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(a) What is meant by the term polarizability? (b) Which of the following atoms would you expect to be most polarizable: \(\mathrm{N}, \mathrm{P}, \mathrm{As}, \mathrm{Sb}\) ? Explain. (c) Put the following molecules in order of increasing polarizability: \(\mathrm{GeCl}_{4}, \mathrm{CH}_{4}\) \(\mathrm{SiCl}_{4}, \mathrm{SiH}_{4}\), and \(\mathrm{GeBr}_{4}\). (d) Predict the order of boiling points of the substances in part (c).

Short Answer

Expert verified
(a) Polarizability refers to the ability of an atom or molecule to distort its electron cloud when subjected to an external electric field. (b) Sb is the most polarizable among N, P, As, and Sb. (c) The order of increasing polarizability for the given molecules is: CH4 < SiH4 < SiCl4 < GeCl4 < GeBr4. (d) The order of boiling points is: CH4 < SiH4 < SiCl4 < GeCl4 < GeBr4.

Step by step solution

01

1. Define Polarizability

Polarizability refers to the ability of an atom or molecule to distort its electron cloud when subjected to an external electric field. In general, polarizability increases with the size of the atom or molecule and the number of electrons present because more electrons mean the electron cloud is easier to distort.
02

2. Compare Polarizability of N, P, As, and Sb

All four elements (N, P, As, and Sb) belong to Group 15 (the nitrogen group) in the periodic table. As we move down the group, the atomic size increases and the number of electrons in the atom also increases. Hence, polarizability increases down the group. So, the order of increasing polarizability is: N < P < As < Sb. Therefore, Sb is the most polarizable.
03

3. Arrange Molecules in Order of Increasing Polarizability

To compare the polarizability of the molecules, we can consider the size and electron cloud of the central atom in the molecules. In general, larger atoms with more electrons are more polarizable. The order of increasing polarizability of the central atoms (C, Si, and Ge) is: C < Si < Ge. Additionally, considering the ligands (Cl and Br) in the molecules, Br has more electrons than Cl, making the molecule with Br more polarizable. Following this reasoning, the order of increasing polarizability of the molecules is: CH4 < SiH4 < SiCl4 < GeCl4 < GeBr4.
04

4. Predict the Order of Boiling Points

Boiling points are primarily affected by the strength of the intermolecular forces between molecules. Greater polarizability usually results in stronger dispersion (London) forces between molecules, leading to higher boiling points. Based on our previous findings concerning polarizability, the order of boiling points should follow the same trend: CH4 < SiH4 < SiCl4 < GeCl4 < GeBr4.

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Key Concepts

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

Atomic Polarizability
Atomic polarizability describes how easily an atom can have its electron cloud distorted due to external electric fields, such as those from nearby ions or dipole moments. This distortion leads to the creation of an instantaneous dipole moment, which is a temporary unequal distribution of charge.

An important aspect to comprehend is that polarizability is directly linked to the atom's size and number of electrons it has. Larger atoms with more outer electrons exhibit greater polarizability because these electrons are further from the nucleus and less tightly held, making them more readily displaced.
Molecular Polarizability
Moving from atoms to molecules, Molecular polarizability extends the concept of electron cloud distortion to the molecular level. When a molecule's electrons are distorted, temporary dipoles across the molecule are created, influencing how the molecule interacts with electric fields and nearby particles.

Consider factors such as molecular shape, electron distribution, and the presence of polar bonds in addition to the size and number of electrons found in the individual atoms that make up the molecule. Non-polar molecules can also be polarizable due to the distortion of their electron cloud.
Intermolecular Forces
Intermolecular forces are the forces that act between molecules. They include London dispersion forces, dipole-dipole interactions, and hydrogen bonding. These forces are critical in determining the physical properties of a substance, such as melting and boiling points.

