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Halogens combine with one another to produce interhalogens such as \(\mathrm{BrF}\), Sketch a possible molecular structure for this molecule and decide if the \(\mathrm{F}-\mathrm{Br}-\mathrm{F}\) bond angles will be less than or greater than ideal.

Short Answer

Expert verified
\( \mathrm{BrF}_3 \) has a T-shaped structure with \( \mathrm{F}-\mathrm{Br}-\mathrm{F} \) bond angles slightly less than 90°.

Step by step solution

01

Identify Molecular Geometry

To find out the molecular geometry of \( \mathrm{BrF}_3 \), we first need to determine the number of bonding pairs and lone pairs around the central atom. Bromine (Br) is the central atom and it typically follows the octet rule. In the case of \( \mathrm{BrF}_3 \), bromine is bonded to three fluorine atoms and has two lone pairs.
02

Use VSEPR Theory

According to the VSEPR theory (Valence Shell Electron Pair Repulsion), the electron pairs around the central atom will arrange themselves to minimize repulsion. With three bonded pairs and two lone pairs, \( \mathrm{BrF}_3 \) will adopt a T-shaped molecular geometry.
03

Analyze Bond Angles

In a T-shaped molecule like \( \mathrm{BrF}_3 \), the ideal bond angle is 90° due to the electron repulsion from the lone pairs. However, the presence of two lone pairs on bromine pushes the bonded fluorine atoms closer together, causing the \( \mathrm{F}-\mathrm{Br}-\mathrm{F} \) bond angle to be slightly less than 90°.
04

Draw the Molecular Structure

To sketch the \( \mathrm{BrF}_3 \) molecule, place bromine in the center. Position the three fluorine atoms such that two are bonded along the horizontal plane and one along the vertical. Include the two lone pairs on bromine, which occupy the equatorial positions of a trigonal bipyramidal shape, but the resulting structure is T-shaped. This factors into the bond angle being less than 90°.

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Key Concepts

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

Molecular Geometry
Molecular geometry describes the three-dimensional arrangement of atoms within a molecule. In the case of interhalogen compounds like \( \mathrm{BrF}_3 \), understanding molecular geometry is crucial for predicting physical and chemical properties.
A molecule's geometry is determined by the number of bonds and lone electron pairs around the central atom. For instance, in \( \mathrm{BrF}_3 \), bromine is the central atom, bonded to three fluorine atoms, with two lone pairs. This configuration leads to a specific geometric arrangement that impacts the molecule's behavior and interactions.
Different molecular geometries include linear, trigonal planar, tetrahedral, and more. Each shape results from the angles between bonds formed by the shared and lone electron pairs, impacting how molecules interact with each other.
VSEPR Theory
The Valence Shell Electron Pair Repulsion (VSEPR) theory is a fundamental model used to predict a molecule's shape.
This theory is based on the idea that electron pairs around an atom, including bonding and lone pairs, repel each other. As a result, these pairs arrange themselves as far apart as possible to minimize repulsion.
  • Bonding pairs are electrons shared between atoms, forming chemical bonds.
  • Lone pairs are non-bonded electron pairs occupying the valence shell of the central atom.

In \( \mathrm{BrF}_3 \), VSEPR theory helps deduce that the molecule adopts a T-shaped structure. Here, bromine has three bonding pairs and two lone pairs. The lone pairs exert force, rearranging the molecule’s geometry to reduce repulsion, which in turn predicts the T-shape of \( \mathrm{BrF}_3 \).
Bond Angles
Bond angles in a molecule refer to the angles formed between adjacent bonds emanating from the central atom. These angles are crucial in determining the overall shape and properties of the molecule. Predicted by VSEPR theory, bond angles are affected by both bonding and lone pairs.
In a idealized geometry, such as planar or linear, bond angles can reach typical values like 120° or 180°. However, the actual angles often differ due to the influence of lone pairs and electron repulsion.
For \( \mathrm{BrF}_3 \), the ideal bond angle for a T-shaped molecule is 90°. However, because of the two lone pairs on the bromine atom, the \( \mathrm{F}-\mathrm{Br}-\mathrm{F} \) bond angles are slightly smaller than 90°, as the lone pairs push the bonded fluorine atoms closer together. This subtle shift is essential for understanding molecular interactions and reactivity.
T-shaped Structure
The T-shaped structure is a specific type of molecular geometry that is both fascinating and instructive. It arises in molecules with a central atom bonded to three atoms, accompanied by two lone pairs.
In this form, the central atom is often positioned at the intersection of the 'T,' with two bonded atoms forming the top bar and one forming the vertical part.
  • For \( \mathrm{BrF}_3 \), bromine forms the center, and the three fluorine atoms assume the T-shape with bromine at the nexus.
  • The T-shape results from the minimization of repulsion, as modeled by VSEPR theory.
  • The two lone pairs of electrons on the bromine repulse the bonding pairs, forcing them to adopt this specific alignment.

Such unique geometry influences the physical and chemical properties of \( \mathrm{BrF}_3 \), dictating how it interacts in reactions and experiences polarizability and dipole moments, critical to understanding its stability and reactivity in different environments.

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Most popular questions from this chapter

Consider the chemistries of the elements sodium, magnesium, aluminum, silicon, and phosphorus. (a) Write a balanced chemical equation depicting the reaction of each element with elemental chlorine. (b) Describe the bonding in each of the products of the reactions with chlorine as ionic or covalent. (c) Draw Lewis electron dot structures for the products of the reactions of silicon and phosphorus with chlorine. What are their electron-pair and molecular geometries?

Give the name and symbol of each element having the valence configuration [noble gas] \(n s^{2} n p^{1}.\)

A The density of lead is \(11.350 \mathrm{g} / \mathrm{cm}^{3},\) and the metal crystallizes in a face-centered cubic unit cell. Estimate the radius of the lead atom.

The steering rockets in the Space Shuttle use \(\mathrm{N}_{2} \mathrm{O}_{4}\) and a derivative of hydrazine, 1,1 -dimethylhydrazine (page 278 ). This mixture is called a hypergolic fuel because it ignites when the reactants come into contact: \(\mathrm{H}_{2} \mathrm{NN}\left(\mathrm{CH}_{3}\right)_{2}(\ell)+2 \mathrm{N}_{2} \mathrm{O}_{4}(\ell) \longrightarrow\) $$ 3 \mathrm{N}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{CO}_{2}(\mathrm{g}) $$ (a) Identify the oxidizing agent and the reducing agent in this reaction. (b) The same propulsion system was used by the Lunar Lander on moon missions in the 1970 s. If the Lander used \(4100 \mathrm{kg}\) of \(\mathrm{H}_{2} \mathrm{NN}\left(\mathrm{CH}_{3}\right)_{2},\) what mass (in kilograms) of \(\mathrm{N}_{2} \mathrm{O}_{4}\) was required to react with it? What mass (in kilograms) of each of the reaction products was generated?

A major use of hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) is in steam boilers in power plants. (a) The reaction of hydrazine with \(\mathrm{O}_{2}\) dissolved in water gives \(\mathrm{N}_{2}\) and water. Write a balanced equation for this reaction. (b) \(\mathrm{O}_{2}\) dissolves in water to the extent of \(3.08 \mathrm{cm}^{3}\) (gas at STP) in \(100 .\) mL of water at \(20^{\circ} \mathrm{C} .\) To consume all of the dissolved \(\mathrm{O}_{2}\) in \(3.00 \times 10^{4}\) I. of water (enough to fill a small swimming pool), what mass of \(\mathrm{N}_{2} \mathrm{H}_{4}\) is needed?

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