Chemistry, asked by ahalyaamir, 8 months ago

Calculate the ionic radii of Na+ and F- in NaF crystal whose inter ionic
distance is equal to 231 pm.

Answers

Answered by raviwanje155

♠ FOLLOW ♠

•It is defined as the distance from the centre of the nucleus of the ion up to which it exerts its influence on the electron cloud of the ion. Ionic radius of uni-univalent crystal can be calculated using Pauling's method from the inter ionic distance between the nuclei of the cation and anion. Pauling assumed that ions present in a crystal lattice are perfect spheres, and they are in contact with each other therefore,

d = rC+ + rA- ------------------ (1)

Where d is the distance between the centre of the nucleus of cation C+ and anion A- and rC+, r A- are the radius of the cation and anion respectively.

Pauling also assumed that the radius of the ion having noble gas electronic configuration (Na+ and Cl-having 1s2 2s2, 2p6 configuration) is inversely proportional to the effective nuclear charge felt at the periphery of the ion.

•

Answered by abhi178

It has given that inter ionic distance is equal to 231 pm.

To find : The ionic radii of Na⁺ and F¯

solution : the structure of NaCl is similar to the structure of NaF.

here inter ionic distance = 231 pm

i.e., r⁺ + r¯ = 231 pm ........(1)

also we know, r⁺/r¯ = 0.414

⇒r⁺ = 0.414 r¯ ........(2) putting it in equation (1)

0.414r¯ + r¯ = 231 pm

⇒1.414r¯ = 231 pm

⇒r¯ = 231/1.414 = 163.36 pm