Question

The separation between neighboring Na+ and Cl- ions in crystal of NaCl is 0.281 nm, whilst the separation between neighboring Na ions in metallic sodium is 0.372 nm. Given that the ionic radius of Cl- is 0.181 nm, determine the radius of the Na+ ions in each case. Explain why the values are different.

Answer 1

Cl- r = 0.181

Crystal NaCl r = 0.281

Metallic NaCl r =0.372

Crystal and Metallic r Na+ = ??

Ionic compounds are formed due to the electrostatic attraction between oppositely charged ion. Ionic compounds consist of

A metal (an element on the lower left side of the periodic table), which loses one or more electrons to become a positively charged ion (cation) and
A non-metal (an element on the upper right side of the periodic table), which gains one or more electrons to become a negatively charged ion (anion).
The charge of many elements can be predicted based on their position on the periodic table.

An atom of sodium has one 3s electron outside a closed shell, and it takes only 5.14 electron volts of energy to remove that electron. The chlorine lacks one electron to fill a shell, and releases 3.62 eV when it acquires that electron (it's electron affinity is 3.62 eV). This means that it takes only 1.52 eV of energy to donate one of the sodium electrons to chlorine when they are far apart. When the resultant ions are brought closer together, their electric potential energy becomes more and more negative, reaching e = -1.52 eV at about 0.94 nm separation.

Knowing the Potencial Energy for NaCl, we can find the radius of a single ion.

E = -k e2 / r
so
r = -k e2 / E

r = -1.44 eV nm / -1.52 eV

r = 0.947 nm

The environment is different in the normal solid state where sodium chloride (common table salt) forms cubical crystals. The ion separation is 0.28 nm, somewhat smaller than that in the metallic form (0.371 nm)