The ability of an element to release its ultraperipheral electrons for the formation of positive ions is expressed in the amount of energy supplied to its atoms sufficient to remove them from the electrons.
This energy is called ionization energy. In simple terms, ionization energy is the energy supplied to an isolated atom or molecule to neutralize its electron with the weakest bound valence layer to form a positive ion.
Its unit is electron volts eV or kJ / mol and is measured in an electric discharge tube in which a fast-moving electron collides with a gaseous element to eject one of its electrons.
The lower the ionization energy (IE), the better the ability to form cations.
For any atom Y, Y +IE →Y++e_
This can be explained with the Bohr atomic model by considering a hydrogen atom in which an electron rotates around a
positively charged nucleus due to its colombic attraction.
Electrons can only have fixed or quantified energy levels. The energy of a drilling model electron is quantified and reported as follows:
E(n)= – 13.6Z2eV/n2
Where Z is the ordinal number and n is the principal quantum number, where n is an integer. For a hydrogen atom, the IE is 13.6 eV.
The IE (eV) is the energy required to bring the electron of n = 1 (ground state or most stable state) to infinity.
By taking the reference 0 (eV) at infinity, the ionization energy can be written as follows:
The concept of IE confirms the proof of Bohr’s model that the electron can rotate around the nucleus in fixed or discrete energy levels
or energy layers represented by the principal quantum number “n”. ,
When the first electron moves away from the neighborhood of the positive nucleus, more energy is needed to remove the next
weakly bound electron as the electrostatic attraction increases, ie, the second energy of ionization is larger than the first.
For example, the first IE of sodium (Na) is given by:
Na+5.139076 eV(IE1)→Na+ +e– And its second ionization energy is
Therefore, IE2 is> IE1 (eV). This is true even if there are a number K of ionizations, then IE1 <IE2 <IE3 ………. <IEK
Metals have low ionization energy. A low IE implies a better conductivity of the element.
For example, the conductivity of silver (Ag, atomic number Z = 47) is 6.30 × 10 7 s / m and its IE is 7.575 eV and for copper
(Cu, Z = 29) 5.76 × 107 s / m and its ionization The energy is 7,726 eV.
In ladders, the low IE causes the electrons in the positively charged array to move and form an electron cloud.
Factors that influence the IE
In the periodic table, the general trend is that the IE increases from left to right and decreases from top to bottom.
Thus, the factors that influence the IE can be summarized below:
The IE decreases with the size of the atom, because with increasing radius of the atom, the colombic
attraction between the nucleus and the outermost electron decreases. and vice versa.
The presence of internal electrons in the sheath protects or weakens the columbic
attraction between the nucleus and the electrons of the valence. As a result, the IE decreases.
The number of internal electrons means more shielding. However, in the case of gold, the IE is greater than silver, even if its size is larger than that of silver.
This is due to the weak shielding provided by the internal orbitals d and f in the case of gold.
The larger the nuclear charge, the more difficult it becomes to ionize the atom,
as the force of attraction between the nucleus and the electrons increases.
The more stable the electronic configuration of the atom,
the more difficult it is to remove an electron, which results in more ionization energy.