Atomic Number as the Basis for the Periodic Table

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In his periodic table Mendeleev had to place certain elements out of the order of increasing atomic mass in order to get them into the proper group. He assumed this was because of errors in calculating atomic masses. However, with improved methods of determining atomic masses and with the discovery of argon (group 0, atomic mass 39.9), which was placed ahead of potassium (group I, atomic mass 39.1), it became clear that a few elements might always remain “out of order”. At the time, these out-of-order placements were justified by chemical evidence. Elements were placed in the groups based on their chemical properties. However, this “out-of-order” placement was easily corrected when atomic numbers (introduced by Moseley) were used instead of atomic masses. For example, placing argon (Z = 18) before potassium (Z = 19) was actually correct and not out-of-place according to their atomic masses. From the standpoint of Moseley’s discovery of atomic number (1913), the periodic law should be re-state as follows:

Similar properties recur periodically when elements are arranged according to increasing atomic number.

Periodic Trends in Atomic Properties

Atomic radius

Because orbital have no specific boundaries, the size of an atom cannot be specified precisely. The easiest way to express the size of an atom is through the measurement of its radius, and there are two ways in which this can be accomplished. For elements that form diatomic molecules, the atomic radius is equal to one-half the internuclear distance. This is called the covalent atomic radius. For example, X-ray crystallography determined that the internuclear distance in Br₂ is 228 pm. The covalent atomic radius for bromine is 114 pm.

For nonmetals that do not form diatomic molecules, the atomic radii are estimated from their various covalent compounds. The radii for metals atoms (called metallic radii) are obtained from half the internuclear distance of two adjacent atoms in metal crystals. Covalent atomic radii are always smaller than if the radii were estimated from the 90% electron density volumes of isolated atoms. This is because, when two atoms approach each other to form covalent bonds, their “electron cloud” interpenetrates. However, these values form a self-consistent data set, and we are more interested in their trends in the periodic table than their absolute values.

In general, the atomic radii decrease in going from left to right across a given period in the periodic table.

This decrease in atomic radii is due to the increasing “effective nuclear charge” in going from left to right, and valence electrons are drawn closer to the nucleus, decreasing the atomic size. For example,

Li > Be > B > C > N > O > F; Na > Mg > Al > Si > P > S > Cl;

Atomic radii increase from top to bottom down a group.

This is because of the increase in orbital size due to increasing principal quantum levels. As one goes down a given group, the valence electrons occupy energy levels with a higher quantum number than the one before. For example,

Li < Na < K < Rb < Cs; F < Cl < Br < I

The radii of cations are smaller and the radii of anions are larger than those of the corresponding atoms from which the ions are derived.

This is because, when an atom loses one of more electrons, repulsions between electrons decreases and effective nuclear attractions on the remaining electrons increase. This results in a decrease in atomic sizes. For a set of isoelectronic cations (ions containing the same number of electrons), the higher the positive charge on the cation, the smaller the size. In the formation of anions, electrons are added to the atoms, which results in an increase of the electron-electron repulsions and expansion of charge cloud expands. For isoelectronic species, the trends of their radii are as follows:

Al³⁺ < Mg²⁺ < Na⁺ < Ne < F^- < O^2- < N^3-; Ca²⁺ < K⁺ < Ar < Cl^- < S^2- ;

Ionization Energy

Ionization energy is the amount of energy required to remove an electron from a gaseous atom or ion:

M(g) à M⁺(g) + e^-; M⁺(g) à M²⁺(g) + e^- ;

Where the atom or ion is assumed to be in its ground state (lowest electronic energy state). The ionization energy values are normally expressed in kJ/mol. For example,

Al(g) à Al⁺(g) + e^- ; I₁ = 580 kJ/mol;

Al⁺(g) à Al²⁺(g) + e^- ; I₂ = 1815 kJ/mol;

Al²⁺(g) à Al³⁺(g) + e^- ; I₃ = 2740 kJ/mol;

