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Explain the following variations in atomic or ionic radii: (a) I>I>I+ (b) Ca2+>Mg2+>Be2+ (c) Fe>Fe2+>Fe3+

Short Answer

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The three cases demonstrate different trends in atomic and ionic radii: (a) I>I>I+: The iodine anion has gained an electron, causing repulsion between electrons and resulting in an increased atomic radius. The iodine cation has lost an electron, reducing electron-electron repulsion and decreasing its radius. (b) Ca2+>Mg2+>Be2+: As we move down Group 2 elements in the Periodic Table, the atomic radius increases due to the addition of electron shells. All three ions have divalent charges but differ in electron shells. (c) Fe>Fe2+>Fe3+: As neutral iron loses electrons to form cations, electron-electron repulsion decreases and effective nuclear charge felt by the remaining electrons increases, causing the cations to have smaller radii.

Step by step solution

01

Understanding Atomic/ Ionic Radius

The radius of an atom/ ion is determined by the number of electron shells it has (the period it belongs to in the Periodic Table), whether it has lost, gained or has not changed the number of electrons compared to the neutral atom (ionization state) and the effective nuclear charge felt by the outermost (valence) electrons.
02

Iodine Anion, Neutral Atom, and Cation

(a) Compare I, I, and I+. All have the same numbers of protons, but differ in the number of electrons. - I has gained one electron, so the repulsion between the extra electron and the other electrons causes the atomic radius to increase. - I is neutral iodine and thus has a smaller radius because it does not have the extra electron of I. - I+ has lost an electron, reducing the electron-electron repulsion, which results in a decrease in radius due to the attractive force from the protons in the nucleus.
03

Divalent Cations of Group 2 Elements

(b) Compare Ca2+, Mg2+, and Be2+. These are all divalent cations, so they have the same charge state, but the atoms from which they are derived differ in their position in the Periodic Table. - Group 2 elements (where Ca, Mg, and Be are located) have a trend of increasing atomic radius as we move down the group due to the addition of electron shells. Hence, Ca2+ has a larger ionic radius than Mg2+, which in turn has a larger radius than Be2+.
04

Iron in Different Oxidation States

(c) Compare Fe, Fe2+, and Fe3+. These all feature the same atomic species, but the number of electrons differs. - Fe is neutral iron and has the most electrons of the three. As electrons are lost to form Fe2+ and Fe3+, the cations are smaller because the electron-electron repulsion decreases and the effective nuclear charge felt by the remaining electrons increases. Therefore, Fe>Fe2+>Fe3+.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ionization State
The ionization state of an atom refers to its electrical charge resulting from the loss or gain of electrons. When an atom gains electrons, it becomes an anion with a negative charge. Conversely, when it loses electrons, it becomes a cation with a positive charge. This process can significantly affect the atomic or ionic radius based on electron-electron repulsion and the number of electron shells.

  • Anions are larger because adding electrons increases repulsion among electrons, causing the outermost shell to expand.
  • Cations are smaller since losing electrons reduces repulsion, allowing the nuclei to pull the remaining electrons closer.
Understanding ionization state explains why I is larger than I, and I+ is the smallest. Similarly, for metals like iron, from Fe to Fe3+, the decrease in electron count leads to a smaller size due to less repulsion and greater attractive forces towards the nucleus.
Effective Nuclear Charge
The effective nuclear charge (Zeff) is a crucial concept that determines the overall force exerted by the nucleus on the electrons in an atom. It's the net positive charge experienced by electrons after accounting for shielding effects from other electrons, especially in inner shells.

As electrons are removed, the Zeff increases because the remaining electrons experience less shielding and thus feel a stronger attraction to the nucleus. For instance:
  • In Fe2+ and Fe3+, as electrons are stripped away, the remaining electrons are pulled closer, reducing atomic size.
  • Even among similar ions like Ca2+, Mg2+, and Be2+, those lower in the periodic table tend to have larger radii due to additional shielding, even as Zeff acts strongly on outer shells.
Thus, the concept of Zeff explains the variation in sizes among ions and contributes to understanding why cations are generally smaller than their neutral atoms.
Periodic Table Trends
The periodic table is a powerful tool for predicting and explaining trends in atomic and ionic sizes. As you move across a period from left to right, atomic numbers increase, but the addition of electrons generally does not increase the size due to increased Zeff pulling electrons closer.

Key trends include:
  • Within the same group, atomic and ionic radii increase from top to bottom because of additional electron shells, even as Zeff might increase slightly.
  • In a period, as seen in Ca2+, Mg2+, Be2+, moving left to right generally results in smaller radii due to stronger attraction from increasing protons.
By understanding these periodic trends, one can predict and explain the relative sizes of ions and atoms, providing a clearer insight into chemical properties and behaviors.

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Most popular questions from this chapter

Chlorine reacts with oxygen to form Cl2O7. (a) What is the name of this product (see Table 2.6)? (b) Write a balanced equation for the formation of Cl2O7(l) from the elements. (c) Would you expect Cl2O7 to be more reactive toward H+(aq) or OH(aq)?(d) If the oxygen in Cl2O7 is considered to have the -2 oxidation state, what is the oxidation state of the Cl ? What is the electron configuration of Cl in this oxidation state?

Consider the isoelectronic ions Cl and K+. (a) Which ion is smaller? (b) Using Equation 7.1 and assuming that core electrons contribute 1.00 and valence electrons contribute nothing to the screening constant, S, calculate Zeff  for these two ions. (c) Repeat this calculation using Slater's rules to estimate the screening constant, S.(d) For isoelectronic ions, how are effective nuclear charge and ionic radius related?

Tungsten has the highest melting point of any metal in the periodic table: 3422C. The distance between W atoms in tungsten metal is 274 pm. (a) What is the atomic radius of a tungsten atom in this environment? (This radius is called the metallic radius.) (b) If you put tungsten metal under high pressure, predict what would happen to the distance between W atoms.

Write a balanced equation for the reaction that occurs in each of the following cases: (a) Potassium metal is exposed to an atmosphere of chlorine gas. (b) Strontium oxide is added to water. (c) A fresh surface of lithium metal is exposed to oxygen gas. (d) Sodium metal reacts with molten sulfur.

Mercury in the environment can exist in oxidation states 0, +1, and +2. One major question in environmental chemistry research is how to best measure the oxidation state of mercury in natural systems; this is made more complicated by the fact that mercury can be reduced or oxidized on surfaces differently than it would be if it were free in solution. XPS, X-ray photoelectron spectroscopy, is a technique related to PES (see Exercise 7.111 ), but instead of using ultraviolet light to eject valence electrons, X rays are used to eject core electrons. The energies of the core electrons are different for different oxidation states of the element. In one set of experiments, researchers examined mercury contamination of minerals in water. They measured the XPS signals that corresponded to electrons ejected from mercury's 4 forbitals at 105eV, from an X-ray source that provided 1253.6eV of energy (1ev=1.602×1019 J) The oxygen on the mineral surface gave emitted electron energies at 531eV, corresponding to the 1s orbital of oxygen. Overall the researchers concluded that oxidation states were +2 for Hg and -2 for O. (a) Calculate the wavelength of the X rays used in this experiment. (b) Compare the energies of the 4f electrons in mercury and the 1s electrons in oxygen from these data to the first ionization energies of mercury and oxygen from the data in this chapter. (c) Write out the ground- state electron configurations for Hg2+ and O2; which electrons are the valence electrons in each case?

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