Ionization And Electron Gain Enthalpy

Ionization enthalpy, also called ionization energy, is defined as the energy associated with removing an electron in a gaseous neutral atom to form a cation. This is called an endothermic process, which means it needs energy to be put in. The energy required for the removal of the first electron is called the first ionization energy; successive ionization energies are required for the removal of succeeding electrons. The factors that come into play in ionization energy are the size of the atom, nuclear charge, and the effect of shielding the electrons. In general, the trend of ionization energy increases down a group because of the increase in the size of the atom, and it decreases down a group owing to a decrease in nuclear charge.

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This Story also Contains

1. Ionization Energy And Electron Affinity
2. Some Solved Examples
3. Summary

Ionization Energy And Electron Affinity

Ionization energy and electron affinity are defined at absolute zero. At any other temperature, heat capacities for the reactants and the products have to be taken into account.

$\begin{aligned} & \mathrm{M}(\mathrm{g}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{e}^{-}(\text {for ionization }) \\ & \mathrm{M}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{M}^{-}(\mathrm{g}) \text { (for electron gain) }\end{aligned}$

at temperature, T is

$\Delta_r H^{\ominus}(T)=\Delta_r H^{\ominus}(0)+\int_0^T \Delta_r C_p^{\ominus} d T$

The value of C_p for each species in the above reaction is 5/2 R (C_V = 3/2R)

So, $\Delta_r C_p^{\ominus}=+5 / 2 \mathrm{R}$ (for ionization) $\Delta_I C_p^{\ominus}=-5 / 2 \mathrm{R}$ (for electron gain)

Therefore,

$\begin{aligned} & \Delta_r H^{\ominus}(\text { ionization enthalpy })=E_0(\text { ionization energy })+5 / 2 \mathrm{RT} \\ & \Delta_r H^{\ominus}(\text { electron gain enthalpy })=-\mathrm{A}(\text { electron affinity })-5 / 2 \mathrm{RT}\end{aligned}$

Recommended topic video on ( Ionization And Electron Gain Enthalpy)

Some Solved Examples

Example 1: The first and second ionization enthalpies of metal are 496 and $4560 \mathrm{~kJ} \mathrm{~mol}^{-1}$, respectively. How many moles of HCl and $\mathrm{H}_2 \mathrm{SO}_4$, respectively, will be needed to react completely with 1 mole of the metal hydroxide?
1)1 and 2

2) 1 and 0.5

3)1 and 1

4)2 and 0.5

Solution

According to the given data of I.E., this element must belong to group 1, and thus it is monovalent and forms a hydroxide of the type M(OH).

$\mathrm{MOH}+\mathrm{HCl} \longrightarrow \mathrm{MCl}+\mathrm{H}_2 \mathrm{O}$

$2 \mathrm{MOH}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{M}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O}$

Thus, from the reaction stoichiometry, it is clear that 1 mole of the metal hydroxide requires 1 mole and 0.5 moles of HCl and H₂SO₄ respectively for complete neutralization.

Example 2: The absolute value of the electron gain enthalpy of halogens satisfies :
1) Cl> F > Br > I

2)F >Cl> Br > I

3)Cl> Br > F > I

4)I > Br >Cl> F

Solution

The absolute value of the electron gain enthalpy of halogens satisfies:

Cl > F > Br > I

Chlorine has a higher electron gain enthalpy than fluorine due to less electron density.

Example 3: The enthalpy change for the conversion of $\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g})$ to $\mathrm{Cl}^{-}(\mathrm{aq})$ is $(-) \mathrm{kJmol}^{-1}$ (Nearest integer) Given :$\Delta_{\mathrm{dis}} \mathrm{H}_{\mathrm{Cl}_{2(\mathrm{~g})}^{\ominus}}^{\ominus}=240 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{\mathrm{eg}} \mathrm{H}_{\mathrm{Cl}(\mathrm{g})}^{\ominus}=-350 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{\mathrm{hyd}} \mathrm{H}_{\mathrm{Cl}_{(\mathrm{g})}^{-}}^{\ominus}=310 \mathrm{~kJ} \mathrm{~mol}^{-1}$

1) 610

2)510

3)400

4)256

Solution

$\begin{aligned} \Delta \mathrm{H}_\gamma^{\circ} & =\frac{1}{2} \times \mathrm{BE}+\Delta \mathrm{H}_{\mathrm{eg}}+\Delta \mathrm{H}_{\mathrm{Hyd}} \\ & =\frac{1}{2} \times 240+(-350)+(-380) \\ & \Rightarrow 120-350-380 \\ & \Rightarrow-610\end{aligned}$

Example 4: For electron gain enthalpies of the elements denoted as $\Delta_{\mathrm{eg}} H$, the incorrect option is :

1)$\Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{Te})<\Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{PO})$

2) $2 . \Delta_{\text {eg }} H(\mathrm{Se})<\Delta_{\text {eg }} H(\mathrm{~S})$

3)$\Delta_{\text {eg }} H(\mathrm{Cl})<\Delta_{\text {eg }} H(\mathrm{~F})$

4)$\Delta_{\text {eg }} H(\mathrm{I})<\Delta_{\text {eg }} H(\mathrm{At})$

Solution

Electron gain enthalpies $\rightarrow$
$\begin{aligned} & \rightarrow \mathrm{S}>\mathrm{Se}>\mathrm{Te}>\mathrm{O} \\ & \rightarrow \mathrm{Cl}>\mathrm{F}>\mathrm{Br}>\mathrm{I}\end{aligned}$

Summary

Ionization enthalpy and electron gain enthalpy are some of the important concepts in understanding the chemical behavior by an element. Ionization enthalpy measures the energy requirement necessary to pull electrons away from an atom. It generally increases across a period due to stronger nuclear attraction. It decreases down a group due to increased atomic size and electron shielding. Electron gain enthalpy, on the other hand, measures the energy change for the addition of an electron to a neutral atom. It could be exothermic or endothermic, reflecting an element's tendency to gain electrons. Nonmetals always have negative electron gain enthalpies, which imply that energy is released when an electron is added.