- 1 First Ionization Energy In Elements
- 2 Second Ionization Energy
- 3 Determine ionization energy
- 4 Important factors of Ionization Energy
- 5 Ionization Energy Chart
- 6 Examples of Ionization Energy
- 7 Ionization Energy Trends
The amount of energy (work) required to remove one electron from the last orbit of the parent atom is term as ionization energy. Let’s talk about first ionization energy and successive Ionization energies.
First Ionization Energy In Elements
It is easy to remove the loosely bonded electron from the outermost orbit. Both first Ionization potential and first ionization energy have the same meaning.
First Ionization: Na (g) + E1→Na+ (g) + e–
It is due to fact that attraction between proton and electron will increase the ionization enthalpy. Increases in the binding force results increase of ionization enthalpy. First Ionization energy(enthalpy) of elements are linked with their electronic configurations.
Generally, In each period, ionization energy maximum at noble gases while minimum energy notice at the alkali metals. When an electron removes from the outermost orbit, then the first ionization energy is found to be maximum in alkali metals.It will achieve 2s22p6configuration,
Second Ionization Energy
Similarly, the energy required for the removal of an electron from the unipositive ion (produced above) is referred to as second ionization potential.
Second Ionization: Na+ (g) + E2→Na2+ (g) + e–
and thus the third, fourth, etc. ionization potentials may be defined. In general, the number of ionization potentials of an element may be as much as its number of electrons.
The second ionization energy, IE2 is always more than the first ionization energy (IE1) because, in the 2nd ionization energy or potential, an electron is to be removed from a positive ion which has the attraction for electrons and thus makes removal of an electron a difficult task. Thus in general
Third IE >>> Second IE > First IE.
Successive Ionization Energies(first ionization energy, second, third…..)
The ionization energy trend changes when moving across period and group. The minimum energy required to remove electron varies and depends upon the number of factors.
Determine ionization energy
Actually, the electrons are held by the nucleus of the metal atoms by a certain force called binding force. In order to escape electrons we to supply the energy to overcome the binding force. This job is performed by the photon which contains minimum energy be called as threshold energy to break down the binding energy. The threshold energy is also known as work function.
Ionization Energy=Work function + kinetic energy
- Energy E = hv.
- Work function E0 =hv0
- kinetic energy K.E = 1/2mv2.
- If the energy of incident Photon is <(less than) threshold energy no electron will be emitted.
- If the incident Photon has the energy =(equal) to the threshold energy electron will just release from the metal surface.
- In case, an incident photon has the energy >(greater than) than threshold energy emit electron will acquire some kinetic energy.
- The kinetic energy of electrons is directly proportional to the frequency of striking Photon and it is quite independent of intensity.
- The number of electrons is ejected per second depends upon the intensity of striking Photon, not upon their frequency.
Important factors of Ionization Energy
Atomic size or radius
The larger the atomic size, smaller is the ionization energy. As the size of the atom increases, the outer electrons lie farther away from the nucleus and thus exert less attraction towards the nucleus and hence can be easily removed.
The larger the number of electrons in the inner shell, the greater is the screening effect on the electrons in the outer (valence) shell which thus experiences less attraction from the nucleus and thus can be easily removed leading to the lower value of ionization potential. Now as we move down a group, the number of inner shells increases and hence the ionization potential tends to decrease.
Nuclear charge Effect in first ionization energy
Nuclear charge is defined as the net nuclear attraction towards the valence shell electrons. More is the effective nuclear charge, more tightly the electrons are held with the nucleus and thus more energy will be required to remove electron leading to higher ionization energy.
Effective nuclear charge (Zeff) = Z-S
Where Z = Nuclear charge ; S = Screening constant.
On moving along the period the charge on nucleus increases as the atomic number increases while the valence shell remains the same and thus effective nuclear charge increases which lead to higher ionization energy. Therefore, ionization energy increases along the period with some disorders, viz. IE of elements of the 3rd group is less than that of the 2nd group elements, and similarly, IE of group 16 elements is less than that of group 15 elements.
An increase in positive charge on the ion increases the effective nuclear charge which in turn increases the ionization energy. On the other hand, an increase in negative charge on the ion decreases the effective nuclear charge which in turn decreases the ionization energy.
Half filled and fully filled electronic configuration
According to Hund’s rule, atoms having half-filled or completely filled orbitals are comparatively more stable and hence more energy is needed to remove an electron from such atoms leading to higher ionization energy than the usually expected value.
Arrangement of electrons(Symmetry)
Symmetry plays a vital role in first ionization energy. If an atom or an ion has s2p6 configuration, its ionization energy is extremely high due to the presence of the so-called octet arrangement (noble gas configuration). This explains why IE2 of Li(lithium) is very very high as compared to IE1.
IE3 >>> IE2 > IE1
Removal of s, p, d and f electrons from the same shell
Since s-orbital is more close t the nucleus than the p-orbital of the same orbit, its electrons experience more attraction than that of p and hence their removal is difficult leading to higher IE. In general, the ionization energy follows the following order s> p > d > f orbital of the same orbit.
s> p > d > f orbital of the same orbit.
Remember: I.E. measured in a unit of an electron volt (eV) per atom or kilojoule per mole or kilocalorie per mole.
1 eV/atom k cal.
Ionization Energy Chart
Examples of Ionization Energy
The ejection of electrons from caesium metal is easy because caesium element has the tendency to hold the electrons loosely because ionization enthalpy is very low.
Question: Be has higher ionization potential than B – Explain?
- Be has fully filled orbital (1s22s2)
- Boron does not have a fully filled orbital (1s22s22p1).
Again s electron is more penetrating than p-electron. So, more energy required to remove an electron from 2s orbital than to 2p orbital. So Be has higher ionization energy.
Ionization Energy Trends
Do look for periodic ionization trend. Before solving any question related with first Ionization Energy. Ionization energy increases across a period with breaks where an electron of a slightly higher energy is removed from the same shell.So, first ionization energy also has some exceptions.
Generally, on moving from left to right along a period, the I.E. increases with increase in atomic number.
The order breaks, where an electron of a slightly higher energy is removed from the same shell.
|I . E.||500||900||800||1086||1403||1314||1681||2081|
Thus the break occurs at ‘B’ and ‘O’.
Reason: Higher stability of half and fully filled orbitals. Transition elements Increase in nuclear charge helps in holding the valence electrons in ns orbitals more firmly while the shielding effect of (n-1) d-electrons on ns electron acts in opposite direction.
The net result is that there is a slight change in I.P.
On moving from top to bottom in a group the value of I.P. decreases. It is due to the following factors.
1. The nuclear charge decreases.
2. An addition of principal energy shell.
3. In transition elements considering the trend, I.P. decreases as we move from 3d to 4d. Considering the trend leaving aside. It is found that I.P increases on moving from 4d to 5d. This is so, as the 4f- electrons due to lanthanide contraction is unable to screen 5d electrons more effectively. Further, due to the increase of +32 unit of nuclear charge, there is an increment in value of I.P.
For example TI(Thallium)
Reason- Because of the filling of 4f orbitals before Tl (81) and Pb (82). We find that it is unable to shield the outer
electrons. Further increase of +32 unit of nuclear charge cause increase of I.P.
Must Read: Electronegativity of elements.
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