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9.4: The Inert-Pair Effect

  • Page ID
    31173
  • Skills to Develop

    • To become familiar with the Inert Pair Effect and the two periodic trends responsible for its origin.

    The inert-pair effect refers to the empirical observation that the heavier elements of groups 13–17 often have oxidation states that are lower by 2 than the maximum predicted for their group. For example, although an oxidation state of +3 is common for group 13 elements, the heaviest element in Group 13, thallium (Tl), is more likely to form compounds in which it has a +1 oxidation state.

    The inert pair effect says that the \(ns^2\) valence electrons of metallic elements, especially the \(5s^2\) and \(6s^2\) pairs that follow the second and third row of transition metals, are less reactive than would be expected based on periodic trends such as effective nuclear charge, atomic sizes, and ionization energies. In, Tl, Sn, Pb, Sb, Bi, and, To some extent, Po do not always show their expected maximum oxidation states. Rather, sometimes they form compounds in which their oxidation states are 2 less than what would be expected.

    There appear to be two major reasons for the inert-pair effect:

    1. increasing ionization energies and
    2. decreasing bond strengths.

    Ionization Energies

    One of the main principles used to explain this effect is the trend in ionization energies going down a particular group. Atomic size generally increases down a group, and because atomic size increases down a group, ionization energy is typically expected to gradually decrease down the group. There is a decrease from B to Al as would be expected, but the ionization energies of Ga, In, and Tl are higher than the trend would usually predict. The best explanation is that the 4s, 5s, and 6s electrons of these elements are not well shielded from the nucleus by the d and f subshells. These electrons are more difficult to ionize because they experience a larger effective nuclear charge than expected.

    The ionization energies increase because filled (n − 1)d or (n − 2)f subshells are relatively poor at shielding electrons in ns orbitals. Thus the two electrons in the ns subshell experience an unusually high effective nuclear charge, so they are strongly attracted to the nucleus, reducing their participation in bonding. It is therefore substantially more difficult than expected to remove these ns2electrons, as shown in Table 9.4.1 by the difference between the first ionization energies of thallium and aluminum. Because Tl is less likely than Al to lose its two ns2 electrons, its most common oxidation state is +1 rather than +3.

    Table \(\PageIndex{1}\): Ionization Energies (I) and Average M–Cl Bond Energies for the Group 13 Elements
    Element Electron Configuration I1 (kJ/mol) I1 + I2 + I3 (kJ/mol) Average M–Cl Bond Energy (kJ/mol)
    B [He] 2s22p1 801 6828 536
    Al [Ne] 3s23p1 578 5139 494
    Ga [Ar] 3d104s24p1 579 5521 481
    In [Kr] 4d105s2p1 558 5083 439
    Tl [Xe] 4f145d106s2p1 589 5439 373

    Source of data: John A. Dean, Lange’s Handbook of Chemistry, 15th ed. (New York: McGraw-Hill, 1999).

    Bond Energies

    Another important principle involved in the inert pair effect is the trend in bond energies going down a group. Bond energies usually decrease down the group. A decrease in bond energies down a group is expected due to increasing atomic size, and consequentially, an increase in bond distance. When the bond distance increases, bond strength decreases because it is more difficult for the two elements to hold together. This trend combined with the higher-than-average ionization energies make up a lot of the reason for the inert pair effect. To give an example, for the elements affected by the inert pair effect, more energy is needed to get the electrons to form bonds, and not enough of the energy is recovered when the bonds are formed to get the elements to their expected oxidation state, so they achieve the normal oxidation state -2.

    Going down a group, the atoms generally became larger, and the overlap between the valence orbitals of the bonded atoms decreases. Consequently, bond strengths tend to decrease down a column. As shown by the M–Cl bond energies listed in Table \(\PageIndex{1}\), the strength of the bond between a group 13 atom and a chlorine atom decreases by more than 30% from B to Tl. Similar decreases are observed for the atoms of groups 14 and 15.

    In moving down a group in the p-block, increasing ionization energies and decreasing bond strengths result in an inert-pair effect.

