B.sc 1st year Book
(Page 3)

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

Ionic radii:

The ionic radius is the radius of an ion (cation or anion) in an ionic crystal. It is defined as the distance length between the atom of the nucleus of an ion and the point-up area between which the nucleus exerts an effect on its (e-)electron cloud’. Since the atoms of the same element do not form ionic bonds hence It is very difficult to measure ionic radii as compared to covalent radii, It is because of the following reasons.
  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.       

  Ionic Radii, Atomic Radii, Determination-of-Ionic-radii-by-lands-method $$ r_{I^-}\:=\frac{d_{I^-}-_{I^-}}{2}=2.16A^{\circ }$$ 

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

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∘.


AC2 = AB2 + BC2

(AB = BC = CD = AD = d1 – 1 – m6.00A)

AC2 = 62 + 62 = 72A

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

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

Whereas in the other alkali metal halides (like Na+1, K+I, Cs+1, etc) which contain larger cations as compared to Li cation: halide ions do not touch each other but cation-anion contact will occur in their crystal lattice as shown in figure 2.04. Hence, the sum of the radii of the cation and the anion would be equal to the inter-ionic distance between the cation and anion thus in the case of K+1the ionic crystal dK+Tis given as:
dk + 1- = rk + r1-
On Putting dK+ – 1 = 3.50 Å  and r1 = 2.12 Å, the ionic radius of K+, fK+ thus calculated as –
rK+ = dK+ -I – r1
   = 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 :

pairs having the same number of electrons as
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.
Cation and anion contact
If the internuclear distance between the two ions c+ and a in the ionic crystal is c+ais dc+a−. which is equal to the sum of the ionic radii of a cation and anion.
I.e dc+ – H = fc+ +  ra–  ……… (i)
where rc+ is the radius of the cation and ra 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 ).
$$ ∴\:r_{ion}\infty \frac{1}{Z^{\cdot }} $$  Where (Z* = Z – σ)
$$ r_{ion}\infty \frac{1}{\left(Z\:-\sigma \right)} $$
$$ \zeta _{\infty }=C\frac{1}{\left(Z-\sigma \right)}=C\frac{1}{Z^{\cdot }} $$…………………. (ii)\
Where C is a proportionality constant whose value depends on the electronic configuration of c+and a−ions, Z∗ and σ are effective nuclear charges and screening constants of Ions respectively and Z is an actual nuclear charge of the ion. Applying equation (ii) to cation and anion, we get:
$$ r_{c^+}=\frac{C}{\left(Z\cdot c^+-\sigma _{c^+}\right)}=\frac{C}{Z^{\cdot }c^+}\:………..\left(iii\right) $$
$$ r_{a^-}=\frac{C}{\left(Z\cdot a^–\sigma _{a^-}\right)}=\frac{C}{Z^{\cdot }a^+}\:………..\left(iv\right) $$
dividing the equation (iii) by (iv) we get
$$ \frac{r_{c^+}}{r_{a^-}}=\frac{\left(Z\:a^–\sigma \:_{a^-}\right)}{\left(Za^+-\sigma _{c^+}\right)}=\frac{Z^{\cdot }a^-}{Z^{\cdot }c^+}\:………..\left(v\right) $$
where Za− and Zc+ are effective nuclear charges of anion and cation respectively and σa− and σc+ are screening constants for anion and cation respectively.
By solving equations (i) and (iii) we can calculate the value of rc+ and ra− by knowing the isoelectronic ions and their radii are given in table 2.5.

Atomic radil of metailic elements (B)

Ionic radii in A of isoelectronic ions.
Ionic radii in A of isoelectronic ions.

Illustration: Let us apply Pauling’s method to calculate the radii of the isoelectronic Na+ and Fions in Na+Fcrystal. 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+ Fions 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+ = (ZNa+ – σNa+)

                     = 11.0 – 4.5 = 6.5

and        Z*F = (ZF – σF )

                    = 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 rNa+ and rf- equal to 0.95 and 1.36 A respectively.
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