step by step solution

01

1. X-Ray Crystallography

: X-Ray crystallography is a technique used to determine the atomic and molecular structure of a crystal. This experiment can help prove that the new compound has a silicon-silicon double bond by showing the differences in bond lengths between single and double bonds. The bond length for a silicon-silicon double bond is shorter than that of a single bond. If the new compound has a shorter bond length than the one typically observed for a silicon-silicon single bond, it provides evidence that the compound has a silicon-silicon double bond.
02

2. Infrared Spectroscopy

: Infrared (IR) spectroscopy is an experimental technique that measures the vibrational frequencies of molecules, which corresponds to the stretching and bending of their bonds. Different types of bonds have characteristic IR absorption frequencies. The presence of a silicon-silicon double bond would exhibit a unique IR absorption frequency compared to a silicon-silicon single bond. Analyzing the IR spectrum of the new compound, we can verify the presence of a silicon-silicon double bond by comparing the observed absorption frequencies with known values for single and double silicon-silicon bonds.
03

3. Nuclear Magnetic Resonance (NMR) Spectroscopy

: NMR spectroscopy is a technique that provides information about the chemical environment and connectivity of atoms in a molecule. The NMR spectrum of a compound with a silicon-silicon double bond would differ from that of a compound with a silicon-silicon single bond. By analyzing the chemical shifts and coupling patterns of silicon NMR signals, we can infer the bonding environment around the silicon atoms. If the NMR signals in the new compound are consistent with the presence of a silicon-silicon double bond, this supports the idea that the new compound has a double bond rather than a single bond. In conclusion, a combination of X-Ray crystallography, IR spectroscopy, and NMR spectroscopy experiments can provide evidence for the presence of a silicon-silicon double bond in the new compound. Comparing results from these experiments to known data for single and double silicon-silicon bonds can confirm the bond type in the synthesized compound.

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.

X-Ray Crystallography
X-Ray Crystallography is a powerful technique utilized to unveil the atomic and molecular makeup of crystals. Chemists highly value it, especially when verifying the presence of silicon-silicon bonds. To detect a silicon-silicon double bond, scientists often rely on comparing bond lengths. Generally, a double bond is shorter than a single bond.
To conduct this experiment, a crystal of the compound is exposed to X-rays. The X-rays are diffracted by the crystals, creating a pattern that can be captured on film or an electronic detector. This pattern is then analyzed to reveal the structure of the crystal, including the lengths of the bonds between atoms.
The crucial clue here is the bond length. If the measured silicon-silicon bond length is shorter than the typical length of a silicon-silicon single bond, this indicates the presence of a double bond. This method offers a visual map of the atomic structure, providing clear evidence to support or refute the existence of a silicon-silicon double bond.
Infrared Spectroscopy
Infrared Spectroscopy (IR) is a key analytical technique used to investigate molecular vibrations, identifying unique absorption bands corresponding to different types of chemical bonds. When looking for a silicon-silicon double bond, this technique proves invaluable because each bond type in a molecule has a characteristic absorption frequency in the IR spectrum.
During an IR experiment, an IR beam passes through the compound, allowing the molecule's bonds to absorb energy at specific frequencies, causing them to vibrate. By recording this absorption, spectra are produced, reflecting distinct peaks based on the bond types present.
  • The stretching frequency of a silicon-silicon double bond differs from that of a single bond.
  • Analyzing these frequencies helps identify the specific type of bond present.
Comparing the experimental data to standard values for absorption of silicon-silicon bonds enables chemists to confirm the presence of a double bond. A molecule containing such a bond will exhibit unique absorption peaks that are distinguishable from those of single bonds.
Nuclear Magnetic Resonance Spectroscopy
Nuclear Magnetic Resonance (NMR) Spectroscopy is another valuable tool for exploring the structural details of molecules. This technique is particularly useful for understanding the chemical environment around atoms, such as the silicon atoms in compounds suspected of containing silicon-silicon double bonds.
NMR works by detecting magnetic properties of certain atomic nuclei. When placed in a magnetic field, nuclei such as those in silicon atoms can absorb radiofrequency radiation, resulting in an NMR spectrum. This spectrum contains peaks that correspond to the chemical shifts of different atoms, revealing their environment and connectivity.
The presence of a silicon-silicon double bond affects these shifts and the coupling patterns in the spectrum.
  • Evaluation of silicon NMR signals helps uncover the bonding scenario.
  • Consistent chemical shifts and coupling patterns associated with double bonds offer strong evidence to suggest their existence.
In essence, by studying the NMR spectrum, chemists can visualize the silicon connectivity and confirm whether a silicon-silicon bond is indeed a double bond.