    The net effect of these two factors—increasing ionization energies and decreasing bond strengths—is that in going down a group in the p-block, the additional energy released by forming two additional bonds eventually is not great enough to compensate for the additional energy required to remove the two ns2 electrons.

    Example \(\PageIndex{1}\)

    Based on the positions of the Group 13 elements in the periodic table and the general trends outlined in this section,

    1. classify these elements as metals, semimetals, or nonmetals.
    2. predict which element forms the most stable compounds in the +1 oxidation state.
    3. predict which element differs the most from the others in its chemistry.
    4. predict which element of another group will exhibit chemistry most similar to that of Al.

    Given: positions of elements in the periodic table

    Asked for: classification, oxidation-state stability, and chemical reactivity

    Strategy:

    From the position of the diagonal line in the periodic table separating metals and nonmetals, classify the group 13 elements. Then use the trends discussed in this section to compare their relative stabilities and chemical reactivities.

    Solution:

    1. Group 13 spans the diagonal line separating the metals from the nonmetals. Although Al and B both lie on the diagonal line, only B is a semimetal; the heavier elements are metals.
    2. All five elements in group 13 have an ns2np1 valence electron configuration, so they are expected to form ions with a +3 charge from the loss of all valence electrons. The inert-pair effect should be most important for the heaviest element (Tl), so it is most likely to form compounds in an oxidation state that is lower by 2. Thus the +1 oxidation state is predicted to be most important for thallium.
    3. Among the main group elements, the lightest member of each group exhibits unique chemistry because of its small size resulting in a high concentration of charge, energetically unavailable d orbitals, and a tendency to form multiple bonds. In group 13, we predict that the chemistry of boron will be quite different from that of its heavier congeners.
    4. Within the s and p blocks, similarities between elements in different groups are most marked between the lightest member of one group and the element of the next group immediately below and to the right of it. These elements exhibit similar electronegativities and charge-to-radius ratios. Because Al is the second member of group 13, we predict that its chemistry will be most similar to that of Be, the lightest member of group 2.

    Exercise \(\PageIndex{1}\)

    Based on the positions of the group 14 elements C, Si, Ge, Sn, and Pb in the periodic table and the general trends outlined in this section,

    1. classify these elements as metals, semimetals, or nonmetals.
    2. predict which element forms the most stable compounds in the +2 oxidation state.
    3. predict which element differs the most from the others in its chemistry.
    4. predict which element of Group 14 will be chemically most similar to a group 15 element.

    Answer:

    1. nonmetal: C; semimetals: Si and Ge; metals: Sn and Pb
    2. Pb is most stable as M2+.
    3. C is most different.
    4. C and P are most similar in chemistry.

    The Relativistic (And more Advanced) Perspective

    The \(1s\) electrons of heavier elements have such high momenta that they move at speeds close to the speed of light which means relativistic corrections become important. This leads to an increase of the electron mass. Since it's know from the quantum mechanical calculations of the hydrogen atom that the electron mass is inversely proportional to the orbital radius this results in a contraction of the \(1s\) orbital. Now, this contraction of the \(1s\) orbital leads to a decreased degree of shielding for the outer \(s\) electrons, which in turn leads to a cascade of contractions of those outer \(s\) orbitals. The result of this relativistic contraction of the \(s\) orbitals is that the valence \(s\) electrons behave less like valence electrons and more like core electrons, i.e. they are less likely to take part in chemical reactions and they are harder to remove via ionization, because the \(s\) orbitals' decreased size lessens the orbital overlap with potential reaction partners' orbitals and leads to a lower energy. So, while lighter \(p\)-block elements (like \(\ce{Al}\)) usually "give away" their \(s\) and \(p\) electrons when they form chemical compounds, heavier \(p\)-block elements (like \(\ce{Tl}\)) tend to "give away" their \(p\) electrons but keep their \(s\) electrons. That's the reason why for example \(\ce{Al(III)}\) is preferred over \(\ce{Al(I)}\), but \(\ce{Tl(I)}\) is preferred over \(\ce{Tl(III)}\).

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