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Chapter 8: Problem 149

Compare the work function for cesium ( \(206 \mathrm{~kJ} / \mathrm{mol}\) ) with its first ionization energy ( \(376 \mathrm{~kJ} / \mathrm{mol}\) ). Explain the difference.

Short Answer

Expert verified
The difference between the work function and first ionization energy of cesium stems from the context in which they are measured. Work function, measuring the energy needed to remove an electron from a metal surface, is less than the first ionization energy, which measures the energy required to remove the most loosely held electron from a neutral gaseous atom, hence the difference in their values.

Step by step solution

01

Understanding the Terms

Firstly, we need to clarify what these terms mean in this context. The work function is the minimal energy required to remove an electron from a metal surface. This concept is mainly employed in the field of quantum mechanics. On the other hand, the first ionization energy is the energy needed to remove the most loosely held electron from a neutral gaseous atom.
02

Comparing the Values

The given values are 206 kJ/mol for the work function and 376 kJ/mol for the first ionization energy. It can be observed that the first ionization energy is higher than the work function.
03

Explaining the Difference

The reason for the observed difference is connected to the nature of the measurements. The work function is measured for electrons on the surface of a metal, while ionization energy is measured for isolated gas-phase atoms. In the solid state, the electron to be removed is actually in a collective energy state represented by bands. Thus, the energy to remove an electron is less than that in an isolated atom where the electron is in a specific, localized energy state. Therefore, it requires more energy to remove an electron from a neutral gaseous atom, resulting in a higher value for first ionization energy than for the work function.

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Most popular questions from this chapter

An element X reacts with hydrogen gas at \(200^{\circ} \mathrm{C}\) to form compound Y. When \(Y\) is heated to a higher temperature, it decomposes to the element \(X\) and hydrogen gas in the ratio of \(559 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) (measured at STP) for \(1.00 \mathrm{~g}\) of \(\mathrm{X}\) reacted. \(\mathrm{X}\) also combines with chlorine to form a compound \(Z,\) which contains 63.89 percent by mass of chlorine. Deduce the identity of \(X\).

Give equations to show that molecular hydrogen can act both as a reducing agent and an oxidizing agent.

Why do elements that have high ionization energies also have more positive electron affinities? Which group of elements would be an exception to this generalization?

A hydrogenlike ion contains only one electron. The energies of the electron in a hydrogenlike ion are given by $$ E_{n}=-\left(2.18 \times 10^{-18} \mathrm{~J}\right) Z^{2}\left(\frac{1}{n^{2}}\right) $$ where \(n\) is the principal quantum number and \(Z\) is the atomic number of the element. Calculate the ionization energy (in \(\mathrm{kJ} / \mathrm{mol}\) ) of the \(\mathrm{He}^{+}\) ion.

Explain why the electron affinity of nitrogen is approximately zero, while the elements on either side, carbon and oxygen, have substantial positive electron affinities.
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