Electron gain enthalpy of an element is equal to the energy released when an electron is added to valence shell of an isolated gaseous atom.

$$A(g)+e^{-} \longrightarrow A^{-}(g) ; \Delta_{\text {eg }} H=\text { negative }$$

Factors affecting electron gain enthalpy

(i) Effective nuclear charge Electron gain enthalpy increases with increase in effective nuclear charge because attraction of nucleus towards incoming electron increases.

(ii) Size of an atom Electron gain enthalpy decreases with increase in the size of valence shell.

(iii) Type of subshell More closer is the subshell to the nucleus, easier is the addition of electron in that subshell.

Electron gain enthalpy (in decreasing order) for addition of electron in different subshell ( $n$-same) is $s>p>d>f$

(iv) Nature of configuration Half-filled and completely-filled subshell have stable configuration, so addition of electron in them is not energetically favourable.

Variation in the periodic table As a general rule, electron gain enthalpy becomes more and more negative with increase in the atomic number across a period. The effective nuclear charge increases from left to right across a period and consequently it will be easier to add an electron to a smaller atom. Electron gain enthalpy becomes less negative as we go down a group because the size of the atom increases and the added electron would be farther from the nucleus. Electron gain enthalpy of O or F is less than that of the succeeding element ( S or Cl ) because the added electron goes to the smaller $n=2$ level and suffers repulsion from other electrons present in this level. For the $n=3$ level ( S or Cl ), the added electron occupies a larger region of space and suffers much less repulsion from electrons present in this level.

Ionisation enthalpy The minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom so as to convert it into a gaseous cation is called its ionisation enthalpy. It is represented by $\Delta_1 H$.

Factors affecting ionisation enthalpy of the elements

lonisation enthalpy depends upon the following factors

(i) Nuclear charge The ionisation enthalpy increases with increase in nuclear charge. This is due to the fact that with increase in nuclear charge, the electrons of the outer shell are more firmly held by the nucleus and thus greater energy is required to pull out an electron from the atom.

e.g., the ionisation enthalpy increases as we move along a period from left to right due to increased nuclear charge.

(ii) Atomic size or radius Ionisation enthalpy decreases as the atomic size increases. As the distance of the outer electrons from the nucleus increases with increase in atomic radius, the attractive force on the outer electron decreases.

As a result, outer electrons are held less firmly and hence lesser amount of energy is required to knock them out. Thus, ionisation enthalpy decreases with increase in atomic size. Ionisation enthalpy is found to decrease on moving down a group

(iii) Penetration effect of the electrons lonisation enthalpy increases as the penetration effect of the electrons increases. It is well known fact that in case of multielectron atoms, the electrons of the s-orbital has the maximum probability of being found near the nucleus and this probability goes on decreasing in case of $p, d$ and $f$-orbitals of the same shell.

In other words, s-electrons of any shell are more penetrating towards the nucleus than p-electrons the same shell. Thus, within the same shell, the penetration effect decreases in the order $s>p>d>f$

e.g., First ionisation enthalpy of aluminium is lower than that of magnesium. This is due to the fact that in case of aluminium $\left(1 s^2 2 s^2 2 p^6 3 s^2 3 p_x^1\right)$, we have to pull out a $p$-electron to form $\mathrm{Al}^{+}$ion whereas in case of magnesium $\left(1 s^2 2 s^2 2 p^6 3 s^2\right)$ we have to remove an $s$-electron of the same energy shell to produce $\mathrm{Mg}^{+}$ ion.

(iv) Shielding or screening effect of inner shell electrons As the shielding or the screening effect of the inner electrons increases, the ionisation enthalpy decreases. Consequently, the force of attraction by the nucleus for the valence shell electrons decreases and hence the ionisation enthalpy decreases.

(v) Effect of arrangement of electrons If an atom contains exactly half filled or completely filled orbitals then such an arrangement has extra stability. Therefore, the removal of an electron from such an atom requires more energy than expected.

e.g., $\mathrm{Be}\left(1 s^2 2 s^2\right)$ has higher ionisation enthalpy than $\mathrm{B}\left(1 s^2 2 s^2 2 p^1\right)$ and $\mathrm{N}\left(1 s^2 2 s^2 2 p_x^6 2 p_y^1 2 p_z^1\right)$ has higher ionisation enthalpy than $\mathrm{O}\left(1 s^2 2 s^2 2 p_x^2 2 p_y^1 2 p_z^1\right)$. In general, as we move from left to right in a period, the ionisation enthalpy increases with increasing atomic numbers.

