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(a) Write down the formula of cryolite. (b) Write down the formula of perovskite. (c) Cryolite is described as possessing a 3 -dimensional structure closely related to that of perovskite. Suggest how this is possible when the stoichiometries of the two compounds do not appear to be compatible.

Short Answer

Expert verified
Cryolite (Na3AlF6) and perovskite (CaTiO3) can have similar 3D structures despite different formulas due to their cubic lattice formations.

Step by step solution

01

Write the Formula of Cryolite

The chemical formula for cryolite is \( \text{Na}_3\text{AlF}_6 \). Cryolite is a mineral composed of sodium, aluminum, and fluoride ions.
02

Write the Formula of Perovskite

The chemical formula for perovskite is \( \text{CaTiO}_3 \). Perovskite is a mineral made up of calcium, titanium, and oxygen atoms.
03

Compare and Analyze Structures

Despite the different formulas, cryolite (\( \text{Na}_3\text{AlF}_6 \)) and perovskite (\( \text{CaTiO}_3 \)) can share a related structure due to the arrangement of their component ions in a cubic lattice form. Both structures can be seen as variations on a theme of corner-sharing octahedra, usually with varying size and charge distribution allowing for 3-dimensional network structures. Cryolite, through lattice substitutions or defects, can adopt similar 3D connectivity.

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Key Concepts

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

Cryolite Formula
Cryolite is an interesting mineral with the chemical formula \( \text{Na}_3\text{AlF}_6 \). It is made up of sodium (Na), aluminum (Al), and fluoride (F) ions. These elements form a specific arrangement where the sodium and aluminum ions are surrounded by fluoride ions.

This arrangement leads to cryolite's unique properties, such as a low melting point, which is particularly useful in aluminum processing. In industrial applications, cryolite is often used in the production of aluminum through electrolysis. The fluoride ions play a crucial role by allowing the aluminum ions to be released more easily during the process.

The structure corresponds to a three-dimensional network where each aluminum atom is surrounded by six fluoride ions, forming an octahedron. The sodium ions fill the gaps within this framework, ensuring a stable structure. This complex structure makes cryolite highly efficient for its specific uses.
Perovskite Structure
Perovskite is another fascinating mineral with the formula \( \text{CaTiO}_3 \). It consists of calcium (Ca), titanium (Ti), and oxygen (O) atoms, arranged in a distinct structure known as the perovskite structure. This structure is characterized by a three-dimensional framework of corner-sharing octahedra.

In the perovskite structure, the titanium ions are surrounded by six oxygen ions, forming an octahedral shape. These octahedra are connected at the corners, creating a continuous network. The calcium ions fill the spaces in between this network, providing structural stability.

It's important to note that the perovskite structure is adaptable and can accommodate a wide variety of elements in its crystal lattice. This adaptability is why perovskites are used in many advanced technologies, such as solar cells and superconductors. Their ability to host various atoms while maintaining structural integrity makes them highly versatile.
3D Network Structures
Despite cryolite and perovskite having different chemical formulas, they share a remarkable structural similarity in their three-dimensional (3D) network arrangements. Both minerals exhibit a pattern of corner-sharing octahedra that contributes to their robust 3D frameworks.

In cryolite, the octahedra are composed of aluminum and fluoride ions, while in perovskite, they consist of titanium and oxygen ions. The key to these structures is how the octahedra connect at the corners, enabling a stable and extensive network.

Such networks are crucial because they allow for the accommodation of different ions and lattice arrangements. These characteristics are what enable cryolite to potentially mimic perovskite's structure through various ionic substitutions or structural defects, aligning their 3D connectivity.

This ability to adapt while maintaining a stable structure makes 3D network structures crucial not just from a chemistry perspective but also for practical applications where strength and stability are required, such as in building materials, electronic devices, and energy technologies.

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Most popular questions from this chapter

(a) \(\mathrm{NMR}\) spectroscopic data for \(\left[\mathrm{HAl}\left(\mathrm{BH}_{4}\right)_{2}\right]_{n}\) are consistent with the compound existing in two forms in solution. One form is probably a dimer and the other, a higher oligomer. Each species possesses one boron environment, and in the \(^{11} \mathrm{B}\) NMR spectrum, each species exhibits a binomial quintet. The chemical shift of the signal for each species in the \(^{27} \mathrm{Al}\) NMR spectrum suggests an octahedral environment for the Al atom. Suggest a structure for the dimer \(\left[\mathrm{HAl}\left(\mathrm{BH}_{4}\right)_{2}\right]_{2}\) which is consistent with these observations and comment on whether the data indicate a static or dynamic molecule. (b) The elemental analysis for an adduct \(\mathbf{A}\) is \(15.2 \%\) B \(75.0 \%\) Cl, \(4.2 \%\) C and \(5.6 \%\) O. The \(^{11}\) B NMR spectrum of A contains two singlets \((\delta-20.7\) and \(+68.9 \mathrm{ppm}\) ) with relative integrals \(1: 3 ;\) the signal at \(\delta-20.7\) ppm is characteristic of a \(\mathrm{B}\) atom in a tetrahedral environment, while that at \(\delta+68.9\) ppm is consistent with trigonal planar boron. In the IR spectrum, there is a characteristic absorption at \(2176 \mathrm{cm}^{-1} .\) Suggest an identity for \(\mathbf{A}\) and draw its structure.

Write a brief account of the bonding and reactivity of borazine which emphasizes the ways in which this compound is similar or dissimilar to benzene.

The ordering of the relative stabilities of adducts \(\mathrm{L} \cdot \mathrm{BH}_{3}\) for some common adducts is, according to L: \(\mathrm{Me}_{2} \mathrm{O}<\mathrm{THF}<\mathrm{Me}_{2} \mathrm{S}<\mathrm{Me}_{3} \mathrm{N}<\mathrm{Me}_{3} \mathrm{P}<\mathrm{H}^{-}\). In addition to answering each of the following, indicate how you could use NMR spectroscopy to confirm your proposals. (a) What happens when \(\mathrm{Me}_{3} \mathrm{N}\) is added to a THF solution of THF.BH \(_{3}\) ? (b) Will \(\mathrm{Me}_{2} \mathrm{O}\) displace \(\mathrm{Me}_{3} \mathrm{P}\) from \(\mathrm{Me}_{3} \mathrm{P} \cdot \mathrm{BH}_{3} ?\) (c) Is \(\left[\mathrm{BH}_{4}\right]^{-}\) stable in THF solution with respect to a displacement reaction? (d) Suggest what may be formed when \(\mathrm{Ph}_{2} \mathrm{PCH}_{2} \mathrm{CH}_{2} \mathrm{PPh}_{2}\) is added to a THF solution of THF. \(\mathrm{BH}_{3},\) the latter remaining in excess.

Suggest likely products for the following reactions: (a) \(\mathrm{BCl}_{3}+\mathrm{EtOH} \rightarrow\) (b) \(\mathrm{BF}_{3}+\mathrm{EtOH} \rightarrow\) (c) \(\mathrm{BCl}_{3}+\mathrm{PhNH}_{2} \rightarrow\) (d) \(\mathrm{BF}_{3}+\mathrm{KF} \rightarrow\)

Explain how, during dimerization, each \(\mathrm{BH}_{3}\) molecule acts as both a Lewis base and a Lewis acid.

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