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

Consider the clements \(\mathrm{O}, \mathrm{Ba}, \mathrm{Co}, \mathrm{Bc}, \mathrm{Br}\), and Se. From this list, select the element that (a) is most electronegative, (b) exhibits a maximum oxidation state of \(+7\), (c) loses an clcctron most rcadily, (d) forms \(\pi\) bonds most readily, (c) is a transition metal, ( \(f\) ) is a liquid at room temperature and pressure.

Explain why \(\mathrm{SO}_{2}\) can be used as a reducing agent but \(\mathrm{SO}_{3}\) cannot.

SOLUTION Analyze We must write balanced nuclear equations in which the masses and charges of reactants and products are equal. Plan We can begin by writing the complete chemical symbols for the nuclei and decay particles that are given in the problem. Solve (a) The information given in the question can be summarized as $$ { }_{80}^{201} \mathrm{Hg}+{ }_{-1}^{0} \mathrm{e} \longrightarrow{ }_{Z}^{A} \mathrm{X} $$ The mass numbers must have the same sum on both sides of the equation: $$ 201+0=A $$ Thus, the product nucleus must have a mass number of 201. Similarly, balancing the atomic numbers gives $$ 80-1=Z $$ Thus, the atomic number of the product nucleus must be 79 , which identifies it as gold (Au): $$ { }_{80}^{201} \mathrm{Hg}+{ }_{-1}^{0} \mathrm{e} \longrightarrow{ }_{79}^{201} \mathrm{Au} $$ (b) In this case we must determine what type of particle is emitted in the course of the radioactive decay: $$ { }_{90}^{231} \mathrm{Th} \longrightarrow{ }_{91}^{231} \mathrm{~Pa}+{ }_{\mathrm{Z}}^{\mathrm{X}} \mathrm{X} $$ From \(231=231+A\) and \(90=91+Z\), we deduce \(A=0\) and \(Z=-1\). According to Table \(21.2\), the particle with these characteristics is the beta particle (electron). We therefore write $$ { }_{90}^{231} \mathrm{Th} \longrightarrow{ }_{91}^{231} \mathrm{~Pa}+{ }_{-1}^{0} \mathrm{e} \text { or } \quad{ }_{90}^{231} \mathrm{Th} \longrightarrow{ }_{91}^{231} \mathrm{~Pa}+\beta^{-} $$

Consider the elements L.i, K, Cl, C, Ne, and Ar. From this list, select the element that (a) is most clectronegative, (b) has the greatest metallic character, (c) most readily forms a positive ion, (d) has the smallest atomic radius, (c) forms \(\pi\) bonds most readily, (f) has multiple allotropes.

Carbon forms an unusual unstable oxide of formula \(\mathrm{C}_{3} \mathrm{O}_{2}\), called carbon suboxide. Carbon suboxide is made by using \(\mathrm{P}_{2} \mathrm{O}_{5}\) to dehydrate the dicarboxylic acid called malonic acid, which has the formula \(\mathrm{HOOC}-\mathrm{CH}_{2}-\mathrm{COOH}\). (a) Write a balanced reaction for the production of carbon suboxide from malonic acid. (b) How many grams of carbon suboxide could be made from \(20.00 \mathrm{~g}\) of malonic acid? (c) Suggest a Lewis structure for \(\mathrm{C}_{3} \mathrm{O}_{2}\) - (Hint: The Lewis structure of malonic acid suggests which atoms are connected to which.) (d) By using the information in Table 8.5, predict the \(\mathrm{C}-\mathrm{C}\) and \(\mathrm{C}-\mathrm{O}\) bond lengths in \(\mathrm{C}_{3} \mathrm{O}_{2}\). (e) Sketch the Lewis structure of a product that could result by the addition of \(2 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) to \(1 \mathrm{~mol}\) of \(\mathrm{C}_{3} \mathrm{O}_{2}\) -

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