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Chapter 7: Problem 67

(a) Why is calcium generally more reactive than magnesium? (b) Why is calcium generally less reactive than potassium?

Short Answer

Expert verified
(a) Calcium is generally more reactive than magnesium because it has a lower second ionization energy, attributed to its larger atomic size and higher electron shielding effect, making it easier for calcium to lose electrons. (b) Calcium is generally less reactive than potassium because potassium has a lower first ionization energy, which results from its lower nuclear charge and larger atomic size, allowing potassium to lose an electron more easily and be more reactive than calcium.

Step by step solution

01

Part (a): Electron configurations

First, let's look at the electron configurations and the periodic table positions of calcium (Ca) and magnesium (Mg). Calcium (Ca): Atomic number 20, Electron configuration: [Ar] 4s虏 Magnesium (Mg): Atomic number 12, Electron configuration: [Ne] 3s虏 Both calcium and magnesium belong to the same group (Group 2) in the periodic table, making them alkaline earth metals.
02

Part (a): Ionization energy

In order to compare reactivity, we can look at the ionization energy, which represents the energy needed to remove an electron from an atom. The lower the ionization energy, the more easily an element loses an electron, and the more reactive it is. Group 2 elements typically lose 2 electrons to form a 虏鈦� ion, so we will compare the second ionization energies of calcium and magnesium.
03

Part (a): Trends in the periodic table

As we move down a group in the periodic table, ionization energies generally decrease due to: 1. An increase in atomic size 2. Electron shielding effect Calcium has a lower second ionization energy than magnesium because it has a larger atomic size and higher electron shielding effect, which makes it easier for calcium to lose electrons and be more reactive than magnesium.
04

Part (b): Electron configurations

Now, let's compare the reactivity of calcium (Ca) and potassium (K). Potassium (K): Atomic number 19, Electron configuration: [Ar] 4s鹿 Note that potassium is in the same period as calcium but belongs to a different group (Group 1) in the periodic table, making it an alkali metal.
05

Part (b): Ionization energy

Alkali metals typically lose 1 electron to form a 鹿鈦� ion, so we will compare the first ionization energies of calcium and potassium.
06

Part (b): Trends in the periodic table

As we move from left to right in a period in the periodic table, ionization energies generally increase due to: 1. An increase in nuclear charge 2. A decrease in atomic size Potassium has a lower first ionization energy than calcium because it has a lower nuclear charge and larger atomic size, making it easier for potassium to lose an electron and be more reactive than calcium.

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Most popular questions from this chapter

An element \(\mathrm{X}\) reacts with oxygen to form \(\mathrm{XO}_{2}\) and with chlorine to form \(\mathrm{XCl}_{4} \cdot \mathrm{XO}_{2}\) is a white solid that melts at high temperatures (above \(1000^{\circ} \mathrm{C}\) ). Under usual conditions, \(\mathrm{XCl}_{4}\) is a colorless liquid with a boiling point of \(58^{\circ} \mathrm{C}\). (a) \(\mathrm{XCl}_{4}\) reacts with water to form \(\mathrm{XO}_{2}\) and another product. What is the likely identity of the other product? (b) Do you think that element \(\mathrm{X}\) is a metal, nonmetal, or metalloid? (c) By using a sourcebook such as the CRC Handbook of Chemistry and Physics, try to determine the identity of element X.

Use electron configurations to explain the following observations: (a) The first ionization energy of phosphorus is greater than that of sulfur. (b) The electron affinity of nitrogen is lower (less negative) than those of both carbon and oxygen. (c) The second ionization energy of oxygen is greater than the first ionization energy of fluorine. (d) The third ionization energy of manganese is greater than those of both chromium and iron.

For each of the following pairs, indicate which element has the smaller first ionization energy: (a) Ti, Ba; (b) \(\mathrm{Ag}, \mathrm{Cu}\); (c) \(\mathrm{Ge}, \mathrm{Cl}\); (d) \(\mathrm{Pb}, \mathrm{Sb}\).

Hydrogen is an unusual element because it behaves in some ways like the alkali metal elements and in other ways like nonmetals. Its properties can be explained in part by its electron configuration and by the values for its ionization energy and electron affinity. (a) Explain why the electron affinity of hydrogen is much closer to the values for the alkali elements than for the halogens. (b) Is the following statement true? "Hydrogen has the smallest bonding atomic radius of any element that forms chemical compounds. If not, correct it. If it is, explain in terms of electron configurations. (c) Explain why the ionization energy of hydrogen is closer to the values for the halogens than for the alkali metals. (d) The hydride ion is \(\mathrm{H}\). Write out the process corresponding to the first ionization energy of the hydride ion. (e) How does the process in part (d) compare to the process for the electron affinity of a neutral hydrogen atom?

Detailed calculations show that the value of \(Z_{\text {eff }}\) for the outermost electrons in \(\mathrm{Si}\) and \(\mathrm{Cl}\) atoms is \(4.29+\) and \(6.12+\), respectively. (a) What value do you estimate for \(Z_{\text {eff }}\) experienced by the outermost electron in both \(\mathrm{Si}\) and \(\mathrm{Cl}\) by assuming core electrons contribute \(1.00\) and valence electrons contribute \(0.00\) to the screening constant? (b) What values do you estimate for \(Z_{\text {eff }}\) using Slater's rules? (c) Which approach gives a more accurate estimate of \(Z_{\text {eff? }}\) ? (d) Which method of approximation more accurately accounts for the steady increase in \(Z_{\text {eff }}\) that occurs upon moving left to right across a period? (e) Predict \(Z_{\text {eff }}\) for a valence electron in \(\mathrm{P}\), phosphorus, based on the calculations for \(\mathrm{Si}\) and \(\mathrm{Cl}\).
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