An oxide of niobium has a unit cell in which there are oxide ions at the middle of each cdge and niobium atoms at the center of each face. What is the empirical formula of this oxide?

Short Answer

Expert verified
The empirical formula of the niobium oxide is NbO.

Step by step solution

01

Count the number of oxide ions per unit cell

As there are 12 edges in a cube and an oxide ion is located at the middle of each edge, each oxide ion is shared between 4 adjacent unit cells. Therefore, the total contribution of oxide ions per unit cell is (12 edges * 1/4 oxide ion per edge) = 3 oxide ions.
02

Count the number of niobium atoms per unit cell

There are 6 faces in a cube and a niobium atom is located at the center of each face, each niobium atom is shared between 2 adjacent unit cells. Hence, the total contribution of niobium atoms per unit cell is (6 faces * 1/2 niobium atom per face) = 3 niobium atoms.
03

Determine the empirical formula

From the previous steps, we have 3 niobium atoms and 3 oxide ions in a single unit cell, giving us the same number of niobium and oxygen atoms. Therefore, the empirical formula, which gives the simplest whole number ratio of atoms, is NbO (niobium(III) oxide).

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Key Concepts

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

Understanding Unit Cells in Crystal Lattices
The concept of a unit cell is fundamental to the study of solid-state chemistry and material science. Think of a unit cell as the smallest chunk or building block of a crystal lattice that, when repeated in three dimensions, creates the entire structure of a crystal. It’s analogous to a three-dimensional pattern that gets repeated like tiles on a floor, but in all directions.

Every unit cell is defined by the positions of its constituent particles, namely atoms, ions, or molecules. These particles are arranged in a precise manner, which is responsible for the physical and chemical properties of the material. For instance, a cubic unit cell, the most simple type of unit cell, has corners and can also have constituents on its faces, edges, and at its center.

In the exercise provided, we examined an oxide of niobium crystallized in such a way that its unit cells have oxide ions at the middle of each edge and niobium atoms at the center of each face. To visualize this, imagine a dice that not only has dots at the corner but also in the middle of each line (edges) and at the center of each side (faces). This arrangement within the unit cell is vital for determining the empirical formula of the compound.
The Role of Oxide Ions in the Structure
Oxide ions, often represented as O2-, play a crucial role in forming the structural backbone of many oxides. These negatively charged ions are oxygen atoms that have gained two extra electrons. In crystal structures, they are typically larger than the metal ions they are associated with and affect the overall geometry and stability of the crystal lattice.

In our exercise, oxide ions occupy the middle of each edge of the unit cell. However, it's important to note that while each unit cell seems to have its own oxide ions, these ions are actually shared with adjacent unit cells. This sharing of ions is a common feature in crystal structures and must be taken into account when counting the actual number of oxide ions per unit cell. By recognizing that each oxide ion is shared by four unit cells, we can calculate that the unit cell effectively contains three oxide ions, as the 12 ions at the edges each contribute a quarter to the count.
Determining the Empirical Formula With Niobium Atoms
Niobium atoms in the given oxide play an equally significant role as the oxide ions in determining the empirical formula of the compound. Niobium is a transitional metal with the capability to form various oxides, with niobium(III) oxide being one possibility.

In terms of its position within the unit cell, each niobium atom is located at the center of a face and thus contributes to the structure of two adjacent unit cells. This sharing means that while there are six faces to a unit cell, effectively there are three niobium atoms per unit cell after considering the sharing.

Once we have correctly counted the niobium atoms and oxide ions in the unit cell, finding the empirical formula becomes a matter of establishing the simplest whole-number ratio of these constituents. In this case, because there is an equal number of niobium atoms and oxide ions, the empirical formula is simply NbO. This balance between the two types of atoms is fundamental for the properties of the material and helps in understanding how atoms combine to form compounds.

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Most popular questions from this chapter

Tetrahedral and octahedral interstitial holes are formed by the spaces left when anions pack in a cubic close-packed array. (a) Which hole can accommodate the larger ions? (b) What is the size ratio of the largest metal cation that can occupy an octahedral hole to the largest that can occupy the tetrahedral hole while maintaining the close-packed nature of the anion lattice? (c) If half the tetrahedral holes are occupied, what will the chemical formula of the compound \(\mathrm{M}_{2} \mathrm{~A}_{2}\) be, where \(\mathrm{M}\) represents the cations and \(\mathrm{A}\) the anions?

Salts can be prepared from organic molecules such as acetic acid and methanol. For example, it is possible to prepare sodium acetate, \(\mathrm{NaCH}_{3} \mathrm{CO}_{2}\), and sodium methoxide, \(\mathrm{NaOCH}_{3}\). How do you expect the forces that hold these compounds together in the solid state to differ from those that hold together salts like sodium chloride or sodium bromide?

Account for the following observations in terms of the type and strength of intermolecular forces. (a) The melting point of xenon is \(-112^{\circ} \mathrm{C}\) and that of argon is \(-189^{\circ} \mathrm{C}\). (b) The vapor pressure of diethyl ether \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{3}\right)\) is greater than that of water. (c) The boiling point of pentane, \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{CH}_{3}\), is \(36.1^{\circ} \mathrm{C}\), whereas that of 2,2 -dimethylpropane (also known as neopentane) is \(9.5^{\circ} \mathrm{C}\).

Aluminum metal has a density of \(2.70 \mathrm{~g} \cdot \mathrm{cm}\) and crystallizes in a lattice with a unit cell edge of \(404 \mathrm{pm}\). (a) What type of cubic unit cell is formed by aluminum? (b) What is the coordination number of aluminum?

Suggest, giving reasons, which substance in each pair is likely to have the higher melting point (Lewis structures may help your arguments): (a) HCl or NaCl; (b) \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OC}_{2} \mathrm{H}_{3}\) (diethyl ether) or \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}\) (butanol); (c) \(\mathrm{HF}\) or \(\mathrm{HCl}\); (d) \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{CH}_{3} \mathrm{OH}\).

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