Chapter 2:- Periodic properties of the elements
B.sc 1st year Book
(Page 5)

Ionization Potential or Ionization Energy

If energy is supplied to an isolated atom from external sources then the promotion of its electron takes place from a lower energy level to a higher energy level. If this process is continued a situation arises when the electron is completely removed from the influence of the attractive force of the nucleus and the atom is converted into a positively charged ion. The energy supplied in this process is known as ‘ionization energy (I E) or ‘Ionization potential(I P). Since in this case energy is required, I E is represented with a positive sign and the process is endothermic.
Thus, the ionization energy is defined as ‘the amount of energy required to convert an isolated gaseous atom into its cation. or The amount of energy needed for an isolated gaseous atom to remove its mast loosely bound electron to produce a cation is called “ionization energy“.
M(g) + Energy(l4) → M+(g) + e-
Unit of IE: Ionization energy is measured in electron volts or kilo calories per mole or kilojoule per mole hence it is represented as eV or Kcal/ mole or KJ/ mole.

Successive ionization energies:

It is noted that electrons are removed in stages one by one from an atom. Thus, an atom may have a number of values of ionization energies viz I1, I2, I3, I4…. which are called first, second, third, fourth,… ionization energy respectively. It is actually equal to its nuclear charge ( Z ). For example for H-atom there is only one value of IE1ieI1; Helium-atom has two values of IE; ie I1 and I2: Lithium has three values of IE, I1, I2, and I3, and so on.

These ionization energies may be defined as :

The amount of energy required to convert an isolated neutral atom into its cation having a single positive charge M′ is called the ‘first ionization energy( I1). The energy required to remove the second electron from the cation, M+ to produce an M2+ ion is called the ‘second ionization energy( I2). Similarly, the energy required to convert M2+ ion into M3+ ion is called third ionization energy (I3), and so on. Thus,
M(g) + (I1) → M+(g) + e-
M+(g) + (I2) → M2+(g) + e-
M2+(g) + (I3) → M3+(g) + e-
Since the removal of an electron from a cation having a high positive charge is relatively more difficult than from a cation having a lower positive charge or its corresponding neutral atom. Hence, for a given element M(g), each successive ionization potential is greater than the previous one, f.e: the magnitude of successive ionization energies has the following order:

I1<I2<I3<I4….As shown in Table 2.9.

 Table 2.9 Successive ionization potential(in eV) I1, I2, I3, I4…. from H to He

The following are the Important factors that affect the magnitude of the ionization energy of the elements:

(i) Ionization energy effect in atomic size:

The larger the size of an atom, the smaller would be its ionization energy. Thus, with the increase in atomic radius, the ionization energy decreases. This is because of the fact that as the size of an atom increases, the electrons of the outermost shell lie. farther away from the nucleus, Hence, according to Coulomb’s law, the attraction between the nucleus and the outermost electron decreases, and it becomes easier to remove an electron from a larger atom than from a smaller atom.
Thus, on moving from top to bottom in a group the ionization energy of the elements decreases with the increase of the atomic size. For example; when we consider the ionization energy of IA group elements we observed that hydrogen has the highest value of ionization energy because of its smallest size whereas cesium; Cs has the lowest value of ionization energy due to its biggest size.
Thus,
$$I.E\:\alpha \frac{1}{atomic\:radii}$$

(ii) Ionization energy effect in nuclear charge:

The greater the charge on the nucleus of an atom greater would be the ionization energies. The reason for this is that when the nuclear charge of an atom is greater, the outer electrons are more tightly held by the nucleus. So more energy would be required to put out an electron from the atom and hence greater would be the value of ionization energy.
Thus, the value of ionization energy gradually increases on moving from left to right in a period with an increase in nuclear charge. For example; in second-period ionization energies increases as we move from Li to Ne (with few exceptions).
In general,

IE directly proportional to the nuclear charge

(iii) Shielding effect or Screening effect:

The larger the shielding effect in an atom, the smaller the ionization energy it is because when the screening effect is greater, the electrons in the valency shell experience less attraction from the nucleus. Consequently. the lower will be the value of ionization energy.
Thus, as we move down a group; the number of inner shells increases which causes a greater screening effect, and hence the ionization energy tends to decrease.
$$i.e\:IE\:\alpha \frac{1}{screening\:effect}$$

Or the effect of removal of s, p, d, and f-electrons from the same energy level or shape of orbitals: In the same principal quantum number (n), the s-electrons are closer to the nucleus than do the p, d or f-electrons, thus more energy will be required to remove s-electrons from an atom than the p, d or f-electrons of the same shell. Hence for a given value of ‘ n ‘, the degree of penetration of electrons will decrease in the following order :

s > p > d > 1
This is the reason why the first ionization energies of boron is less than that of beryllium.
 4Be : 1s2, 2s2 (Electrons removes from 2s-orbital) 2B : 1s2, 2s2, 2p1 (Electrons removes from 2p-orbitals)

Similarly, the first ionization potential of aluminum is less than magnesium

 12Mg : [Ne]3s2 13Al : [Ne]3s2 3p1

(v) Half-filled or full-filled orbitals:

According to Hund’s rule, atoms having half-filled or full-filled p, d, or f-orbitals are comparatively more stable than others. Hence
it is more difficult to pull out the outer electrons from the atom. This means that the ionization energy of an atom having half-tiled or full-filed p, d, or f-orbitals is relatively higher than expected normally from its position in the periodic table. The relative order of the stability of p3,p6,d5, and d10 configurations is :
d5 < p3 < d10 < p6

For example, Be and N in the second period Mg and P in the third period, and As and Ca in the fourth period have slightly higher ionization energies than those normally expected. This is explained on the basis of extra stability of the completely tilled s-orbital in Be, Mg, and Ca and half-tilled p-orbitals in N, P, and As if second third, and fourth periods respectively.
Noble gases have the highest ionization energy values in their respective periods. It is because they have a very stable octet of electrons in their outermost shell.

(vi) Charge on the cation:

Ionization energy increases with an increase in positive charge on the cation. It is because a cation having a higher positive charge removes its outer electron with more difficulty. It is therefore for a given atom the order of different values of ionization energies is:
I1 < I2 < I3 < I4 …..
(b) In a group:
In general, the ionization energies decrease on moving from top to bottom in a group. If I evident from the following l.E.’s data given for group la elements this is because of the following reasons :
(i) As we move down a group, the atomic size and nuclear charge both increase simultaneously. But it is observed that the increase in the nuclear charge, is more effective which causes the nucleus of an atom to attract outer electrons strongly. Hence the magnitude of ionization energy gradually decreases on moving from top to bottom in a group. In the case of the first group these values are as follows:
 Group IA : H U Ni K Pb Cs I.E (in eV) 13.5 5.4 5.1 4.3 4.2 3.9

(ii) There is an increase in the number of inner shell electrons that shield the valence electrons from the nucleus Hence, with the increase of the magnitude of shielding reflected, the ionization energy decreases.

In the comparison of first and second ionization energies ( 11 and 12) for group 1 A elements, it is observed that the removal of the second electron from the first group element requires more energy and it is found that the value is 7 to 14 times greater than the first 1. E. The reason is that the first group elements have only one electron in their outermost orbit In the case of second 1 . E., a large amount of energy is required to remove electrons from the completely filled orbitals. It is evident from the first and second I.E. values of alkali metals as shown below.
 Element : Li Na K Rb Cs First l.E. (eV): 5.4 5.1 4.3 4.2 3.9 Second I.E. (eV): 75.6 47.4 32.0 28.0 25.0

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