Factors governing ionization energy
Factors governing ionization energy. The ionization energy depends upon the following factors:
(a) Size of atom or ion. The ionization energy decreases with the increasing size of atom. The larger the size of atom, lesser is the ionization energy. This is due to the fact that electrons are tightly held in smaller atoms whereas in large atoms, electrons are held quite loose, i.e., lesser energy is required for removal of electrons from larger atoms than the smaller one. Hence ionization energy is lower for larger atoms and higher for smaller atoms.
The I.E of Be (At. No.4) is greater than that of Li (At. No.3) because the nuclear charge of Be (Z=4) is greater than Li(Z=3). Higher the nuclear charge, greater would be the force of attraction between nucleus and outermost electron. Hence, the first I.E. of Be is than that of Li.
The I.E. of Be is more than that of B though the nuclear charge of boron atom (Z=5) is greater than that of beryllium atom (Z = 4). This can be explained as follows:
Boron atom (Z = 5; 1s2 2s2 2px1 2py02pz0) is having one unpaired electron in the 2p-subshell. Be-atom (Z = 4; 1s2 2s2) is having paired electrons in the 2s- subshell.
As the fully filled 2s-subshell in Be-atom is more stable than B-atom due to symmetry, more energy would be needed to remove an electron from Be-atom. Hence, Be has high I. E.
3. The I.E. of carbon (At. No.6) more than that of boron (At. No.5)
Reason: Carbon (Z = 6; 1s2 2s2 2px1 2py1 2pz0) is having more nuclear charge than boron (Z = 5; 1s2 2s2 2px1 2py0 2pz0). In both the cases, one has to remove electron from same 2p-subshell. Carbon is having more nuclear charge than boron. Therefore the nucleus of carbon, attracts the outer 2p-electron more firmly than does boron. Thus, first I.E. of carbon would be more than that of boron.
(b) Magnitude of nuclear charge. The higher the nuclear charge of protons in the nucleus, the higher is the ionization energy. Because of the higher nuclear charge, the electrons are bound with more force and hence higher energy will be required for their removal. For instance, magnesium has higher nuclear charge (12 protons) as compared to sodium (11 protons). Hence ionization energy in case of magnesium is higher as compared to sodium.
Similarly the I.E. of fluorine is more than that of oxygen. It can be explained as follows.
(i) F (Z = 9; 1s2 2s2 2px2 2py2 2pz1) is having more nuclear charge than oxygen (Z = 8; 1s2 2s2 2px2 2py12pz1). In both the cases, the electron has to be removed from the same 2p-subshell. As fluorine is having more nuclear charge than oxygen, it means that the nucleus of fluorine will attract the outer 2p-electrons more firmly than oxygen. Hence, first I.E. of fluorine would be more than that of oxygen.
(c) Effect of number of electrons in the inner shells. (Screening or shielding effect). The attractive force exerted by the nucleus on the most loosely bound electron is atleast partially counterbalanced by the repulsive forces exerted by the electrons present in the inner shells. The electron to be removed is thus shielded from the nucleus by the electrons in the inner shell. Thus, the electron in the valence shell experiences less attraction from the nucleus. Hence the ionisation energy will be low. This is another reason why ionization energy decreases in moving down a group.
(d) Effect of shape of orbital. The shape of orbital also influences the ionization potential. As s-electrons remain closer to the nucleus than p-,d-, and f-electrons of the same valence shell, the ionization energy decreases in the order given below:
For instance, the first ionization energy of aluminium is lower than that of magnesium.
The electronic configuration of magnesium is [Ne]3s2 and that of aluminium is [Ne] 3s2 3p1. Thus, one has to remove 3p-electron in case of aluminium and 3s electron in the case of magnesium. But it is easier to remove the p electron than the s-electron. Thus, the first ionization energy of aluminium is lower than that of magnesium.
(e) Effect of arrangement of electrons. The more stable the electronic arrangement, the greater is the ionization energy. As the noble gases have the stablest electronic arrangements, they show maximum ionization energy.
The I.E of Ne is greater than that of F. It can be explained as follows:
The nuclear charge of Ne (Z = 10) is more than that of F (Z = 9). Greater
the nuclear charge, greater would be the force of attraction between nucleus and outermost electron. Hence, the first I.E. of neon would be greater than that of fluorine.