The London dispersion force, a type of force based on temporary fluctuations of the electron distribution in molecules, shows a strong dependence on polarizability. The more polarizable a molecule is, the stronger the induced dipoles and, consequently, the stronger the intermolecular attractions.
Periodic Trends in Polarizability
In the periodic table, atomic and molecular polarizability display distinct trends. As one moves down a group, polarizability increases due to the increase in atomic size. This trend is due to the additional electron shells that are added as elements become heavier, allowing for a greater ease of electron cloud distortion.

Additionally, polarizability tends to decrease as one moves from left to right across a period, which can be attributed to the increase in nuclear charge. This increase holds the electron cloud more tightly to the nucleus, reducing the ease of distortion.
Boiling Points and Polarizability
The boiling point of a substance is profoundly influenced by the strength of its intermolecular forces. Substances with high molecular polarizability generally have higher boiling points since the induced dipoles lead to stronger intermolecular attractions.

This relationship explains why, with similar structures, heavier molecules will typically boil at higher temperatures. As molecular weight and polarizability increase, the dispersion forces become more significant, raising the energy required to break these intermolecular interactions and, hence, the boiling point.

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Most popular questions from this chapter

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) melts at \(-114{ }^{\circ} \mathrm{C}\) and boils at \(78{ }^{\circ} \mathrm{C}\). Its density is \(0.789 \mathrm{~g} / \mathrm{mL}\). The enthalpy of fusion of ethanol is \(5.02 \mathrm{~kJ} / \mathrm{mol}\), and its enthalpy of vaporization is \(38.56 \mathrm{~kJ} / \mathrm{mol}\). The specific heats of solid and liquid ethanol are \(0.97 \mathrm{~J} / \mathrm{g}-\mathrm{K}\) and \(2.3 \mathrm{~J} / \mathrm{g}-\mathrm{K}\), respectively. (a) How much heat is required to convert \(25.0 \mathrm{~g}\) of ethanol at \(25^{\circ} \mathrm{C}\) to the vapor phase at \(78^{\circ} \mathrm{C} ?\) (b) How much heat is required to convert \(5.00 \mathrm{~L}\) of ethanol at \(-140^{\circ} \mathrm{C}\) to the vapor phase at \(78^{\circ} \mathrm{C} ?\)

The fluorocarbon compound \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}\) has a normal boiling point of \(47.6^{\circ} \mathrm{C}\). The specific heats of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}(l)\) and \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}(g)\) are \(0.91 \mathrm{~J} / \mathrm{g}-\mathrm{K}\) and \(0.67 \mathrm{~J} / \mathrm{g}-\mathrm{K}\), respectively. The heat of vaporization for the compound is \(27.49 \mathrm{~kJ} / \mathrm{mol}\). Calculate the heat required to convert \(50.0 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}\) from a liquid at \(10.00{ }^{\circ} \mathrm{C}\) to a gas at \(85.00{ }^{\circ} \mathrm{C}\).

Explain why the heat of fusion of any substance is generally lower than its heat of vaporization.

For each of the following pairs of substances, predict which will have the higher melting point, and indicate why: (a) HF, \(\mathrm{HCl} ;\) (b) C (graphite), \(\mathrm{CH}_{4}\); (c) \(\mathrm{KCl}, \mathrm{Cl}_{2}\); (d) \(\mathrm{LiF}, \mathrm{MgF}_{2}\).

In a certain type of nuclear reactor, liquid sodium metal is employed as a circulating coolant in a closed system, protected from contact with air or water. Much like the coolant that circulates in an automobile engine, the liquid sodium carries heat from the hot reactor core to heat exchangers. (a) What properties of the liquid sodium are of special importance in this application? (b) The viscosity of liquid sodium varies with temperature as follows: $$ \begin{array}{ll} \hline \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & \text { Viscosity }\left(\mathrm{kg} \mathrm{m}^{-1} \mathrm{~s}^{-1}\right) \\ \hline 100 & 7.05 \times 10^{-4} \\ 200 & 4.50 \times 10^{-4} \\ 300 & 3.45 \times 10^{-4} \\ 600 & 2.10 \times 10^{-4} \\ \hline \end{array} $$ What forces within the liquid sodium are likely to be the major contributors to the viscosity? Why does viscosity decrease with increasing temperature?

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