It is a well-known fact that the electrons in an atom are attracted by the positively charged nucleus. In order to remove electron from an atom, energy has to be supplied to it to
overcome the attractive force. This energy is referred to as ionization energy. Thus, Ionization energy may be defined as the amount of energy required to remove the most loosely bound electron from the isolated gaseous atom in its ground state.
A (g) + Energy (I.E.) A+ (g) + e-
Ionization energy is a very important property which gives an idea about the tendency of an atom to form a gaseous positive ion.
Ionization energy is expressed in terms of kilo Joules per mole of atoms (kJ moz-1)
SUCCESSIVE IONIZATION ENERGIES
Once the first electron has been removed from the gaseous atom, it is possible to remove second and successive electrons from positive ions one after the other. For example,
The amounts of energies required to remove most loosely bound electron from unipositive, dipositive, tripositive ions of the element in gaseous state are called second, third, fourth ionization energies respectively.
The second, third, fourth, etc. ionization energies are collectively known as successive ionization energies. It may be noted that:
The variation in the values of successive ionization energies can be explained, in general, as follows:
After the removal of first electron, the atom changes into monopositive ion. In the ion, the number of electrons decreases but the nuclear charge remains the same as in the parent atom. As a result of this, effective nuclear charge per electron increases. The remaining electrons are, therefore, held more tightly by the nucleus. Thus, more energy is required to remove the second electron. Hence, the value of second ionization energy is greater than the first.
Similarly, the removal of second electron results in the formation of divalent positive ion and the attraction between the nucleus and the remaining electrons increases further. This
accounts for the progressive increase in the values of successive ionization energies.
If the removal of first, second or third electron also results in the removal of the valence shell, the next electron (which , belongs to lower energy shell) will require very large amount
of energy for its removal. For example, in sodium, the removal of first electron leads to the removal of third shell completely. Therefore, for sodium, I.E.2 >> LET
The first, second and third ionization energies of some elements are given in Table 6.10.
Table 6.10. I.E.l‘ I.E.2 and I.E.3 Values
of Some Elements
FACTORS ON WHICH ION IZATION EN ERGY DEPENDS
The ionization energy depends upon the following factors:
1 . Size of the atom;
2. Magnitude of nuclear charge;
3. Screening effect of the inner electron;
4. Penetration effect of the electrons; and
5. Electronic configuration.
1. Size of the Atom. The attractive force between the electron and the nucleus is inversely proportional to the distance between them. Therefore, as the size of the atom increases, the outermost electrons are less tightly held by the nucleus. As a result, it becomes easier to remove the electron. Therefore, ionization energy decreases with increase in atomic size.
2. Magnitude of Nuclear Charge. The attractive force between the nucleus and the electrons increases with the increase in nuclear charge provided their main energy shell remains the same. This is because, the force of attraction is directly proportional to the product of charges on the nucleus and that on the electron. Therefore, with the increase in I nuclear charge, it becomes more difficult to remove an electron and, therefore, ionization energy increases.
3. Screening Effect of the Inner Electrons. In multi electron atoms, the electrons present in the outermost shell do not experience the complete nuclear charge because of
repulsive interaction of the intervening electrons. Thus, the outermost electrons are shielded or screened from the nucleus by the inner electrons. This is known as screening effect.
If the number of electrons in the inner shells is large, the screening effect will be large. As a result, the attractive interactions between the nucleus and outermost electrons will be less. Consequently, ionization energy will decrease. Thus, if other factors do not change, an increase in the number of inner electrons tends to decrease the ionization energy.
4. Penetration Effect of the Electrons. It is a well-known fact that in case of multi-electron atoms, the electrons in the s-orbital have the maximum probability of being found near the nucleus and this probability goes on decreasing in case of p, d and f orbitals. In other words, s-electrons are more penetrating towards the nucleus than p-electrons. The penetration power decreases in a given shell (same value of n) in the order: s > p > d > f
Now, if the penetration of the electron is more, it experiences less shielding effect by the inner electrons and will be held firmly. Consequently, ionization energy will be high. This means· that ionization energy increases with increase in penetration power of the electrons. Thus, for the same shell, it is easier to remove the p-electrons in comparison to the s-electrons.
5. Electronic Configuration. It has been noticed that certain electronic configurations are more stable than the other. The atom having a more stable configuration has less tendency to lose the electron and consequently, has high value of ionization energy.. For example:
- The noble gases have stable configuration (ns2np6). They have highest ionization energies within their respective periods.
- The elements like N (ls2, 2s2, 2p1, 2py, 2p1) and P ( l s 2 2s 2 2p 6 3 s2 3p 1 3p 1 3p 1) have configurations in which orbitals belonging to same sub-shell are exactly half-filled. Such configurations are quite stable and consequently, require more energy for the removal of electron. Hence, their ionization energies are relatively high.