Al³⁺(g) à Al⁴⁺(g) + e^- ; I₄ = 11, 600 kJ/mol;

The first ionization energy (I₁) is the energy needed to remove an electron from a neutral atom, and this would be the electron occupying the highest energy orbital. For aluminum ([Ne]3s²3p¹), it will be the electron in the 3p orbital, and its removal yields Al⁺ ion with electron configuration [Ne]3s². The next two electrons are removed from the 3s orbital, and the fourth from the 2p orbital. The removal of second electron requires about three times as much energy as the first one. This is because, the electrons are at a lower electronic energy state (more stable state) and after the removal of the first electron, the effective nuclear charge on the remaining electrons increases and the removal the subsequent electrons becomes more difficult.

It is noted that, the ionization of the fourth electron in aluminum requires more than quadruple the amount of energy needed to remove the third electron. This electron is removed from the 2p orbital, which is in the inner quantum level or “core” shell, thus requiring much more energy. The pattern in which ionization energies of a given element increase provides evidences of the existence of quantized energy levels in atoms.

The trends of ionization energy in the periodic table are as follows:

Ionization energy generally increases from left to right across periods and decrease from top to bottom down groups.

We see some discontinuities in the ionization energy in going across a period. For example, in Period 2, discontinuities occur in going from beryllium to boron and from nitrogen to oxygen. This anomaly can be explain in terms of electron shielding as well as electron-electron repulsion effects. The decrease in ionization energy in going from beryllium to boron is due to the fact that electrons in the filled 2s orbital provide some shielding for electrons in the 2p orbital from the nuclear charge. The decrease in ionization energy in going from nitrogen to oxygen is due to the extra electron repulsions in the doubly occupied 2p orbital in oxygen atom. In Period 3, discontinuities occur in going from magnesium to aluminum and from phosphorus to sulfur.

Electron Affinity

Electron affinity is the energy change associated with the addition of an electron to a gaseous atom:

X(g) + e^- à X^-(g)

Electron affinities generally have negative values because the process is exothermic – the added electrons are experiencing a net nuclear attractions when entering the atoms.

Electron affinity generally increases (become more negative or more exothermic) from left to right across periods and decreases (become less negative) top to bottom down groups, although certain anomalies are noted. Both nuclear attractions and electron repulsions appear to influence electron affinity. For example, nitrogen atom does not form stable, isolated N^-(g) ion, whereas its neighbors, carbon and oxygen, form stable C^-(g) and O^-(g) ions. These facts reflect the difference in their electron configurations. Nitrogen has the configuration 1s² 2s² 2p³, in which the 2p orbital are half-filled (a stable configuration). Adding one more electron to the 2p orbital would decrease the electronic stability due to increasing electron repulsions. Carbon has the configuration [He]2s²2p², where one of the 2p orbital is empty. An electron added to carbon would occupy this empty 2p orbital, which would not result in a significant electron repulsion. While oxygen contains one more proton than nitrogen; although adding an electron to oxygen would increase the electron repulsion, this will be countered by the higher nuclear attraction in oxygen.

Electron affinities of the halogens are given below:

Atoms: F Cl Br I___

E.A. (kJ/mol): -328 -349 -325 -295

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While we see a general trend of decreasing electron affinity going down the group, electron affinity increases (becomes more negative) from fluorine to chlorine. The 2p subshell in fluorine is much smaller compared to the 3p subshell in chlorine. Having eight electrons in the 2p subshell would create a much stronger electron-electron repulsions compared to having the same number of electrons in the 3p subshell. Although an electron added to fluorine atom would experience a greater nuclear attraction, hence the negative energy change, the strong electron repulsion encountered in fluorine reduces this exothermic energy change. The decreasing trend observed in bromine and iodine is due to the fact that the added electrons are experiencing weaker nuclear attractions as a result of larger atomic size.

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