Electron affinity is a property specifying the "appetite" of an element for gaining electrons. Elements, such as fluorine and oxygen that lack only one or two electrons to complete shells can achieve a lower energy state by absorbing an external electron. For instance, in uniting an electron with a neutral chlorine atom, completing its n = 3 shell and forming a CI ion, 3.61 eV of energy is liberated. Suppose an electron is detached from a sodium atom, whose ionization energy is 5.14 eV.Then transferred to a (faraway) chlorine atom.

(a) Must energy on balance be put in by an external agent, or is some energy actually liberated? If so How much?

(b) The transfer leaves the sodium with a positive charge and the chlorine with a negative. Energy can now be extracted by allowing these ions to draw close forming a molecule. How close must they approach to recover the energy expended in part (a)?

(c)The actual separation of the atoms in a NaCl molecule is 0.24 nm. How much lower in energy is the molecule than the separated neutral atoms?

Step by step solution

01

Definition of electron affinity

Electron affinity is the measure of the energy produced when an electron is added to a neutral atom to create an anion.

02

Step-2(a) Calculation of Energy Input

The Energy input=Energy Output

Energy input+liberated Energy by Chlorine=Ionisation Energy of Sodium

Thus, Energy Input=Ionisation Energy of Sodium-Liberated Energy by chlorine..

$5.14\text{\hspace{0.17em}eV}-3.61\text{\hspace{0.17em}eV}=1.53\text{\hspace{0.17em}eV}$

Thus The Energy input is, $1.53\text{\hspace{0.17em}eV}$.

03

Step-3 (b.) Calculation of Minimum distance between Ions

The potential energy between the Sodium ion and the chlorine ion is given by

$\mathrm{U}=-\frac{{\mathrm{e}}^2}{4\mathrm{\pi }{\in }_0\mathrm{a}}$

To, Calculate the Separation in order to release $1.53\text{\hspace{0.17em}eV}$

$\mathrm{a}=-\frac{{\mathrm{e}}^2}{4\mathrm{\pi }{\in }_0\mathrm{U}}$

Substitute,$\mathrm{e}=1.6\times {10}^{-19}$ ,${\in }_0=8.85\times {10}^{-12}{\text{\hspace{0.17em}C}}^{\text{2}}{\text{/N.m}}^2$, and$\mathrm{U}=1.53\text{\hspace{0.17em}eV}$in the above equation.

Hence, Ion need to be at least as close at 0.94 nm so that it can release as much energy as it spends when forming Ions.

04

Step-4(c) Calculation of value of Sodium energy

$\mathrm{U}=-\frac{{\mathrm{e}}^2}{4\mathrm{\pi }{\in }_0\mathrm{a}}$

Substitute,$\mathrm{e}=1.6\times {10}^{-19}$,${\in }_0=8.85\times {10}^{-12}{\text{\hspace{0.17em}C}}^{\text{2}}{\text{/N.m}}^2$, and$\mathrm{a}=0.24\text{\hspace{0.17em}nm}$in the above equation.

Thus Sodium Energy’s is lowered by

$(5.99\text{\hspace{0.17em}eV}-1.53\text{\hspace{0.17em}eV})=4.46\text{\hspace{0.17em}eV}$

Hence, The molecule Sodium Chlorine’s energy is lowered by $4.46\text{\hspace{0.17em}eV}$..

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