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Chapter 7: Problem 15

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

Short Answer

Expert verified
The n=3 electrons in Krypton (Kr) will experience a greater effective nuclear charge, with \(Z_{eff}(Kr) = 26\), compared to Argon (Ar), where \(Z_{eff}(Ar) = 8\). This stronger attraction between the electrons and nucleus in Kr results in the n=3 electrons being closer to the nucleus than those in Ar.

Step by step solution

01

Understand the concept of effective nuclear charge

Effective nuclear charge refers to the net positive charge experienced by electrons in an atom. Because electrons in inner shells shield the electrons in outer shells from the full positive charge of the nucleus, the effective nuclear charge experienced by outer shell electrons is less than the total positive charge of the protons in the nucleus.
02

Calculate the effective nuclear charge for the n=3 shell electrons in Ar and Kr

The formula for calculating effective nuclear charge (Zeff) is given as follows: \[Z_{eff} = Z - S\] Where \(Z\) is the atomic number (number of protons) and \(S\) is the shielding constant, which represents the shielding effect of inner shell electrons. Argon (Ar) has an atomic number of 18, meaning it has 18 protons. Since the electrons in question are in the n=3 shell, they will experience significant shielding from the 10 electrons in the n=1 and n=2 shells. For simplicity, we'll assume that the shielding constant, S, is equal to the number of inner shell electrons. Thus, for Ar: \[Z_{eff}(Ar) = 18 - 10 = 8\] Krypton (Kr) has an atomic number of 36, meaning it has 36 protons. It also has 10 electrons in the n=1 and n=2 shells, which cause a shielding effect on the electrons in the n=3 shell. Therefore, for Kr: \[Z_{eff}(Kr) = 36 - 10 = 26\]
03

Compare the effective nuclear charges and determine which electrons will be closer to the nucleus

Comparing the calculated effective nuclear charges, we can see that: \[Z_{eff}(Kr) > Z_{eff}(Ar)\] This means electrons in the n=3 shell in Krypton experience a greater effective nuclear charge than those in Argon. A greater effective nuclear charge results in a stronger attraction between the electrons and the nucleus. As a result, the n=3 electrons in Kr will be closer to the nucleus than those in Ar.

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Most popular questions from this chapter

An element \(X\) reacts with oxygen to form \(\mathrm{XO}_{2}\) and with chlorine to form \(\mathrm{XCl}_{4} \mathrm{XO}_{2}\) is a white solid that melts at high temperatures (above \(\left.1000^{\circ} \mathrm{C}\right)\). 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 \(X\) is a metal, nonmetal, or metalloid? Explain. (c) By using a sourcebook such as the CRC Handbook of Chemistry and Physics, try to determine the identity of element \(\mathrm{X}\).

(a) During the period from about 1800 to about 1865 , the atomic weights of many elements were accurately measured. Why was this important to Mendeleev's formulation of the periodic table? (b) What property of the atom did Moseley associate with the wavelength of X-rays emitted from an element in his experiments? (c) Why are chemical and physical properties of the elements more closely related to atomic number than they are to atomic weight?

(a) How is the concept of effective nuclear charge used to simplify the numerous electron-electron repulsions in ? many-electron atom? (b) Which experiences a greater effective nuclear charge in a Be atom, the 1 s electrons or the 2 s electrons? Explain.

For each of the following statements, provide an explanation (a) \(\mathrm{O}^{2-}\) is larger than \(\mathrm{O} ;\) (b) \(\mathrm{S}^{2-}\) is larger than \(\mathrm{O}^{2-}\) : (c) \(S^{2-}\) is larger than \(K^{+}\); (d) \(\mathrm{K}^{+}\) is larger than \(\mathrm{Ca}^{2+}\).

Chlorine reacts with oxygen to form \(\mathrm{Cl}_{2} \mathrm{O}_{7} .\) (a) What is the name of this product (see Table 2.6)? (b) Write a balanced equation for the formation of \(\mathrm{Cl}_{2} \mathrm{O}_{7}(l)\) from the elements. (c) Under usual conditions, \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) is a colorless liquid with a boiling point of \(81^{\circ} \mathrm{C}\). Is this boiling point expected or surprising? (d) Would you expect \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) to be more reactive toward \(\mathrm{H}^{+}(a q)\) or \(\mathrm{OH}^{-}(a q) ?\) Explain.
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