The ionisation enthalpies keep on decreasing regularly as we move down a group from one element to the other.

There are numerous physical properties of elements such as melting points, boiling points, heats of fusion and vaporisation, energy of atomisation, etc., which show periodic variations.

The cause of periodicity in properties is the repetition of similar outer electronic configurations after certain regular intervals. e.g., all the elements of 1st group (alkali metals) have similar outer electronic configuration, i.e., $n s^1$.

$$\begin{aligned} &{ }_3 \mathrm{Li}=1 s^2, 2 s^1 \\ &{ }_{11} \mathrm{Na}=1 s^2, 2 s^2, 2 p^6, 3 s^1 \\ &{ }_{19} \mathrm{~K}=1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 4 s^1 \\ &{ }_{37} \mathrm{Rb}=1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^{10}, 4 s^2, 4 p^6, 5 s^1 \\ &{ }_{55} \mathrm{Cs}=1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^{10}, 4 s^2, 4 p^6, 4 d^{10}, 5 s^2, 5 p^6, 6 s^1 \\ &{ }_{87} \mathrm{Fr}=1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^{10}, 4 s^2, 4 p^6, 4 d^{10}, 4 f^{14} \\ & 5 s^2, 5 p^6, 5 d^{10}, 6 s^2, 6 p^6, 7 s^1 \end{aligned}$$

Therefore, due to similar outermost shell electronic configuration all alkali metals have similar properties. e.g., sodium and potassium both are soft and reactive metals. They all form basic oxides and their basic character increases down the group. They all form unipositive ion by the lose of one electron.

Similarly, all the elements of 17th group (halogens) have similar outermost shell electronic configuration, i.e., $n s^2 n p^5$ and thus possess similar properties.

$$\begin{aligned} & { }_9 \mathrm{~F}=1 s^2, 2 s^2, 2 p^5 \\ & { }_{17} \mathrm{Cl}=1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^5 \\ & { }_{35} \mathrm{Br}=1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^{10}, 4 s^2, 4 p^5 \\ & { }_{53} I=1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^{10}, 4 s^2, 4 p^6, 4 d^{10}, 5 s^2, 5 p^5 \\ & { }_{85} \mathrm{At}=1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^{10}, 4 s^2, 4 p^6, 4 d^{10}, \\ & 4 f^4, 5 s^2, 5 p^6, 5 d^{10}, 6 s^2, 6 p^5 \end{aligned}$$

All the elements of group IA (or I), i.e., alkali metals have the similar outer electronic configuration, i.e., $n s^1$ where $n$ refers to the number of principal shell. These electronic configurations are given below

Hence, placement of all these elements in group 1 of the periodic table because of similarity in electronic configuration and all the elements have similar properties.

The main drawbacks of Mendeleef's periodic table are

(i) Some elements having similar properties were placed in different groups whereas some elements having dissimilar properties were placed in the same group.

e.g., alkali metals such as $\mathrm{Li}, \mathrm{Na}, \mathrm{K}$, etc., (IA group) are grouped together with coinage metals such as $\mathrm{Cu}, \mathrm{Ag}, \mathrm{Au}$ (IB group) though their properties are quite different. Chemically similar elements such as $\mathrm{Cu}(\mathrm{IB}$ group) and Hg (IIB group) have been placed in different groups.

(ii) Some elements with higher atomic weights are placed before the elements with lower atomic weights in order to maintain the similar chemical nature of elements.

i.e., $\quad { }_{18}^{39.9} \mathrm{Ar}$ and ${ }_{19}^{39.1} \mathrm{~K} ;{ }_{27}^{58.9} \mathrm{Co}$ and ${ }_{28}^{58.7} \mathrm{Ni}$, etc.

(iii) Isotopes did not find any place in the periodic table. However, according to Mendeleef's classification, these should be placed at different places in the periodic table. (All the above three defects were however removed when modern periodic law based on atomic number was given).

(iv) Position of hydrogen in the periodic table is not fixed but is controversial.

(v) Position of elements of group VIII could not be made clear which have been arranged in three triads without any justification.

(vi) It could not explain the even and odd series in IV, V and VI long periods.

(vii) Lanthanides and actinides which were discovered later on have not been given proper positions in the main frame of periodic table.