Question

Which species in each pair has the greater polarizability? Explain. (a) \(\mathrm{Ca}^{2+}\) or \(\mathrm{Ca}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{3}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (c) \(\mathrm{CCl}_{4}\) or \(\mathrm{CF}_{4}\)

Short answer

(a) \( \text{Ca} \). (b) \( \text{CH}_3\text{CH}_2\text{CH}_3 \). (c) \( \text{CCl}_4 \).

Step-by-step solution

Understand Polarizability

Polarizability refers to how easily the electron cloud of a species can be distorted. It generally increases with increasing size of the species and decreasing effective nuclear charge.

Compare \(\text{Ca}^{2+}\) and \( \text{Ca} \)

\( \text{Ca} \) is a neutral atom, while \( \text{Ca}^{2+} \) is a cation with a +2 charge. The cation \( \text{Ca}^{2+} \) has lost two electrons, resulting in a smaller size and a higher effective nuclear charge, making its electron cloud less distortable compared to \( \text{Ca} \). Therefore, \( \text{Ca} \) has greater polarizability.

Compare \( \text{CH}_3\text{CH}_3 \) and \( \text{CH}_3\text{CH}_2\text{CH}_3 \)

\( \text{CH}_3\text{CH}_2\text{CH}_3 \) (propane) has a larger electron cloud compared to \( \text{CH}_3\text{CH}_3 \) (ethane) due to having more atoms and more electrons. A larger electron cloud is more easily distorted, so \( \text{CH}_3\text{CH}_2\text{CH}_3 \) has greater polarizability.

Compare \( \text{CCl}_4 \) and \( \text{CF}_4 \)

\( \text{CCl}_4 \) contains chlorine atoms which are larger and more easily polarizable than the fluorine atoms in \( \text{CF}_4 \). Hence, \( \text{CCl}_4 \) has greater polarizability.

Key concepts

Electron Cloud Distortion

Polarizability is fundamentally about how easily an electron cloud can be distorted. Think of an electron cloud as a soft, fluffy ball of negative charge around an atom or molecule. When another nearby charge (positive or negative) comes close, it can pull on or push against this fluffy ball, distorting its shape. The easier it is to stretch or squish this electron cloud, the higher the polarizability of the species. This concept is very important in understanding the behavior of atoms and molecules in different chemical environments.

Let's relate this to question (a) in the exercise. Calcium ion, \( \text{Ca}^{2+} \), has lost two electrons, so its electron cloud is more compact and harder to distort compared to the neutral calcium atom (\( \text{Ca} \)). Hence, the neutral \( \text{Ca} \) has greater polarizability. The same reasoning is applied to larger molecules like ethane (\( \text{CH}_3 \text{CH}_3 \)) and propane (\( \text{CH}_3 \text{CH}_2 \text{CH}_3 \)). Propane, having more atoms and a bigger electron cloud, is easier to distort, hence it has greater polarizability. Similarly, chlorine atoms in \( \text{CCl}_4 \) are larger than fluorine atoms in \( \text{CF}_4 \), leading to greater polarizability for \( \text{CCl}_4 \).

Effective Nuclear Charge

Effective nuclear charge is the net positive charge experienced by electrons in an atom. It takes into account the actual nuclear charge (due to protons in the nucleus) and the shielding effect caused by electrons between the nucleus and the electron of interest. Mathematically, effective nuclear charge (\( Z_{\text{eff}} \)) can be approximated by the formula:

\[ Z_{\text{eff}} = Z - S \]
where \( Z\) is the number of protons and \( S \) is the shielding constant.

In the context of polarizability, a higher effective nuclear charge means that the electrons are held more tightly by the nucleus. This leads to a more compact electron cloud, which is harder to distort. For example, \( \text{Ca}^{2+} \) has a higher effective nuclear charge than \( \text{Ca} \) because it has lost electrons, reducing the shielding effect and making the remaining electrons feel a stronger pull from the nucleus.

In pairs like \( \text{CCl}_4 \) vs. \( \text{CF}_4 \), \( \text{CCl}_4 \) has chlorine atoms with more electrons, but the effective nuclear charge felt by the outermost electrons in chlorine is less pronounced due to larger size and shielding effect. Thus, the electrons are less tightly bound and more polarizable in \( \text{CCl}_4 \).

Ion Size Effect

The size of an ion plays a crucial role in its polarizability. Larger ions have more diffused electron clouds, which means their electron cloud can be more easily distorted. This is because the electrons are farther away from the nucleus and are less tightly held.

Considering \( \text{Ca}^{2+} \) and \( \text{Ca} \), \( \text{Ca}^{2+} \) is smaller since it has lost two electrons, causing a contraction of the electron cloud. This makes \( \text{Ca}^{2+} \) less polarizable compared to the neutral \( \text{Ca} \) atom. The same principle applies to molecules like ethane and propane. Propane, being a larger molecule, has a larger spread of its electron cloud, making it more easily polarizable.

In \( \text{CCl}_4 vs. \text{CF}_4 \), chlorines are larger than fluorines. This means \( \text{CCl}_4 \) has a larger, more diffused electron cloud which is easier to distort compared to \( \text{CF}_4 \). Thus, the ion size significantly affects the ease with which the electron cloud can be distorted, influencing the polarizability of the species.