Problem 86 - Chemistry The Central Science

Question

The following observations are made about two hypothetical elements \(A\) and \(B :\) The \(A-A\) and \(B-B\) bond lengths in the elemental forms of \(A\) and \(B\) are 2.36 and \(1.94 \hat{A},\) respectively. A and \(\mathrm{B}\) react to form the binary compound \(\mathrm{AB}_{2},\) which has a linear structure (that is \(\angle \mathrm{B}-\mathrm{A}-\mathrm{B}=180^{\circ} ) .\) Based on these statements, predict the separation between the two B nuclei in a molecule of \(\mathrm{AB}_{2}\) .

Step-by-Step Solution

Answer

The separation between the two B nuclei in a molecule of AB2 is approximately 4.30 Å.

Step 1: Draw the linear AB2 molecule structure

Draw a straight line representing the A-B bonds. Place the A atom in the center and the two B atoms at the ends of the line. This gives us the linear structure of AB2, with a B-A-B angle of \(180^{\circ}\).

Step 2: Create a triangle with the B atoms and the A atom

To measure the distance between the B atoms, we will create a triangle consisting of the two B atoms and the A atom. Since the angle A between the B atoms is \(180^{\circ}\), we have an isosceles triangle with one side being the length AB, and the other two sides being the length BB.

Step 3: Use the bond length in AB2 to find the AB side of the triangle

In the given exercise, the A-A bond length is 2.36 Å, and the B-B bond length in the elemental form is 1.94 Å. The bond lengths in the elemental forms can be used to predict the bond length in the compound AB2. Here, we can assume that the bond length AB in the compound is approximately an average of the A-A and B-B bond lengths: AB = (2.36 Å + 1.94 Å) / 2 = 2.15 Å So, the AB side of the triangle is 2.15 Å.

Step 4: Determine the length of the other two sides of the isosceles triangle

Since the triangle we have formed is an isosceles triangle, the other two sides (Beside) have the same length as the AB side. Therefore, Beside = 2.15 Å.

Step 5: Calculate the distance between the two B nuclei using the triangle properties

Now, we can use the triangle properties to calculate the distance between the two B nuclei. In an isosceles triangle with the two equal sides and angle A, the distance between the two vertices opposite the equal sides can be calculated using the law of cosines: \(BB = \sqrt{AB^2 + AB^2 - 2(AB)(AB)\cos A}\) Since the angle A between the B nuclei is \(180^{\circ}\), the cosine value of A is -1: \(BB = \sqrt{2.15^2 + 2.15^2 - 2(2.15)(2.15)(-1)}\) \(BB = \sqrt{4.6225 + 4.6225 + 9.245}\) \(BB = \sqrt{18.490}\) BB = \(\approx 4.30\) Å

Step 6: Write down the final answer

The separation between the two B nuclei in a molecule of AB2 is approximately 4.30 Å.

Key Concepts

Bond Length Determination

Understanding how to determine bond lengths is a fundamental aspect of studying molecular structures. Bond length is the average distance between the nuclei of two bonded atoms within a molecule. It is a measure of the distance at which the energy minimum occurs between the two atoms forming a bond.

To predict bond lengths in new molecules, such as the hypothetical compound AB2 from the exercise, we usually start with known bond lengths of similar molecules or the elemental forms of the atoms involved. However, these are approximations because bond lengths can vary with the type and number of bonded atoms, and the molecular environment.

For the compound AB2, with a known linear structure, we inferred that the bond length between A and B (AB) could be approximated by averaging the bond lengths of A-A and B-B in their elemental forms. This simplification is based on the principle that a bond will have a length that is roughly between that of the interacting elements. Mathematically, this estimation is expressed as:\[ AB = \frac{{A-A + B-B}}{2} = \frac{{2.36 \text{{ Å}} + 1.94 \text{{ Å}}}}{2} = 2.15 \text{{ Å}} \]

This approach provides a foundational step in solving for the distance between the B nuclei in the AB2 molecule.

Molecular Geometry

The shape of molecules, known as molecular geometry, plays a crucial role in the physical and chemical properties of substances. It's defined by the spatial arrangement of atoms in a molecule and can significantly influence reactivity, color, magnetism, and biological activity.

In our scenario, the molecular geometry of AB2 is described as linear, which means all atoms lie in a straight line with bond angles of \(180^\circ\). This specific angle leads to a geometric shape where the A atom acts as a central vertex of an isosceles triangle, with the B atoms forming the two base vertices. Linear molecules are characterized by their symmetrical shape, which allows us to apply isosceles triangle properties to calculate unknown distances when other dimensions are given or can be estimated.

By identifying the molecule's geometric layout we were able to effectively turn our problem into a question of classical geometry, allowing us to use simple mathematical equations to find the solution.

Isosceles Triangle Properties

An isosceles triangle is a type of triangle with at least two sides of equal length. These equal sides are known as legs, while the third side is known as the base. An important property of an isosceles triangle is that the angles opposite the equal sides are also equal.

In this context, we converted the structure of AB2 into an isosceles triangle, with the molecule's linear geometry implying a vertex angle of \(180^\circ\) at atom A. This angle is particularly helpful, because when it is \(180^\circ\), the law of cosines simplifies dramatically due to the fact that the cosine of \(180^\circ\) is -1. We can use this property to find the separation between the B nuclei (BB), considered the 'base', by applying the following adaptation of the law of cosines for our specific isosceles triangle setting:\[ BB = \sqrt{2(AB^2) - 2(AB)(AB)(-1)} \]This equation results in a distance calculation that's straightforward due to the linear configuration of the molecule. The use of this geometric approach simplifies complex chemical structures into manageable mathematical problems, making the concepts more accessible to learners.