i) `text(Atomic Radius)` : The values of ionisation potential of an element decreases as its atomic radius increases. This is because the elecrostatic force of attraction between the nucleus and the outermost electron decreases as the distance between them increases. So, the energy required for the removal of electron will comparatively be less.

`text(ionisation potential) prop 1/ text(Atomic radius)`

ii) `text(Effective Nuclear Charge)` : The greater the effective charge on the nucleus of an atom, the more difficult it would be to remove an electron from the atom because electrostatic force of attraction between the nucleus and the outermost electron increases. So, greater energy will be required to remove the electron.

Ionisation potential a Effective nuclear charge `(Z_(eff))`

iii) `text(Penetration Effect of Orbitals)` : The order of energy required to remove electron from `s`, `p`, `d`-and `f`-orbitals of a shell is `s > p > d > f` because the distance of the electron from the nucleus increases. For example - The value of ionisation potential of `Be (Z= 4, 1s^2 quad 2s^2)` and `Mg(Z=12, 1s^2 quad 2s^2 quad 2p^6 quad 3s^2)` are more than the `I.P.` of `B (Z=5, 1s^2 quad 2s^2 quad 2p_x^1)` and `Al (Z= 13, 1s^2 quad 2s^2 quad 2p^6 quad 3s^2 quad 3p_x^1)` because the penetration power of `2s` and `3s` electrons is more than `2p` and `3p` orbitals respectively. More energy will be required to separate the electrons from `2s` and `3s` orbitals.

iv) `text(Shielding or Screening Effect)` : The shielding or screening effect increases if the number of electrons in the inner shells between the nucleus and the outermost electrons increases. This results in decrease of force of attraction between the nucleus and the outermost electron and lesser energy is required to separate the electron. Thus the value of `I.P.` decreases.

Ionisation potential `prop 1/ text(Shielding effect)`

v) `text(Stability of Half-filled and Fully-filled Orbitals)` : According to Hund's rule the stability of half filled or completely filled degenerate orbitals is comparatively high. So comparatively more energy is required to separate the electron from such atoms.

`text(For example)`

a) Removal of electron is comparatively difficult from the half filled configuration of `N (Z=7, 1s^2 quad 2s^2 quad 2p_x^1, p_y^1, p_z^1)`.

b) The ionisation potential of inert gases is very high due to most stable `s^2 p^6` electronic configurations.

i) `text(For Normal Elements)` : On moving from left to right in a period, value of ionisation potential of elements increases because effective nuclear charge also increases.

`text(Exceptions)` :

a) In a period, the ionization energy of `IIA` group elements is more than the elements of `IlIA` because penetration power of `s`-orbitals electrons. The value of ionization energy of `Be` `(1s^2 quad 2s^2)` is more than `B (1s^2 quad 2s^2 quad 2p_x^1)` because the penetration power of `2s`-electrons of `Be` is more than the `2p_x` electrons of `B`.

b) In a period, the ionization energy of `VA` elements is more than the elements of `VIA` because the half filled `p^3` configuration of `VA` elements is comparatively of higher stability. `VIA` group elements `(p^4)` have the tendency to acquire comparatively more stable `(p^3)` configuration by the loss of one electron. Ionization energy `N (1s^2 quad 2s^2 quad 2p_x^1 p_y^1 p_z^1) > O (1s^2 quad 2s^2 quad 2p_x^2 p_y^1 p_z^1)`. See fig.

Thus `P > S`, `As > Se`

But the value of `I.P. `of `Sb` (`VA`) &`Te` (`VIA`) and `Bi` (`VA`) & `Po` (`VIA`) are according to general rule i.e.

`Sb (VA)` `<` `Te (VIA)`

`Bi (VA) < Po (VIA)`

On moving from top to bottom in a group the value of `I.P.` decreases because the atomic size increases.

