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Resonance structures for urea are represented below: The IR spectrum of free urea has absorptions at 3500 and \(3350\left(\nu\left(\mathrm{NH}_{2}\right)\right), 1683(\nu(\mathrm{CO}))\) and \(1471 \mathrm{cm}^{-1}\) \((\nu(\mathrm{CN})) .\) Urea can bond to metal ions through either an \(N-\) or \(O\) -donor atom. When urea bonds through the \(\mathrm{O}\) atom, the contribution from resonance form \(\mathbf{A}\) decreases. In the IR spectrum of [Pt(urea) \(\left._{6}\right] \mathrm{Cl}_{2}\) bands at 3390,3290,3130,3030,1725 and \(1395 \mathrm{cm}^{-1}\) are assigned to the vibrational modes of metal-bound urea. Suggest why these data suggest the formation of \(\mathrm{Pt}-\mathrm{N}\) rather than \(\mathrm{Pt}-\mathrm{O}\) bonds.

Short Answer

Expert verified
The increase in C=O wavenumber and significant C-N shift suggest Pt-N bonds form rather than Pt-O bonds.

Step by step solution

01

Understand Resonance Structures

Resonance structures refer to different ways of drawing the same molecule, showing the delocalization of electrons. In urea, there are resonance structures wherein either the nitrogen or oxygen might share electron pairs with the carbon. The key resonance forms involve a double bond between carbon and either nitrogen (through lone pair donation) or oxygen.
02

Analyze the IR Spectrum

IR spectroscopy helps identify the presence of particular bonds in a molecule by measuring the vibrational frequencies. Different functional groups absorb at characteristic wavenumbers. For free urea, the primary absorption bands are for N-H stretches at 3500 and 3350 cm鈦宦, a C=O stretch at 1683 cm鈦宦, and a C-N stretch at 1471 cm鈦宦.
03

Compare Metal-Complex IR Spectrum

For [Pt(urea)鈧哴Cl鈧, the IR spectrum shows shifts in absorption bands: N-H at 3390, 3290, and new bands for C=O at 1725 cm鈦宦, and C-N at 1395 cm鈦宦. These shifts imply changes in bonding due to coordination with the platinum metal ion.
04

Explain Shifts Indicating N-Coordination

The C=O stretch moves from 1683 to 1725 cm鈦宦. If the urea coordinated through oxygen, the C=O bond would weaken, lowering its wavenumber. The observed increase suggests that urea does not bond through oxygen. The significant shift in C-N from 1471 to 1395 cm鈦宦 supports N-coordination as C-N bonds experience more changes.
05

Conclude Coordination Type

Since the spectral data indicate no weakening of the C=O bond (but rather a strengthening), and significant interaction with the C-N bonds, it suggests that the bonding occurs through nitrogen rather than oxygen.

