Ionization energy: Wikis

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The term ionization energy (EI) (of an atom or molecule) is most commonly used to refer to the energy required to remove (to infinity) the outermost electron in the atom or molecule when the gas atom or molecule is isolated in free space and is in its ground electronic state. This quantity was formerly called ionization potential, and was at one stage measured in volts. The name "ionization energy" is now strongly preferred. In atomic physics the ionization energy is measured using the unit "electronvolt" (eV). In chemistry, it is more normal to give the value, in kJ/mol (or formerly kcal/mol) of the related quantity that (strictly) should be called the "molar ionization energy" but is often just called "ionization energy" [1].

More generally, the nth ionization energy is the energy required to strip off the nth electron after the first n − 1 electrons have been removed. It is considered a measure of the "reluctance" of an atom or ion to surrender an electron, or the "strength" by which the electron is bound; the greater the ionisation energy, the more difficult it is to remove an electron. The ionization energy is, thus, an indicator of the reactivity of an element. Elements with a low ionization energy tend to be reducing agents and form cations, which in turn combine with anions to form salts.

The term "ionization energy" is sometimes used as a name for the work needed to remove (to infinity) the topmost electron from an atom or molecule adsorbed onto a surface. However, due to interactions with the surface, this value differs from the ionisation energy of the atom or molecule in question when it is in free space. So, in the case of surface-adsorbed atoms and molecules, it may be better to use the more general term "electron binding energy", in order to avoid confusion. Both these names are also sometimes used to describe the work needed to remove an electron from a "lower" orbital (i.e., not the topmost orbital) to infinity, both for free and for adsorbed atoms and molecules; in such cases it is necessary to specify the orbital from which the electron has been removed.

Contents

Values and trends

Main article: Molar ionization energies of the elements

Generally the (n+1)th ionization energy is larger than the nth ionization energy. Always, the next ionization energy involves removing an electron from an orbital closer to the nucleus. Electrons in the closer orbitals experience greater forces of electrostatic attraction; thus, their removal requires increasingly more energy.

Some values for elements of the third period are given in the following table:

Successive molar ionization energies in kJ/mol
(96.485 kJ/mol = 1 eV/particle)
Element First Second Third Fourth Fifth Sixth Seventh
Na 496 4,560
Mg 738 1,450 7,730
Al 577 1,816 2,881 11,600
Si 786 1,577 3,228 4,354 16,100
P 1,060 1,890 2,905 4,950 6,270 21,200
S 999.6 2,260 3,375 4,565 6,950 8,490 27,107
Cl 1,256 2,295 3,850 5,160 6,560 9,360 11,000
Ar 1,520 2,665 3,945 5,770 7,230 8,780 12,000
BE vs Z.jpg

Large jumps in the successive molar ionization energies occur when passing noble gas configurations. For example, as can be seen in the table above, the first two molar ionization energies of magnesium (stripping the two 3s electrons from a magnesium atom) are much smaller than the third, which requires stripping off a 2p electron from the very stable neon configuration of Mg2+.

Periodic trend for ionization energy. Each period begins at a minimum for the alkali metals, and ends at a maximum for the noble gases.

Ionization energy is also a periodic trend within the periodic table organization. Moving left to right within a period or upward within a group, the ionization energy generally increases. As the atomic radius decreases, it becomes harder to remove an electron that is closer to a more positively charged nucleus.

Electrostatic explanation

Atomic ionization energy can be predicted by an analysis using electrostatic potential and the Bohr model of the atom, as follows.

Consider an electron of charge -e and an atomic nucleus with charge +Ze, where Z is the number of protons in the nucleus. According to the Bohr model, if the electron were to approach and bind with the atom, it would come to rest at a certain radius a. The electrostatic potential V at distance a from the ionic nucleus, referenced to a point infinitely far away, is:

V = \frac{Ze}{a} \,\!

Since the electron is negatively charged, it is drawn inwards by this positive electrostatic potential. The energy required for the electron to "climb out" and leave the atom is:

E = eV = \frac{Ze^2}{a} \,\!

This analysis is incomplete, as it leaves the distance a as an unknown variable. It can be made more rigorous by assigning to each electron of every chemical element a characteristic distance, chosen so that this relation agrees with experimental data.

It is possible to expand this model considerably by taking a semi-classical approach, in which momentum is quantized. This approach works very well for the hydrogen atom, which only has one electron. The magnitude of the angular momentum for a circular orbit is:

 L = |\mathbf r \times \mathbf p| = rmv = n\hbar

The total energy of the atom is the sum of the kinetic and potential energies, that is:

 E = T + U = \frac{p^2}{2m_e} - \frac{Ze^2}{r} = \frac{m_e v^2}{2} - \frac{Ze^2}{r}

Velocity can be eliminated from the kinetic energy term by setting the Coulomb attraction equal to the centripetal force, giving:

 T = \frac{Ze^2}{2r}

Now the energy can be found in terms of Z, e, and r. Using the new value for the kinetic energy in the total energy equation above, it is found that:

E = - \frac{Ze^2}{2r}

Solving the angular momentum for v and substituting this into the expression for kinetic energy, we have:

\frac{n^2 \hbar^2}{rm_e} = Ze^2

This establishes the dependence of the radius on n. That is:

r(n) = \frac{n^2 \hbar^2}{Zm_e e^2}

At its smallest value, n is equal to 1 and r is the Bohr radius a0. Now, the equation for the energy can be established in terms of the Bohr radius. Doing so gives the result:

 E = - \frac{1}{n^2} \frac{Z^2e^2}{2a_0} = - \frac{Z^213.6eV}{n^2}

Quantum-mechanical explanation

According to the more complete theory of quantum mechanics, the location of an electron is best described as a probability distribution. The energy can be calculated by integrating over this cloud. The cloud's underlying mathematical representation is the wavefunction which is built from Slater determinants consisting of molecular spin orbitals. These are related by Pauli's exclusion principle to the antisymmetrized products of the atomic or molecular orbitals.

In general, calculating the nth ionization energy requires calculating the energies of Zn + 1 and Zn electron systems. Calculating these energies exactly is not possible except for the simplest systems (i.e. hydrogen), primarily because of difficulties in integrating the coorelation terms. Therefore, approximation methods are rountinely employed, with different methods varying in complexity (computational time) and in accuracy compared to empirical data. This has become a well-studied problem and is routinely done in computational chemistry. At the lowest level of approximation, the ionization energy is provided by Koopmans' theorem.

Vertical ionization energy and adiabatic ionization energy

The geometry of a molecular ion may be different from the neutral molecule. The measured ionization energy can refer to the vertical ionization energy, in which case the ion has the same geometry as the neutral molecule, or to the adiabatic ionization energy, in which case the ion has its lowest energy, relaxed geometry. This is illustrated in the figure. For a diatomic molecule the only geometry change possible is the bond length. The figure shows an ion with a slightly longer bond length than the neutral molecule. The harmonic potential energy surfaces are shown in green (neutral) and red (ion) with vibrational energy levels. The vertical ionization energy is always greater than the adiabatic ionization energy [2].

See also

References

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Simple English

Ionization energy is the energy needed to take an electron from an atom. The atom is not connected to any other atoms. The chemical elements to the left of the periodic table have a much lower ionization energy. The ones to the right have a much higher ionization energy. The ionization energy increases as each electron is removed.


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