VARIATION OF ION IZATION ENERGY IN THE PERIODIC TABLE
Let us now, study the variation of ionization energy across the period and along the group of representative elements on :be basis of above factors.
Variation Across the Period
In general, the value of ionization energy increases with e increase in atomic number across the period. This can be attributed to the fact that moving across the period from left right,
(i) nuclear charge increases regularly;
(ii) addition of electrons occurs in the same energy level;
(iii) atomic size decreases.
Thus, due to the gradual increase in nuclear charge and simultaneous decrease in atomic size, the attractive force between the nucleus and the electron cloud increases. Consequently the electrons are more and more tightly bound to the nucleus. This results in the gradual increase in ionization across the period. The ionization energies of the ments of 2nd periods are given in Fig. 6.9.
On carefully examining the values given in Fig. 6.9, wed some exceptions within the period. These can be explained on the basis of other factors governing ionization energy .
(i) There is an increase in ionization energy from Li to Be. This is due to increased nuclear charge and smaller atomic size.
Fig. 6.9. Variation of ionization energies
among elements of 2nd period.
(ii) There is a decrease in the value of ionization energy from Be to B inspite of increased nuclear charge. The effect of increased nuclear charge is cancelled by (a) greater penetration of 2s electron as compared to 2p electron: (b) better shielding of 2p electrons by the inner electrons. Thus. 2p electron of boron is relatively less tightly held by its nucleus in comparison to 2s electrons of Be .
(iii) There is a regular increase in ionization energy from B to C to N. It is again due to gradual increase of nuclear charge and decrease of atomic size.
(iv) There is slight decrease in ionization energy from N to O. It is attributed to the relatively stable configuration of the nitrogen due to a half filled 2p-orbital. In the nitrogen atom three 2p-electrons are present in different atomic orbitals (Hund’s rule) whereas in the oxygen atom, two of the four 2p-electrons are present in the same 2p-orbital resulting in an increased electron-electron repulsion. Therefore, it is easier to remove one of the 2p-electron from oxygen than it is, to remove one of the 2p-electrons from nitrogen.
(v) There is an expected increase in ionization energy from O to F to Ne.
Variation in a Group
The values of ionization energies of elements decrease regularly with the increase in atomic number within a group.
The values of ionization energies of the elements of group 1 have been represented graphically in Fig. 6.10.
Fig. 6.10. Variation of ionization energy among elements of group 1.
The decrease in the value of ionization energy within the group can be explained on the basis of net effect of the following factors:
As we move down the group there -is:
(i) A gradual increase in the atomic size due to progressive addition of new energy shells;
(ii) Increase in the shielding effect on the outermost electron due to in’2rease in the number of inner electrons.
The nuclear charge also increases but the effect of increased nuclear charge is cancelled by the increase in atomic size and the shielding effect. Consequently, the nuclear hold on the valence electron decreases gradually and ionization energy also decreases. The variation of first ionization energy as a function of atomic number for elements with atomic number up to 60 have been shown in Fig. 6. 1 1 . The maxima in the curve represents noble gases which means that the
ionization energies of noble gases are highest- within their periods. The minima in the curve represent alkali metals which implies that ionization energies of alkali metals are the lowest
within their respective periods.
Fig. 6.11. Ionization energies of elements up to Z = 60.
Example 6.8 From each set, choose the atom which has the largest ionization energy and explain your answer
(i) F, 0, N. (ii) Mg, P, Ar.
(iii) B, Al, Ga.
Solution . (i) F has the highest ionization energy among F, 0 and N because it has smallest size and highest nuclear charge. In general, ionization energy increases as we go from left to right in a period.
(ii) Ar (a noble gas) has the highest ionization energy among the elements Mg, P and Ar because it has stable electronic configuration and maximum nuclear charge.
(iii) B, AI and Ga belong to group- 13. Among these elements B has the largest ionization energy because on moving down a group, from top to bottom, ionization energy decreases. B is the first element of group 13.
Example 6.9. Out of Na+ and Ne which has higher ionization energy ? Explain why.
Solution. Na+ has higher ionization energy than Ne. Na+ and Ne are isoelectronic species. However, the nuclear charge in Na+ is more than in Ne. Hence, the electrons are more tightly held in Na+ and it has higher ionization energy.
Example 6. 10. How do you explain that 31 Ga has slightly higher ionization energy than 13Al, although it occupies lower position in the group ?
Solution. 13AJ : 1s2, 2s2, 2p6, 3s2, 3pl
13Ga : ls2, 2?, 2p6, 3s2, 3p6, 3d10, 4s2, 4p1
In Ga, the 10 electrons present in 3d-sub-shell do not shield the outer electrons from the nucleus effectively. As a result, effective 1uclear charge in Ga increases. This explains why ionization energy of Ga is slightly more than that of 13Al.