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Ionization Energy



Ionization Energy

The process of removing an electron from an isolated atom to form a positive ion is called
ionisation. Energy will be required to remove an electron from the atom against the force of attraction
of the nucleus.

The ionisation energy (IE) of an element: is defined as the energy needed to remove a single
electron from an atom of the element in the gaseous state. That is,


Since one, two or more electrons may be removed from the same atom, one after the other, we
have as many ionisation energies of the element.

The First ionisation energy (IE1), is the energy needed to remove the first electron from the
gaseous atom M to form M+ ion.

The Second ionisation energy (IE2), is the energy needed to remove a second electron, from the gaseous M+ ion to form M2+ ion.

Higher ionisation energies can be defined in the same way. We can depict the first, second and third ionisation energies in the form of equations as:


Ionization energies are sometimes called Ionization potentials. Ionization energies are usually
expressed in electron volts (eV) per atom, or in kilojoules per mole of atoms (kJ mol–1). For
conversion, 1eV atom–1 = 96.48 kJ mol–1.

Measurement of Ionisation Energies

The amount of energy required to detach an electron from an atom can be measured by supplying the required energy as thermal energy, electrical energy, or radiant energy. Thus ionisation energies can be determined from the spectrum of the element or by any of the two methods detailed below.

(1) The Electrical method

The apparatus used is shown in the following figure:


The electrically heated tungsten wire emits electrons.

The grid can be charged positively to different voltages which we read with a voltmeter.

The plate opposite the grid has a small negative charge.

When the potential to the grid is zero, no current flows between the grid and the plate. However if we give sufficient potential to the grid, the electrons emitted by the tungsten wire are accelerated towards the grid, pass through it and ionise the atoms between grid and plate. The electron ejected by each atom is attracted to grid and positive ion is attracted to plate.

A current thus passes between grid and  plate which is shown up by an ammeter.

The minimum grid voltage that just produces a current is called ionization potential.

If V be the ionization potential, the ionization energy (IE) is calculated as:

(2) Photo-ionisation Method

The gaseous atoms are introduced into a chamber containing two electrically charged plates



As neutral atoms, they do not conduct electricity and no current flows between the plates.
When radiant energy (hν) is supplied to the gaseous atoms, ionisation will occur and electric current will flow.

The frequency of the radiation used is gradually increased. The minimum frequency necessary to cause ionisation of the gaseous atoms, as shown by the flow of an electric current is noted. From this frequency the ionisation energy is calculated.

Order of Successive Ionization Energies

The second ionisation energy (IE2) is larger than the first ionisation energy (IE1) because it is
more difficult to detach an electron from a +ve ion than a neutral atom.

The third ionisation energy (IE3) is still larger as the third electron has to be detached from a 2 + ion. Thus in general successive ionisation energies increase in magnitude. That is,

IE1 < IE2 < IE3 < IE4 , and so on.

For illustration, the first four ionisation energies for sodium and magnesium are listed below:

Principal Trends in Ionization Energies

A graph of the first ionisation energies against atomic number (Z) for the first 18 elements of the Periodic Table is shown in the following figure:

figure (1)
The important trends as illustrated by the graph are:
(1) Ionisation energies increase across a period. e.g., Li to Ne.
(2) Ionisation energies decrease down a group e.g., Li, Na, K.
(3) There are regular discontinuities in the increase trend across a period e.g., Be to B, and N to O.

Increase across a Period

As we pass from left to right in a period, the first ionisation energy shows a steady increase. Thus in Period 2 from Li to N, we have:


Explanation

The outer-shell electrons in the elements of the same period are arranged in the same shell. For example, the build up of electrons in Period 2 from Li to B is shown in the following figure:


Moving from Li to B, the positive charge on the nucleus increases whereas the distance between the nucleus and valence electrons decreases. Therefore more energy is required to remove an electron as we go from left to right in the Period. Since the number of screening electrons remains the same, they do not upset the increase trend.

Decrease down a Group

In the elements of a vertical Group of the Periodic table, the number of outer shell electrons is the same. But the following changes are noted from top to bottom.

(1) The principal quantum number n containing the valence electrons increases.

(2) The nuclear charge (At. No.) increases.

(3) The number of electrons in the inner shells (shielding electrons) increases.

The net result of these changes is that the first ionisation energies down a group record a progressive decrease. Thus for Group IA we have:


Let us explain the above decrease trend by taking example of lithium and sodium. They have the atomic structures.


Lithium and sodium both have one outer-shell electron. The number of shielding electrons in sodium is 10 while in lithium it is 2. If we assume that the inner shell electrons provide hundred percent screening, the core charge attracting the outer-shell electron would be :


Thus the same net charge (+ 1) attracts the outer-shell electrons to the core. But the distance of the outer electron from the nucleus is greater in Na (n = 3) than in Li (n = 1). Therefore the force of attraction between the outer electron and the core will be less in Na than in Li. That explains the lower IE of Na compared to Li. By the same line of argument, the decrease trend in IE from element to element while going down a Group can be justified.

Regular Discontinuities

As already discussed, the first ionisation energies increase across a period. But this increase
trend is upset at the third and sixth element in a period.

As clear from graph in Fig.(1), there are breaks at B and O which occupy the third and fifth positions respectively in the 2nd period. The IE1 of B is less than that of Be and the IE1 of O is less than that of N.

Explanation

(a) The electronic configuration of Be and B are :


The 2p orbital electron of B is already higher in energy than the 2s orbital electron. Therefore the
removal of electron from B requires less energy and its IE1 is lower.

(b) The electronic configuration of N and O is :


The 2p orbitals may be represented as:


 Whenever two electrons occupy a particular orbital, they repel each other. As a result it is easier to remove one of the paired 2p electrons from O than it is to remove an unpaired electron from N atom. Thus IE1 of O is lower than that of N.

Reference: Essentials of Physical Chemistry /Arun Bahl, B.S Bahl and G.D. Tuli / multicolour edition.



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