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As we move across a period of the periodic table, why do the sizes of the transition elements change more gradually than those of the representative elements?

Short Answer

Expert verified
The sizes of transition elements change more gradually than those of representative elements across a period because the electron shielding by the inner d and f electrons is less effective than the shielding by s and p electrons. This results in a small increase in effective nuclear charge, leading to a more gradual decrease in atomic size for transition elements. In contrast, representative elements experience a significant increase in effective nuclear charge due to constant shielding effect and an increase in the number of protons in the nucleus, resulting in a more noticeable decrease in their atomic size.

Step by step solution

01

Understanding Atomic Size

Atomic size generally refers to the distance between an atom's nucleus and its outermost electrons. Moving across a period in the periodic table, atomic size usually decreases. This is because of the increase in effective nuclear charge (the net positive charge experienced by the outermost electrons) due to the addition of protons in the nucleus.
02

Factors Affecting Atomic Size

There are two main factors affecting the atomic size of an element: 1. Effective nuclear charge (Z_eff): As the number of protons in the nucleus increases, the attractive force between the nucleus and the electron increases, pulling the electrons closer to the nucleus and decreasing the atomic size. 2. Electron shielding: Inner shell electrons repel outer shell electrons, reducing the effective nuclear charge experienced by the outer shell electrons. This leads to a decrease in the force of attraction between the nucleus and the outer electrons and increases the atomic size.
03

Electron Configurations of Transition Elements

Transition elements are located in the d-block of the periodic table and have partially filled d-subshells in their electron configuration. The electron shielding by the inner f and d electrons is less effective than the shielding by the s and p electrons. As a result, the effective nuclear charge experienced by the outermost electrons increases marginally, leading to a small decrease in atomic size.
04

Electron Configurations of Representative Elements

Representative elements are located in the s and p blocks of the periodic table. When moving across a period, electrons are added to the same energy level. The shielding effect remains almost constant while the effective nuclear charge increases considerably due to the increase in the number of protons in the nucleus. This results in a significant decrease in atomic size.
05

Conclusion

The sizes of transition elements change more gradually than those of representative elements when moving across a period of the periodic table because the electron shielding by the inner d and f electrons is less effective than the shielding by s and p electrons. This leads to a small increase in effective nuclear charge, resulting in a more gradual decrease in atomic size for transition elements. On the other hand, representative elements experience a significant increase in effective nuclear charge due to constant shielding effect and increase in the number of protons in the nucleus, leading to a more noticeable decrease in their atomic size.

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Most popular questions from this chapter

The first ionization energy of the oxygen molecule is the energy required for the following process: $$ \mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{2}^{+}(g)+\mathrm{e} $$ The energy needed for this process is \(1175 \mathrm{~kJ} / \mathrm{mol}\), very similar to the first ionization energy of Xe. Would you expect \(\mathrm{O}_{2}\) to react with \(\mathrm{F}_{2}\) ? If so, suggest a product or products of this reaction.

Write electron configurations for the following ions, and determine which have noble-gas configurations: (a) \(\mathrm{Cr}^{3+}\), (b) \(\mathrm{N}^{3-}\) (c) \(\mathrm{Sc}^{3+}\), (d) \(\mathrm{Cu}^{2+}\), (e) \(\mathrm{Tl}^{+}\), (f) \(\mathrm{Au}^{+}\).

Why does xenon react with fluorine, whereas neon does not?

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.

One way to measure ionization energies is photoelectron spectroscopy (PES), a technique based on the photoelectric effect. em (Section 6.2) In PES, monochromatic light is directed onto a sample, causing electrons to be emitted. The kinetic energy of the emitted electrons is measured. The difference between the energy of the photons and the kinetic energy of the electrons corresponds to the energy needed to remove the electrons (that is, the ionization energy). Suppose that a PES experiment is performed in which mercury vapor is irradiated with ultraviolet light of wavelength \(58.4 \mathrm{~nm}\). (a) What is the energy of a photon of this light, in \(\mathrm{eV}\) ? (b) Write an equation that shows the process corresponding to the first ionization energy of \(\mathrm{Hg}\). (c) The kinetic energy of the emitted electrons is measured to be \(10.75 \mathrm{eV}\). What is the first ionization energy of \(\mathrm{Hg}\), in \(\mathrm{kJ} / \mathrm{mol} ?\) (d) With reference to Figure \(7.11\), determine which of the halogen elements has a first ionization energy closest to that of mercury.

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