Reasoning Problem 1887 An atom of Na can combine with an ato

Reasoning Problem 18.87 An atom of Na can combine with an atom of CI to form the molecule Naci. It is a good approximation to view this as a two-step process (see the igure below) in which the Na atom gives up an electron to the Cl atom, forming a positive ion (Nat) and a negative ion (CI) These ions are then attracted by the electric force, forming the stable molecule Na cr. To calculate the changes in electric potential energy associated with these processes, assume that the outermost electron in Na is r 0.19 nm from the nucleus (this is the atomic radius of Na) and that the separation or the ions in Na\"Cis 0.24 nm. Na Cl 0.10 nm 0.19 nm Na+CI 0.24 nm (a) What is the energy required to remove an electron and create Na? (b) The energy gained by adding an electron to Ci to make Cr is called the electron affinity, and has been measured by chemists to be 349 kJ/mol of Cl atoms. Express this value in eVlatom (c) What is the electric potential energy of Na cr? (d) combine your answers trom parts (a) through (c) to estimate the energy gained by making Na cr Give some reasons why your (ay combine your answers trom parts (a) through ( to estmate the energy gained by making Narcr Give some reasons answer difters from the measured dissociation energy of this molecule. 4.26 eV In particular, what aspect of your calculation in part (a) causes it to overestimate the ionization energy of Na? the lonization energy or Na? Part 1- Recognize the Principles Part 2 -Identify the Relationships

Solution

(a) In sodium atom

- K_e * e² / r the effective potential of nucleus on the outer most electron as can be consider as the valence electron is attracted by a proton at a distance 0.19 nm

= 8.98 x 10⁹ x (1.6 x 10^-19)² / 0.19 x 10^-9 = 12.09 x 10^-19 Joule

= 12.09 x 10^-19 /1.6 x 10^-19 = 7.56 eV

(b) 349 KJ / mole = 349 x 10³ x (1/1.6 x 10^-19) / (6.02 x 10²³ ) = 3.6 x 10¹⁸ eV/atom

(3) effective cahrge on Na ion = +1e

effective cahrge on Cl ion = -1e

Electric potential energy = - K_e * e² / r

= 8.98 x 10⁹ x (1.6 x 10^-19)² / 0.24 x 10^-9 = 9.57 x 10^-19 Joule

= 9.57 x 10^-19/1.6 x 10^-19 = 5.98 eV

(d) Change in the potential energy when NaCl forms

= 5.98 - 7.56 = - 1.58 eV