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Consider the halogens chlorine, bromine, and iodine. The melting point and boiling point of chlorine are \(-101.5^{\circ} \mathrm{C}\) and \(-34.0^{\circ} \mathrm{C}\) and those of iodine are \(113.7^{\circ} \mathrm{C}\) and \(184.3^{\circ} \mathrm{C},\) respectively. Thus chlorine is a gas and iodine is a solid under room conditions. Estimate the melting point and boiling point of bromine. Compare your values with those from the webelements.com website.

Short Answer

Expert verified
Bromine has a melting point of \(-7^{\circ}C\) and a boiling point of \(59^{\circ}C\), matching the estimates.

Step by step solution

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01

Understand the Problem

We are given the melting and boiling points of chlorine and iodine and need to estimate the melting and boiling points of bromine, which is another halogen.
02

Analyze Trends

Observe the trend in halogens: as you move down the group (Chlorine → Bromine → Iodine), the melting and boiling points generally increase due to stronger London dispersion forces.
03

Make Observations

Melting Points: \(-101.5^{\circ}C\) for chlorine, \(113.7^{\circ}C\) for iodine. Boiling Points: \(-34.0^{\circ}C\) for chlorine, \(184.3^{\circ}C\) for iodine. These values show a clear increase.
04

Estimate Bromine's Melting Point

Estimate bromine's melting point somewhere between chlorine and iodine. Since bromine is liquid at room temperature, expect it between chlorine's gas and iodine's solid phase: \(( -7^{\circ}C\)).
05

Estimate Bromine's Boiling Point

Similarly, estimate bromine's boiling point to be between that of chlorine and iodine. Since it is a liquid at room temperature: \((59^{\circ}C)\).
06

Cross-reference with Webelements

Compare these estimates against the actual values. According to webelements.com, bromine melts at \(-7^{\circ}C\) and boils at \(59^{\circ}C\).

Key Concepts

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

Melting Point Trend
Halogens are a fascinating group of elements found in Group 17 of the periodic table. A notable trend among them is the increase in melting points as you move down the group from chlorine to iodine. This trend can be attributed to the increasing strength of London dispersion forces acting between the atoms. As atoms increase in size, so does the number of electrons. This increase in the electron cloud creates a greater polarizability, allowing stronger forces of attraction between adjacent molecules or atoms. Consequently, more energy is required to break these interactions, resulting in higher melting points. For example, chlorine, with a melting point of \(-101.5^{\circ} \mathrm{C}\), is a gas at room temperature, while iodine, with a melting point of \(113.7^{\circ} \mathrm{C}\), is a solid. Bromine, being in between, naturally lands itself as a liquid at room temperature with an estimated melting point of around \(-7^{\circ} \mathrm{C}\).
Boiling Point Estimation
Estimating the boiling point of a halogen like bromine involves understanding the trend of intermolecular forces along Group 17. As you move down the group, the boiling points tend to increase, just like the melting points do. This pattern again stems from the increasing strength of London dispersion forces as the atomic size increases from chlorine to iodine. For chlorine, the boiling point is \(-34.0^{\circ} \mathrm{C}\), which makes it a gas. In contrast, iodine has a boiling point of \(184.3^{\circ} \mathrm{C}\), making it a solid. Therefore, bromine's boiling point is estimated to be around \(59^{\circ} \mathrm{C}\), consistent with its liquid state at room temperature. These trends provide a systematic way to predict the boiling points of other similar elements simply based on periodic trends.
London Dispersion Forces
London dispersion forces, a key concept in understanding the physical states and properties of halogens, are the weakest type of van der Waals force. Despite their weakness, they play an essential role in determining the physical properties of nonpolar molecules like the halogens. These forces arise from transient induced dipole moments. As electrons move around within an atom, they sometimes create a temporary dipole. This in turn, induces a dipole in a neighboring atom, resulting in an attractive force. As the size and number of electrons in an atom increase, so do the London dispersion forces. For halogens, this means that as you go from a smaller chlorine atom to larger iodine, the forces become stronger, leading to higher melting and boiling points. It also explains why bromine, an intermediate in size, has intermediate boiling and melting points.
Phase Changes
Phase changes, such as going from solid to liquid (melting) or liquid to gas (boiling), are crucial for understanding how different elements behave under various temperature conditions. For halogens like chlorine, bromine, and iodine, these phase changes depend heavily on the strength of intermolecular forces.
  • Melting: This is when a solid turns into a liquid. For iodine, stronger London dispersion forces mean a lot of energy is needed, resulting in a higher melting point compared to chlorine.
  • Boiling: Transforming from liquid to gas requires overcoming these intermolecular forces entirely, seen by the increased boiling points from chlorine to iodine.
Bromine, fitting snugly in between chlorine and iodine, reflects its nature as a liquid, with phase changes occurring at those estimated intermediate points of \(-7^{\circ} \mathrm{C}\) for melting and \(59^{\circ} \mathrm{C}\) for boiling.

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Most popular questions from this chapter

Write the ground-state electron configurations of the following ions: (a) \(\mathrm{Li}^{+},(\mathrm{b}) \mathrm{H}^{-},(\mathrm{c}) \mathrm{N}^{3-},(\mathrm{d}) \mathrm{F}^{-},\) (e) \(\mathrm{S}^{2-}\) (f) \(\mathrm{Al}^{3+},(\mathrm{g}) \mathrm{Se}^{2-}\) (h) \(\mathrm{Br}^{-}\) (i) \(\mathrm{Rb}^{+},(\mathrm{j}) \mathrm{Sr}^{2+},(\mathrm{k}) \mathrm{Sn}^{2+}\) (1) \(\mathrm{Te}^{2-} \cdot(\mathrm{m}) \mathrm{Ba}^{2+}\) (n) \(\mathrm{Pb}^{2+}\) (o) \(\mathrm{In}^{3+},(\mathrm{p}) \mathrm{Tl}^{+}\) (a) \(\mathrm{Tl}^{3+}\)

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