Ionic radii by Lande’s Method and Pauling’s Method

Ionic radii:

1. Though it is possible to measure inter-nuclear distance in an ionic crystal very accurately by X-ray diffraction there is no universally accepted formula for measuring the radius of anions or cations. There are several different sets of ionic radii which have been estimated.
2. Correction should be made if the charge on an ion is changed e.g. Fe³⁺ and Fe²⁺.
3. Correction should be made for the coordination number and geometry of an ion.
4. The loans should be spherical which is true for ions of s and p-block with noble gas configuration but not true for transitional elements.
5. It is also not accurate in the case of delocalization of d-electrons where they give rise to 10 metallic conduction or form cluster compounds.

Thus, the size of an ion can be determined only when the size of its opposite ion is known. The following methods are used to measure the ionic radii.

(a) Lande’s Method (1920):

It was its first attempt to determine ionic radii by Lande in 1920. He determined intemuclear. distance of an ionic crystal in which the anion is very large as compared to the cation because there Is anion-anion contact in the crystal lattice For example: in lithium iodide ionic crystal, Li* ion is very small {Li+=0.63 A) and F ion is large (1=2.16 A) due to which there is anion-anion contact in the crystal lattice as shown in fig 2.03a.       

  ^{_{$$ r_{I^-}\:=\frac{d_{I^-}-_{I^-}}{2}=2.16A^{\circ }$$}} 

^{_{$$ d_{Li^+}_{-\:I^-}\:=r_{I^+}+_{Li^+}=0.60A^{\circ } $$}}

By X-ray, crystallography it is found that the edge length AB of L+1 – unit cell is equal to 6.00 A, and △ABC is an equilateral triangle. The simple geometry allows us to fix the diagonal of the square as 2η−+2r+11. It is possible when the angle formed by the diagonal in the corner is equal to 45∘.

Therefore

AC² = AB² + BC²

(AB = BC = CD = AD = d₁ – 1 – m6.00A)

AC² = 6² + 6² = 72A

$$ AC\:=\sqrt{72}=2r_1+2r_1\:=4r_1 $$

$$ r_{1=}\frac{1}{4}\sqrt{72}=2.12A $$

dk + 1- = rk + r1-

r_K+ = d_{K+ -I} – r_1^–

   = 3.52 -2.12

= 1.40 Å

(b) Pauling’s Method :
Pauling’s method has been utilized for calculating ionic radii of such logic crystals which have monovalent isoelectronic ions e.g. Na+F−, K+Cr, Rb+Br, and Cs++– e. the iconic pairs having the same number of electrons as shown below :

This method majorly is based on the following assumptions :
(i) The cation, c⁺, and anion, a^– of the ionic crystal, c+a−are considered to be charged spheres and very closely packed to each other.

If the internuclear distance between the two ions c⁺ and a^– in the ionic crystal is c⁺a⁻is d_c+a−. which is equal to the sum of the ionic radii of a cation and anion.

I.e dc+ – H– = fc+ +  ra–  ……… (i)

where r_c+ is the radius of the cation and r_a– is the radius of the anion.

(ii). If an ionic crystal, c+a- contains isoelectronic Ionic pair which have electronic configuration ns² or ns²np⁶, the radius of the cation or anion is inversely proportional to their effective nuclear charge Zeff (or Z ).

where Z^∗_a− and Z^∗_c+ are effective nuclear charges of anion and cation respectively and σ_a− and σ_c+ are screening constants for anion and cation respectively.

By solving equations (i) and (iii) we can calculate the value of r_c+ and r_a− by knowing the isoelectronic ions and their radii are given in table 2.5.

Illustration: Let us apply Pauling’s method to calculate the radii of the isoelectronic Na⁺ and F⁻ ions in Na⁺F⁻crystal. In this ionic crystal, the internuclear distance between Na+ and F−ions;  dNa⁺-F = 2.13A, and nuclear charges are 11 and 9 respectively. Since Na+ F–ions both have a stable electronic configuration of neon (1s² 2s² 2p⁶) for which using Slater’s rule. screening constant (σ)is found to be :

σ = 2(0.85) + 8(0.35) = 1.70 + 2.80

  = 4.50

Hence,  Z*_Na⁺ = (Z_Na⁺ – σ_Na⁺)

                     = 11.0 – 4.5 = 6.5

and        Z*_F^– = (Z_F^– – σ_F^– )

                    = 9.0 – 4.5 = 4.5

On putting the above values in Pauling’s Equation, we get

After solving equations (i) and (ii) we can get the values of r_Na+ and r_f- equal to 0.95 and 1.36 A respectively.

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