`text(Exceptions)` :

a) In group `IIIA` the ionization potential of `Al` (`13`) is equal to the ionization potential of `Ga` (`31`). Before `Ga` (`31`) the electrons are filled in `3d`-orbitals of ten transition elements. These `3d` orbital electrons do not completely shield the `4p` electron. So the increase of `+18` units in nuclear charge results in the greater increase of effective nuclear charge. Due to increase in nuclear charge the `I.P.` increases which counter balance the decrease in `I.P.` due to the increase in number of shells.

b) The values of `I.P.` of `Tl` (`81`) and `Pb` (`82`) of sixth period is more than the `I.P.` values of `In` (`49`) and `Sn` (`50`) of same groups in period fifth. This is because of the electrons are filled in `4f`-orbitals before `Tl` (`81`) and `Pb` (`82`) which do not completely shield the outer electrons. Thus increase in `+32` units in nuclear charge results in the increase of ionisation potential values.

ii) `text(For Transition Elements)` : On moving from left to right in a transition series :

a) As the atomic number increases the effective nuclear charge also increases. Hence the `I.P.` increases.

b) The shielding effect of (`n-1`)`d` electrons over `ns` electrons increases with the addition of electron in (`n- 1`)`d` orbitals. Hence the `I.P.` decreases.

c) The increased values of `I.P.` due to the increase of effective nuclear charge almost balances the decreased value of `I.P.` due to increase in shielding effect. There is a very small increase in the values of `I.P.` or it may be said that `I.P.` almost remains the same.

d) In first transition series from `Sc` to `Cr` the value of `I.P.` increases because effect of increase in effective nuclear charge is more than the shielding effect `I.P.` values of `Fe`, `Co`, `Ni` and `Cu` are almost same. Due to `d^(10) s^2` configuration of `Zn`, the first `I.P.` increases.

`text(On moving from top to bottom in a group in transition series)`

a) In a group on moving from first to second transition series, the values of `I.P.` decreases because atomic size increases.

b) In moving from second to third transition series the value of `I.P.` somewhat increases except `IIIB` group [`Y`(`39`) `->` `La`(`57`)]. This is because of `14` electrons are filled in `4f`-orbitals of lanthanides which do not shield the `5d`-electrons effectively. Thus the increase in `+32` units in nuclear charge results in the increase of `I.P.`, on moving from left to right this effect decreases and becomes negligible in the later part.

i) Metallic or electropositive character of elements increases as the value of ionisation potential decreases. So in a group moving from top to bottom metallic or electropositive character increases because `I.P.` value decreases. In a period moving from left to right the values of `I.P.` increases so metallic or electropositive character decreases. Non-metallic character increases.

ii) The relative reactivity of the metals increases with the decrease in `I.P.` values. The `I.P.` values of `IA` and `IIA` metals are comparatively low. So they are comparatively more reactive. The `I.P.` values of inert gases are very high. So they are almost unreactive.

In a group moving from top to bottom the reactivity of metal atoms increases because their `I.P.` value decreases.

iii) The reducing power of elements increases as the values of `I.P.` decreases because tendency to lose the electron increases. The reducing power increases going down a group because the `I.P.` value decreases. `Li` is exception in `lA` group. The reducing power of `Li` is highest in its own group. The order of reducing power of `lA` elements is as under

`Li > Cs > Rb > K > Na`

iv) Determination of oxidation state or valency electrons of an element -

a) If the difference of two consecutive `I.P.'s` of an element is `16` `eV` or more, the lower oxidation state is stable. For e.g. the difference of first and second `I.P.` of `Na` is `42.4` `eV`, which is, more than `16` `eV`. So `Na^(+)` will be stable. It can also be explained from its electronic configuration

`Na (11) = 1s^2 quad 2s^2 quad 2p^6 quad 3s^1`

`Na^(+) = 1s^2 quad 2s^2 quad 2p^6`

Neutral `Na` atom has the tendency to acquire the stable `s^2 p^6` configuration by the loss of one electron. Due to `s^2 p^6` configuration of `Na^(+)`, the further separation of electron is difficult. So `lA` group metals form mono-positive ions.

b) If the difference of two consecutive `I.P.s` of an element is `11.0` `eV` or less, the higher oxidation state is stable. For e.g. the difference of first and second `I.P.` of `Mg` is `7.4` `eV` which is less than `11.0` `eV`. So `Mg^(2+)` will be stable. It can also be explained on the basis of its electronic configuration.

The electronic configuration of `Mg^(2+)` is stable `s^2 p^6` configuration.

`Mg^(2+) = 1s^2 quad 2s^2 quad 2p^6`

So `IIA` group elements form dipositive ions.

c) The difference of first and third `I.P.` of `Al` is `12.8` `eV` which is more than `11` `eV`. Therefore first oxidation state of `Al` i.e. `Al^(+)` must be stable. In gaseous state `Al^(+)` is stable. This is due to the proportionate distribution of lattice energy and the difference of second and third `I.P.s` `9.6` `eV` `< 11` `eV`.