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Key Concepts

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

Resonance Structures
Resonance structures are essential in understanding how molecules like urea distribute and share their electrons. Picture urea as a molecule with multiple personalities, where resonance structures show different possible "personalities" urea might "wear." Electrons can be shared between different atoms, creating alternating double and single bonds. For urea, these resonance structures illustrate how electrons can be delocalized across the carbon, nitrogen, and oxygen atoms.
These structures are not real barriers but rather depict potential electron distributions. They affect how the molecule behaves chemically, especially how it coordinates with metals. When forming metal complexes, the specific resonance structure impacts which atom acts as the electron donor鈥攐xygen or nitrogen, determining how urea bonds to metal ions like platinum.
Vibrational Modes
In chemistry, vibrational modes refer to the ways molecules stretch, bend, or twist. Think of it like a dance, where each bond moves in its own rhythm. These movements are very small but crucial because they absorb specific infrared (IR) light frequencies. The IR spectrum acts like a fingerprint, showing us what bonds are in the molecule.
For free urea, we detect different vibrational modes corresponding to specific bonds. The nitrogen-hydrogen (N-H) bonds have their unique vibrations at 3500 and 3350 cm鈦宦, while the carbon-oxygen (C=O) bond vibrates at 1683 cm鈦宦. Each of these vibrations tells us about the molecule's structure and is crucial when comparing free urea to its metal-coordinated counterparts.
Metal-Urea Coordination
Metal-urea coordination focuses on how urea interacts with metal ions, like platinum, to form complexes. Urea is flexible and can bond through either its nitrogen or oxygen atoms. This choice significantly influences the IR spectral shifts.
When urea bonds with metals through nitrogen, it forms a strong link called a coordinate bond. This bond type impacts the vibrational modes we can detect. The alternative鈥攂onding through oxygen鈥攚ould differently affect these vibrational frequencies. The practical implication of this bonding choice is evident in how dramatically the vibrational modes shift when urea forms a complex, revealing the likely coordination pathway.
Spectral Shifts in IR
Spectral shifts in IR spectroscopy give insights into what happens when urea forms complexes with metals. Shifting frequencies indicate changes in bond strengths and interactions. For [Pt(urea)鈧哴Cl鈧, key shifts in C=O and C-N frequencies are observed, signaling how urea coordinates to platinum.
Normally, if urea bonded through oxygen, the C=O bond would weaken, lowering the vibration frequency. However, in this complex, the C=O frequency increases from 1683 to 1725 cm鈦宦, which implies strengthening, not weakening. It strengthens the argument for nitrogen coordination, shown by the decrease in C-N frequency from 1471 to 1395 cm鈦宦. These shifts are like clues, guiding chemists to recognize more about molecular bonding dynamics in metal complexes.

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Most popular questions from this chapter

\(\mathrm{NaBH}_{4}\) contains the tetrahedral \(\left[\mathrm{BH}_{4}\right]^{-}\) ion. Although \(\mathrm{NaBH}_{4}\) hydrolyses slowly in water, it is possible to obtain a clean \(^{1} \mathrm{H}\) NMR spectrum of the compound in \(\mathrm{D}_{2} \mathrm{O}\). Naturally occurring boron consists of two isotopes: \(^{11} \mathrm{B}, 80.1 \%, I=\frac{3}{2},\) and \(^{10} \mathrm{B}, 19.9 \%, I=3\) Assuming that a sharp, well-resolved spectrum is obtained, sketch the expected \(400 \mathrm{MHz}^{1} \mathrm{H}\) NMR spectrum (including a scale) if the signal for the protons occurs at \(\delta-0.2 \mathrm{ppm},\) and the values of \(J_{11} \mathrm{B}_{1 \mathrm{H}}=80.5 \mathrm{Hz}\) and \(J_{10 \mathrm{B} 1 \mathrm{H}}=27.1 \mathrm{Hz} .\) How would the spectrum differ if it were recorded at \(100 \mathrm{MHz} ?\)

On being heated, the fullerene solvate \(\mathrm{C}_{60} \cdot x \mathrm{CHBr}_{3}\) loses solvent in a two-step process. The final weight loss is \(41 \% .\) Account for these data and determine \(x\).

How might you use \(^{31} \mathrm{P}\) NMR spectroscopy to distinguish between \(\mathrm{Ph}_{2} \mathrm{PH}\) and \(\mathrm{Ph}_{3} \mathrm{P} ?\)

The \(^{31} \mathrm{P}\) NMR spectrum of \(\mathrm{PMe}_{3}\) consists of a binomial decet \((J \quad 2.7 \mathrm{Hz})\) (a) Account for this observation. (b) Predict the nature of the \(^{1} \mathrm{H}\) NMR spectrum of PMe \(_{3}\)

(a) Explain why the \(^{19} \mathrm{F}\) NMR spectrum of \(\left[\mathrm{PF}_{6}\right]^{-}\) appears as a doublet. (b) The \(^{31} \mathrm{P}\left\\{^{1} \mathrm{H}\right\\} \mathrm{NMR}\) spectrum of \(t r a n s-\left[\mathrm{PtI}_{2}\left(\mathrm{PEt}_{3}\right)_{2}\right]\) (4.20) shows a three-line pattern, the lines in which have relative integrals of \(\approx 1: 4: 1 .\) What is the origin of this pattern?

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