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Chapter 7: Problem 38

Write equations that show the process for (a) the first two ionization energies of lead and (b) the fourth ionization energy of zirconium.

Short Answer

Expert verified
a) The first two ionization energies of lead (Pb) are represented by these equations: \(1^{st}\) Ionization energy: \(Pb(g) -> Pb^{+}(g) + e^{-}\) \(2^{nd}\) Ionization energy: \(Pb^{+}(g) -> Pb^{2+}(g) + e^{-}\) b) The fourth ionization energy of zirconium (Zr) is represented by this equation: \(4^{th}\) Ionization energy: \(Zr^{3+}(g) -> Zr^{4+}(g) + e^{-}\)

Step by step solution

01

a) First ionization energy of lead

To write the equation for the first ionization energy of lead, we need to remove one electron from the neutral lead atom: Pb(g) -> Pb^(+)(g) + e^(-) Here, X is the lead (Pb) atom, and n is 1 (first ionization energy).
02

b) Second ionization energy of lead

To write the equation for the second ionization energy of lead, we need to remove another electron from the Pb^+ ion obtained in step a: Pb^(+)(g) -> Pb^(2+)(g) + e^(-) Here, X is the lead (Pb) atom and n is now 2 (second ionization energy).
03

c) Fourth ionization energy of zirconium

To write the equation for the fourth ionization energy of zirconium, we need to remove four electrons from the neutral zirconium atom: Zr^(3+)(g) -> Zr^(4+)(g) + e^(-) Here, X is the zirconium (Zr) atom and n is 4 (fourth ionization energy).

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Most popular questions from this chapter

Which will experience the greater effective nuclear charge, the electrons in the \(n=3\) shell in Ar or the \(n=3\) shell in \(\mathrm{Kr}\) ? Which will be closer to the nucleus?

When magnesium metal is burned in air (Figure 3.6), two products are produced. One is magnesium oxide, \(\mathrm{MgO}\). The other is the product of the reaction of \(\mathrm{Mg}\) with molecular nitrogen, magnesium nitride. When water is added to magnesium nitride, it reacts to form magnesium oxide and ammonia gas. (a) Based on the charge of the nitride ion (Table 2.5), predict the formula of magnesium nitride. (b) Write a balanced equation for the reaction of magnesium nitride with water. What is the driving force for this reaction? (c) In an experiment, a piece of magnesium ribbon is burned in air in a crucible. The mass of the mixture of \(\mathrm{MgO}\) and magnesium nitride after burning is \(0.470 \mathrm{~g}\). Water is added to the crucible, further reaction occurs, and the crucible is heated to dryness until the final product is \(0.486 \mathrm{~g}\) of \(\mathrm{MgO}\). What was the mass percentage of magnesium nitride in the mixture obtained after the initial burning? (d) Magnesium nitride can also be formed by reaction of the metal with ammonia at high temperature. Write a balanced equation for this reaction. If a 6.3-g Mg ribbon reacts with \(2.57 \mathrm{~g} \mathrm{NH}_{3}(g)\) and the reaction goes to completion, which component is the limiting reactant? What mass of \(\mathrm{H}_{2}(g)\) is formed in the reaction? (e) The standard enthalpy of formation of solid magnesium nitride is \(-461.08 \mathrm{~kJ} / \mathrm{mol}\). Calculate the standard enthalpy change for the reaction between magnesium metal and ammonia gas.

Moseley established the concept of atomic number by studying X rays emitted by the elements. The \(X\) rays emitted by some of the elements have the following wavelengths: $$ \begin{array}{lc} \hline \text { Element } & \text { Wavelength } \\ \hline \mathrm{Ne} & 14.610 \\ \mathrm{Ca} & 3.358 \\ \mathrm{Zn} & 1.435 \\ \mathrm{Zr} & 0.786 \\ \mathrm{Sn} & 0.491 \\ \hline \end{array} $$ (a) Calculate the frequency, \(\nu\), of the \(\mathrm{X}\) rays emitted by each of the elements, in Hz. (b) Plot the square root of \(\nu\) versus the atomic number of the element. What do you observe about the plot? (c) Explain how the plot in part (b) allowed Moseley to predict the existence of undiscovered elements. (d) Use the result from part (b) to predict the \(\mathrm{X}\)-ray wavelength emitted by iron. (e) A particular element emits X rays with a wavelength of \(0.980 \AA\). What element do you think it is?

Write the electron configurations for the following ions, and determine which have noble-gas configurations: (a) \(\mathrm{Ru}^{3+}\), (b) \(\mathrm{As}^{3-}\), (c) \(\mathrm{Y}^{3+}\), (d) \(\mathrm{Pd}^{2+}\), (e) \(\mathrm{Pb}^{2+}\), (f) \(\mathrm{Au}^{3+}\).

Silver and rubidium both form \(+1\) ions, but silver is far less reactive. Suggest an explanation, taking into account the ground-state electron configurations of these elements and their atomic radii.
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